text:
\section{The hilbertian ball and their relatives}\label{sectionball}

Consider $\kappa$ an uncountable cardinal, identified with the set of ordinals less than $\kappa$. The closed ball of the Hilbert space $\ell_2(\kappa)$ equipped with the weak topology can be identified with the following compact subspace of $\mathbb{R}^\kappa$ (by squaring each coordinate and keeping the sign):

$$B(\kappa) := \left\{x\in\mathbb{R}^\kappa : \sum_{i\in\kappa}|x_i|\leq 1\right\}$$

The closed subspaces of $B(\kappa)$ are the \emph{uniform Eberlein compacta}. This class is well known and studied. Some standard facts are the following:

\begin{itemize}
\item $B(\kappa)$ is a Fr\'echet-Urysohn space, that is every point in the closure of a set $A\subset B(\kappa)$ is the limit of a sequence of elements of $A$ (this is a consequence of the fact that every element of $B(\kappa)$ has countable support)

\item The density character and the weight of any closed subspace of $B(\kappa)$ coincide. In particular, any separable subspace is metrizable (also a consequence of the fact that every element of $B(\kappa)$ has countable support).

\item Every closed subset of $B(\kappa)$ contains a $G_\delta$ point (take a point with maximal norm).

\item $B(\kappa)$ is sequentially compact: every sequence contains a convergent subsequence (a $G_\delta$-point in the set of cluster points of the sequence provides a convergent subsequence).

\item Every continuous image of a closed subset of $B(\kappa)$ is homeomorphic to a subspace of $B(\kappa)$ \cite{BenRudWag}.
\end{itemize}

These are properties of the class of subspaces of $B(\kappa)$ rather than properties of $B(\kappa)$ itself (actually, these properties are shared by the more general class of Eberlein compacta, though the proofs require much deeper facts). A study of specific topological properties of the space $B(\kappa)$ that distinguish it from its subspaces has been made in recent papers by the authors \cite{unitball,AviKal,KalendaBall,finiteproducts}. The results include ($\kappa$ is an uncountable cardinal):

\begin{itemize}
\item The space $B(\kappa)$ is a continuous image of $A(\kappa)^\mathbb{N}$, where $A(\kappa)$ is the one point compactification of the discrete space of cardinality $\kappa$ \cite{unitball}. It was shown by Bell \cite{Bell} that not all subspaces of $B(\kappa)$ have this property.

\item In \cite{AviKal} a topological invariant called \emph{fiber order} is introduced that --roughly speaking-- associates to a non-metrizable compact set $K$ and a point $x$ a preordered metrizable compact space $\mathbb{F}_x(K)$ and an ordered set $\mathbb{O}_x(K)$ (which is the canonical quotient of $\mathbb{F}_x(K)$). We proved that $\mathbb{O}_x(B(\kappa))$ is a singleton for $x$ in the sphere and $\mathbb{O}_x(B(\kappa))$ is order-isomorphic to the interval $[0,1]$ for other $x$. This allows to check that $B(\kappa)$ is not homeomorphic to its finite powers and to several other spaces.

\item $B(\kappa)$ does not map continuously onto any product $K\times L$ where both $K$ and $L$ are nonmetrizable \cite{finiteproducts}.
\end{itemize}

There is still much to discover about this space. One open problem is the following:

\begin{problem}\label{productwithinterval}
Is $B(\kappa)$ homeomorphic to $B(\kappa)\times K$ for some metrizable compact $K$? For $K = [0,1]$?
\end{problem}

A type of question that can lead the research for understanding this compact space is to know whether it is homeomorphic to some other compact spaces closely related to it. We provide now a family of such spaces. Given two functions $a,b:\kappa\longrightarrow [-1,1]$ satisfying $-1\leq a(i) < b(i) \leq 1$ for all $i\in\kappa$, we denote
$$B(\kappa,a,b) = \left\{x\in \mathbb{R}^\kappa : \sum_{i\in \kappa}|x_i|\leq 1\text{ and } a(i)\leq x_i \leq b(i) \text{ for all }i\in \kappa\right\}$$

This is a more general family of compact spaces than that consisting of spaces $B(M,N)$ defined in \cite{KalendaBall}, that would correspond to $B(M,a,b)$ where when $b$ is the constant function equal to 1 and $a$ is the function taking the value 0 on $N$ and -1 on $M\setminus N$. Geometrically, for a fixed $i_0\in\kappa$, the set $\{x\in B(\kappa) : a\leq x(i_0)\leq b\}$ corresponds to the intersection of the ball with a \emph{band}, i.e. the region between two hyperplanes, so $B(\kappa,a,b)$ is the intersection of $B(\kappa)$ with a certain family of bands.\\

To avoid trivialities, we will assume that
$$\sum_{i\in\kappa} (a(i)^+ + b(i)^-)<1,$$
where we denote $t^+ = max(0,t)$ and $t^{-} = \max(0,-t)$. We remark that $B(\kappa,a,b)$ is empty if this sum is strictly greater than $1$ and that it is a singleton if the sum equals $1$.

As we shall see later, it is not difficult to check that every space $B(\kappa,a,b)$ is a retract of each other. All the properties of $B(\kappa)$ enumerated above hold for all compacta $B(\kappa,a,b)$, even the computation of the fiber orders $\mathbb{O}_x(K)$ (just the same proof given in \cite{AviKal} is valid for any of these spaces).

\begin{problem}
Are all compact spaces $B(\kappa,a,b)$ homeomorphic to $B(\kappa)$? Determine the classification up to homeomorphism of the spaces $B(\kappa,a,b)$.
\end{problem}

First we remark that we can restrict ourselves to functions $a$ and $b$ satisfying $-1\leq a(i)\leq 0 <b(i) \leq 1$ and $|a(i)|\leq |b(i)|$ for all $i<\kappa$. We call such pairs \emph{admissible}.

\begin{prop}
Every compact space $B(\kappa,a,b)$ is homeomorphic to a space $(B,\kappa,a',b')$ where $a'$, $b'$ are admissible.
\end{prop}

{\sc Proof.}\quad  Suppose that for some $i$ we have either $a(i)>0$ or $b(i)<0$. Denote
$$A=\{i\in\kappa: a(i)>0\},\qquad B=\{i\in\kappa:b(i)<0\}$$
and set $s=\sum_{i\in A} a(i)-\sum_{i\in B} b(i)$. Then for each $x\in B(\kappa,a,b)$ we have
$$|x_i|\le \begin{cases} 1-s,& i\in\kappa\setminus(A\cup B),\\
1-s+a(i), & i\in A, \\ 1-s-b(i), & i\in B.\end{cases}$$
Therefore we can without loss of generality suppose that
\begin{gather}
|a(i)|\le 1-s,|b(i)|\le 1-s \mbox{ for }i\in\kappa\setminus(A\cup B),\\
b(i)\le 1-s+a(i) \mbox{ for } i\in A,\\
a(i)\ge -(1-s-b(i)) \mbox{ for }i\in B.
\end{gather}
Then the mapping which assigns to each $(x_i)_{i\in\kappa}$ the point $(x'_i)_{i\in \kappa}$ such that
$$x'_i=\begin{cases} \frac{x_i-a(i)}{1-s}, & i\in A, \\ \frac{x_i-b(i)}{1-s}, & i\in B, \\ \frac{x_i}{1-s}, & x\in\kappa\setminus(A\cup B),\end{cases}$$
is a homeomorphism of $B(\kappa,a,b)$ onto $B(\kappa,a',b')$, where
$$a'(i)=\begin{cases} 0 & i\in A, \\ \frac{a(i)-b(i)}{1-s} & i\in B, \\
\frac{a(i)}{1-s}& i\in\kappa\setminus(A\cup B), \end{cases} \qquad
b'(i)=\begin{cases} \frac{b(i)-a(i)}{1-s} & i\in A, \\ 0 & i\in B, \\
\frac{a(i)}{1-s}& i\in\kappa\setminus(A\cup B). \end{cases}$$
Then $-1\le a'(i)\le 0\le b'(i)\le 1$ for each $i\in\kappa$. Further, for each $i$ such that $|b'(i)|<|a'(i)|$ we can redefine the pair $(a'(i),b'(i))$ by
$(-b'(i),-a'(i))$. In this way we get an admissible pair $(a'',b'')$ such that $B(\kappa,a,b)$ is homeomorphic to $B(\kappa,a'',b'')$.
\quad{$\blacksquare$}\medskip

 For $t\in [-1,1]$, by abuse of notation, we will denote by $t$ the constant function equal to $t$ and by $t^\ast$ the function on $\kappa$ that it is equal to $t$ except in one point, where it takes value $0$. We can focus on several particular cases:

\begin{itemize}
\item $B(\kappa) = B(\kappa,-1,1)$ is the ball
\item $B^+(\kappa) = B(\kappa,0,1)$ is the positive cone of the ball
\item $B^\frac{1}{2}(\kappa) = B(\kappa,-1^\ast,1)$ is half a ball
\item $B(\kappa, -t,t)$ for $t\in (0,1)$
\item $B(\kappa,0,t)$ for $t\in (0,1)$
\end{itemize}

\begin{problem}\label{fewlistproblem}
Are any two from the compact spaces above homeomorphic?
\end{problem}

This problem is connected with Problem~\ref{productwithinterval}. It is proven in \cite{KalendaBall} that $B^+(\kappa) = B^+(\kappa)\times [0,1]$ and $B^{\frac{1}{2}}(\kappa) = B^{\frac{1}{2}}(\kappa)\times [0,1]$. These results are obtained by using the notions of \emph{cone} $\Delta(K)$ and \emph{double cone} $\Diamond(K)$ of a compact space $K$. The space $\Delta(K)$ is obtained from $K\times [0,1]$ by gluing $K\times \{1\}$ to a singleton, and $\Diamond(K)$ by gluing $K\times\{1\}$ to a singleton and $K\times\{0\}$ to another singleton. Fix $i\in\kappa$ and suppose that the function $a$ satisfies $a(j)\in\{-1,0\}$ for all $j$ and the function $b$ satisfies $b(j)=1$ for all $j\neq i$. We have that:

\begin{itemize}
\item $B(\kappa,a,b) = B(\kappa\setminus\{i\},a,1)\times [0,1]$ if $b(i)<1$
\item $B(\kappa,a,b) = \Delta(B(\kappa\setminus\{i\},a,1))$ if $b(i) = 1$ and $a(i)>-1$;
\item $B(\kappa,a,b) = \Diamond(B(\kappa\setminus\{i\},a,1)$ if $a(i)=-1$.
\end{itemize}

On the other hand, $\Delta(\Delta(K)) = \Diamond(\Delta(K)) =\Delta(\Diamond(K)) = \Delta(K)\times [0,1]$ for any compact space $K$ \cite{KalendaBall}. These facts allow to prove a number a homeomorphic relations between the spaces $B(\kappa,a,b)$ of the form $B(M,N)$ stated in \cite[Theorem 9]{KalendaBall} (the symbol $\sim$ means homeomorphism):
\begin{enumerate}
\item If both $N$ and $M\setminus N$ are infinite, then $B(M,N)\sim \Delta(B(M,N)) \sim \Diamond(B(M,N))\sim B(M,N)\times [0,1]$.
\item If $M\setminus N$ is finite, then $B(M,N)\sim B^+(M) = \Delta(B^+(M)) \sim \Diamond(B^+(M)) \sim B^+(M)\times [0,1]$.
\item If $N$ is finite, then $B(M,N) \sim B^{\frac{1}{2}}(M) \sim \Delta(B^{\frac{1}{2}}(M)) \sim \Diamond(B^{\frac{1}{2}}(M)) \sim B^{\frac{1}{2}}(M) \times [0,1]$.

\end{enumerate}

\begin{prop}\label{retractofeachother}
Each of the spaces $B(\kappa,a,b)$ is a retract of each other.
\end{prop}

{\sc Proof.}\quad  It is easy to see that (provided $(a,b)$ is an admissible pair) every $B(\kappa,a,b)$ is a retract of $B(\kappa)$, the retraction sending each $(x_i)_{i<\kappa}$ to $(\min\{\max\{a,x_i\},b\})_{i<\kappa}$.

Second, we shall prove that $B(\kappa)$ is a retract of $B^+(\kappa)$.
We start by noticing that $B^+(\kappa)$ is homeomorphic to $B^+(\kappa\times\{0,1\})$. We define a continuous surjection of $B^+(\kappa\times\{0,1\})$ onto $B(\kappa)$ by
$$(x_{i,q})_{i<\kappa,q\in\{0,1\}}\mapsto (x_{i,0}-x_{i,1})_{i<\kappa}.$$
This surjection has a continuous right inverse $u:B(\kappa)\longrightarrow B^+(\kappa\times\{0,1\})$  defined by
$$u(x)_{i,0} = x_{i}^+,\qquad u(x)_{i,1} = x_{i}^-,\qquad i<\kappa.$$

In order to complete the proof, it remains to show that $B^+(\kappa)$ is a retract of each $B(\kappa,a,b)$. So take $B(\kappa,a,b)$ and consider a partition $\kappa = \bigcup_n J_n$ into infinite or empty sets such that $b(i)>\frac{1}{n}$ if $i\in J_n$. Let $\kappa = \bigcup_{i<\kappa}F_i$ be a partition of $\kappa$ indicated in $\kappa$ with the property that for every $i$ there exists $n$ with $F_i\subset J_n$ and $|F_i|=n$. We define a continuous surjection $f:B(\kappa,a,b)\longrightarrow B^+(\kappa)$ by
$$f(x)_i = \sum_{j\in F_i} x_i^+, \qquad i<\kappa.$$
We shall define a  continuous right inverse $u:B^+(\kappa)\longrightarrow B(\kappa,a,b)$ for $f$. Fix $x\in B^+(\kappa)$ and $i<\kappa$. Enumerate $F_i = \{j_1,\ldots,j_n\}$. We define $u(x)$ on $F_i$ in the following way:
$$u(x)_{j_1} = \min\{x_i,b(j_1)\},
u(x)_{j_2} = \min\{x_i-u(x)_{j_1},b(j_2)\},\ldots,u(x)_{j_n} = \min
\{x_i - \sum_{l=1}^{n-1}u (x)_{j_l},b(j_n)\}.$$
\quad{$\blacksquare$}\medskip

We shall focus now in one particular instance of Problem~\ref{fewlistproblem}:

\begin{problem}
Are $B(\kappa)$ and $B^+(\kappa)$ homeomorphic?
\end{problem}

The nice thing about this \emph{nonseparable} problem is that is equivalent to a \emph{separable} problem. In the following theorem $\|x\| := \sum_{i\in\kappa} |x_i|$ (notice that, if we identify $B(\kappa)$ with the ball of $\ell_2(\kappa)$ by means of the homomorphism described above, this is the square
of the Euclidean norm; for this theorem that square is irrelevant). Also, for $x\in [-1,1]^{\aleph_0}$, $\|x\|_\infty = \sup\{|x_n| : n\in\aleph_0\}$.

\begin{thm}\label{equivalentproblems} The following statements are equivalent.
\begin{enumerate}
\item For every uncountable cardinal $\kappa$, $B(\kappa)$ is homeomorphic to $B^+(\kappa)$.
\item There is an uncountable cardinal $\kappa$ such that $B(\kappa)$ is homeomorphic to $B^+(\kappa)$.
\item There exists a homeomorphism $f:B(\aleph_0)\longrightarrow B^+(\aleph_0)$ such that $\|f(x)\| < \|f(y)\|$ whenever $x,y\in B(\aleph_0)$, $\|x\| < \|y\|$.
\item There exists a homeomorphism $f:B(\aleph_0)\longrightarrow B^+(\aleph_0)$ such that $\|f(x)\| = \|x\|$ for all $x\in B(\aleph_0)$
\item There exists a homeomorphism $f:[-1,1]^{\aleph_0}\longrightarrow [0,1]^{\aleph_0}$ in the product topologies, such that $\|f(x)\|_\infty = \|x\|_\infty$ for all $x$.
\end{enumerate}
\end{thm}

{\sc Proof.}\quad  $(1\Rightarrow 2)$ is evident.

$(2\Rightarrow 3)$ requires using the technique of fiber orders from~\cite{AviKal}. As it is computed there, for $x\in K$ the fiber preorder structure $\mathbb{F}_x(B(\kappa))$ is either trivial (a singleton) or it is the compact space $B(\aleph_0)$ endowed with the preorder relation ($x\leq y \iff \|x\|\leq \|y\|$). Just the same computation shows that $\mathbb{F}_x(B^+(\kappa))$ is either trivial or it is the compact space $B^+(\aleph_0)$ endowed with the order ($x\leq y \iff \|x\|\leq \|y\|$). Hence, if $B(\kappa)$ and $B^+(\kappa)$ are homeomorphic, then there must exist an order-preserving homeomorphism between nontrivial fiber preorders $\mathbb{F}_x(B(\kappa))$ and $\mathbb{F}_y(B^+(\kappa))$.

For $(3\Rightarrow 4)$ it is enough to prove the following statement:

Claim: For every strictly increasing bijection $\phi:[0,1]\longrightarrow [0,1]$ there exists a homeomorphism $g:B^+(\aleph_0)\longrightarrow B^+(\aleph_0)$ such that $\|g(x)\| = \phi(\|x\|)$.

The implication follows from the claim because if $(3)$ holds, then there is a strictly increasing bijection $\psi:[0,1]\longrightarrow [0,1]$ such that $\|f(x)\| = \psi(\|x\|)$; apply the claim to $\phi = \psi^{-1}$ and consider $g\circ f$. For proving the claim we define $g(x_0,x_1,x_2,\ldots) = (y_0,y_1,y_2,\ldots)$ as follows: $y_0 = \phi(x_0)$, $y_n = \phi(\sum_{0}^n x_i) - \sum_0^{n-1}y_i$.

Finally, for $(4\Rightarrow 1)$, consider a partition of $\kappa$ into countable sets. For each of these countable sets $S$ we have a norm-preserving homeomorphism $B(S)\longrightarrow B^+(S)$. Gluing all these homeomorphisms together we get a (norm-preserving also) homeomorphism $B(\kappa)\longrightarrow B^+(\kappa)$

It remains to prove the equivalence of $4$ and $5$. This follows from the following theorem, which is a consequence of results of Dijkstra and van Mill \cite{DijMil}.

\begin{thm}\label{dijmil} There exists a homeomorphism $f:[-1,1]^{\aleph_0}\longrightarrow B(\aleph_0)$ such that $f([0,1]^{\aleph_0})=B^+(\aleph_0)$ and $\|f(x)\| = \|x\|_\infty$ for every $x\in [-1,1]^{\aleph_0}$.
\end{thm}

{\sc Proof.}\quad  It follows from \cite[Theorem 5.2]{DijMil} that there is an onto homeomorphism $g:\R^{\aleph_0}\longrightarrow\R^{\aleph_0}$ such that:

\begin{eqnarray}
\label{DM1} \forall x\in \R^{\aleph_0}: \|g(x)\|_1=\|x\|_{\infty} \qquad \mbox{(including the infinite values)} \\
\label{DM2} g\mbox{ is sign-preserving, i.e. }\forall x\in \R^{\aleph_0}\forall n\in\aleph_0: g(x)_n=0 \mbox{ or }\mbox{\rm sign\,} g(x)_n=\mbox{\rm sign\,} x_n
\end{eqnarray}

Let $f$ be the restriction of $g$ to $[-1,1]^{\aleph_0}$. Then $f$ is a norm-preserving homeomorphism
of $[-1,1]^{\aleph_0}$ onto $B(\aleph_0)$. This follows immediately from (\ref{DM1}).
It remains to show that $f([0,1]^{\aleph_0})=B^+(\aleph_0)$.

First notice that it follows from (\ref{DM2}) that for every $x\in\mathbb{R}^{\aleph_0}$ and every $n\in\aleph_0$:

\begin{equation}
		\label{DM3} (x_n\ge 0\Rightarrow g(x)_n\ge0) \text{ and } (x_n=0\Rightarrow g(x)_n=0).
\end{equation}

Hence it is clear that $f([0,1]^{\aleph_0})\subset B^+(\aleph_0)$. It remains to prove the reverse inclusion. As finitely supported vectors are dense in $B^+(\aleph_0)$ and $f([0,1]^{\aleph_0})$ is compact, it is enough to show that the range contains all finitely supported vectors in $B^+(\aleph_0)$.
To prove it we introduce the following notation. If $F\subset\aleph_0$ is finite, we set
\begin{align*}
U(F)&=\{x\in [0,1]^{\aleph_0} : x_i=0\mbox{ for }i\notin F\},\\
V(F)&=\{x\in B^+({\aleph_0}) : x_i=0\mbox{ for }i\notin F\}
\end{align*}
It follows from (\ref{DM1}) and (\ref{DM3}) that $f(U(F))\subset V(F)$ for each finite $F\subset\aleph_0$. If we show that $f(U(F))=V(F)$ for each $F$, the proof will be completed.

We will show it by induction on the cardinality of $F$. If $F$ is a singleton, then it is obvious.
Indeed, if $e_i$ denotes the $i$-th canonical vector in $R^{\aleph_0}$, it follows from (\ref{DM1}) and (\ref{DM3}) that $g(te_i)=te_i$ for each $t\in\R$.

Further, let $n\in\mathbb{N}$ be such that the equality holds for each $F$ of cardinality at most $n$. Suppose that $F$ has cardinality $n+1$. Fix $t\in(0,1]$ and set
\begin{alignat*}{3}
C&= \{x\in U(F): \|x\|_\infty=t\},&  \qquad C_i&=\{x\in C: x_i=0\}\mbox{ for }i\in F,\\
D&= \{x\in V(F): \|x\|_1=t\},& D_i&=\{x\in D: x_i=0\}\mbox{ for }i\in F.
\end{alignat*}
Then clearly $f(C)\subset D$. Moreover, by the induction hypothesis $f(C_i)=D_i$ for each $i\in F$.
We observe that $C$ is homeomorphic to $B_n$, the $n$-dimensional euclidean ball. Indeed, the function $\phi(x) = (x_i - \frac{1}{n+1}\sum_{j\in F}x_j)_{i\in F}$ provides a homeomorphism from $C$ onto the ball of radius $t$ of the space $\{u\in\mathbb{R}^F : \sum_{j\in F}u_j = 0\}$ endowed with the norm $\|u\|' = \sup\{|u_i-u_j| : i\neq j\}$. On the other hand, the pair $(D,\bigcup_{i\in F} D_i)$ is homeomorphic to the pair $(B_n,S_{n-1})$, where $S_{n-1}$ denotes the sphere of $B_n$. This is because $\Delta = D\setminus\bigcup_{i\in F}D_i$ is a convex open subset of the affine space $\{x\in\mathbb{R}^{\aleph_0} : x_i=0\mbox{ for } i\not\in F,\ \sum_{j\in F}x_j = t\}$ and $D=\overline{\Delta}$. Finally, $f(C)$ is a subset of $D$ which is homeomorphic to $D$ and contains $\bigcup_{i\in F}D_i$. Thus $f(C)=D$ (as the only subset of $B_n$ which is homeomorphic to $B_n$ and contains $S_{n-1}$ is the whole set $B_n$).

As $t\in(0,1]$ was arbitrary, we get $f(U(F))=V(F)$ which completes the proof.
\quad{$\blacksquare$}\medskip

\section{Spaces of continuous functions on some Eberlein compacta}\label{sectionCK}

In the preceeding section we dealt with the topological structure of $B(\kappa)$, and now we shall deal with the structure of the Banach space of continuous functions $C(B(\kappa))$. The first remark is the following:

\begin{prop}\label{CBkisomorphic}
Each of the Banach spaces $C(B(\kappa,a,b))$ is isomorphic to $C(B(\kappa))$.
\end{prop}

In order to prove that, one uses Pe\l czy\' nski's decomposition method:

\begin{prop}[{\rm Pe\l czy\'nski}]
Let $X$ and $Y$ be Banach spaces. Suppose that $X$ is isomorphic to a complemented subspace of $Y$ and $Y$ is isomorphic to a complemented subspace of $X$. Suppose also that $X$ is isomorphic to $c_0(X)$, the $c_0$-sum of contably many copies of $X$. Then $X$ and $Y$ are isomorphic.\\
\end{prop}

Proof of Proposition \ref{CBkisomorphic}: If $K$ is a retract of a compact space $L$, then $C(K)$ is a complemented subspace of $C(L)$, so from Proposition~\ref{retractofeachother} we conclude that the spaces $C(B(\kappa,a,b))$ are complemented in each other. A result of Argyros and Arvanitakis \cite{ArgArv} that $c_0(C(B(\kappa)))$ is isomorphic to $C(B(\kappa))$ completes the proof.\quad{$\blacksquare$}\medskip

The situation is nevertheless unclear to us for some compact spaces that we know that are not homeomorphic, but we do not know about the spaces of continuous functions:

\begin{problem}\label{CBkn}
Is $C(B(\kappa))$ isomorphic to $C(B(\kappa)^n)$ for $n\leq\omega$? Is $C(B(\kappa))$ isomorphic to $C(P(K))$, where $P(K)$ is any of the spaces of probability measures considered in \cite{AviKal}: $P(A(\kappa)^n)$, $P(\sigma_n(K))$...?
\end{problem}

Motivated by the classical Milutin's theorem, it was an open problem for long time whether every Banach space of continuous functions $C(K)$ is isomorphic to a space $C(L)$ with $L$ a zero-dimensional compactum. A counterexample to this question was first found by Koszmider \cite{Koszmider}. The key property of this example, and those constructed later in a similar spirit, is that it is an \emph{indecomposable Banach space}, i.e., it contains no nontrivial complemented subspaces. However all the Banach spaces we are considering here are weakly compactly generated, hence they are very rich in complemented subspaces. Therefore, the problem is still worth being investigated under restrictions (like being WCG) that make indecomposable counterexamples impossible.

\begin{problem}\label{CBk}
Is $C(B(\kappa))$ isomorphic to some $C(L)$ with $L$ a zero-dimensional compact space?
\end{problem}

Even the following weaker version of the problem stated by Argyros and Arvanitakis is open \cite{ArgArv}:

\begin{problem} Is there a nonseparable weakly compact convex set $K$ in a Banach space and a zero-dimensional compact space $L$ such that $C(K)\cong C(L)$?
\end{problem}

Hence Problem \ref{CBk} is open also for any of the spaces mentioned in Problem~\ref{CBkn}, and in particular for $K = B(\kappa)^n$ with $n$ finite or countable. Argyros and Arvanitakis \cite{ArgArv} were able to construct a weakly compact star-shaped set $K$ containing a ball such that $C(K)$ is isomorphic to $C(L)$ with $L$ zero-dimensional. Now, we can propose a specific candidate for $L$. If $C(L)$ should be isomorphic to $C(B(\kappa))$, $L$ has to be uniform Eberlein and not scattered. We propose $L = A(\kappa)^\mathbb{N}$, where $A(\kappa)$ is the one-point compactification of a discrete set of cardinality $\kappa$.\\

\begin{problem}
Is $C(B(\kappa))$ or some $C(B(\kappa)^n)$ isomorphic to $C(A(\kappa)^\mathbb{N})$?\\
\end{problem}

Let us have a look at what we know concerning the conditions of Pelczynski's decomposition method when dealing with these problems. First thing is to know if $X\cong c_0(X)$. In the litterature we find:

\begin{itemize}
\item $C(A(\kappa)^\mathbb{N})$ is isomorphic to $c_0(C(A(\kappa)^\mathbb{N})$ \cite{countableproducts}.
\item $C(B(\kappa))$ is isomorphic to $c_0(C(B(\kappa))$ \cite{ArgArv}.\\
\end{itemize}

The second thing is to know which spaces are isomorphic to complemented subspaces of other spaces. In the case of spaces of continuous functions $C(K)$, the usual tool is that of averaging and extension operators. An \emph{extension operator} for a continuous injection $v:K\longrightarrow L$ is an operator $T:C(K)\longrightarrow C(L)$ such that $T(f)(v(x)) = f(x)$ for all $x\in K$ and $f\in C(K)$. An \emph{averaging operator} for a continuous surjection $u:L\longrightarrow K$ is an operator $T:C(L)\longrightarrow C(K)$ such that $T(u\circ f) = f$ for all $f\in C(K)$. In either of the two case, the existence of such an operator $T$ implies that $C(K)$ is complemented in $C(L)$. Often one finds averaging or extension operators that are regular, that is: positive, of norm one, and sending constant functions to constant functions of the same value. Let us see a few remarks about what this technique can say and what it cannot say for our purposes.

\begin{prop}$C(A(\kappa)^\mathbb{N})$ is isomorphic to a complemented subspace of $C(B(\kappa)^\mathbb{N})$.\end{prop}

{\sc Proof.}\quad  Consider the the canonical embedding $u:A(\kappa)\longrightarrow B(\kappa)$ that sends every element $i\in\kappa$ to the $x\in B(\kappa)$ that is zero everywhere except at $i$ where $x_i=1$. This embedding has a regular extension operator: To every continuous function $f\in C(A(\kappa))$ we can associate its extension $T(f)\in C(B(\kappa))$ defined as $T(f)(x) = f(\infty)\left(1-\sum_{i\in\kappa}|x_i|\right) + \sum_{i\in\kappa} |x_i|f(i)$. By \cite[Proposition 4.7]{Pelczynski}, the countable power $u^\mathbb{N}:A(\kappa)^\mathbb{N}\longrightarrow B(\kappa)^\mathbb{N}$ also admits a regular extension operator, hence $C(A(\kappa)^\mathbb{N})$ is isomorphic to a complemented subspace of $C(B(\kappa)^\mathbb{N})$.\quad{$\blacksquare$}\medskip

\begin{prop}
If $m<n\leq\omega$, then $C(B(\kappa)^m)$ is a complemented subspace of $C(B(\kappa)^n)$.
\end{prop}

{\sc Proof.}\quad  $B(\kappa)^m$ is a retract of $B(\kappa)^n$, and this provides an embedding with regular extension operator. \quad{$\blacksquare$}\medskip

The problem with the powers of $B(\kappa)$ arises in finding $C(B(\kappa)^n)$ complemented in $C(B(\kappa)^m)$ for $m<n$. We cannot hope to get this using averaging operators since $B(\kappa)^m$ does not map continuously onto $B(\kappa)^n$ by \cite{finiteproducts}. Next proposition shows that neither regular extension operators can be used.

\begin{prop}
If $m<n\leq\omega$, there does not exist any embedding $B(\kappa)^n\longrightarrow B(\kappa)^m$ with a regular extension operator.
\end{prop}

{\sc Proof.}\quad  Suppose $u:B(\kappa)^n\longrightarrow B(\kappa)^m$ is such an embedding. A regular extension operator would provide a continuous function $g: B(\kappa)^m\longrightarrow P(B(\kappa)^n)$ with $gu(x) = \delta_x$ for $x\in B(\kappa)^n$ \cite{Pelczynski}. Since $B(\kappa)^n$ can be viewed as a compact convex set, we can consider the barycenter map $h:P(B(\kappa)^n)\longrightarrow B(\kappa)^n$ which is continuous, affine, and satisfies $h(\delta_x) = x$. The composition $hg:B(\kappa)^m\longrightarrow B(\kappa)^n$ is continuous and onto, since $hg(u(x))= x$. However such onto maps do not exist for $m<n$ by \cite{finiteproducts}.\quad{$\blacksquare$}\medskip

\begin{prop}
The continuous surjection $f:A(\kappa)^\mathbb{N}\longrightarrow B(\kappa)$ constructed in \cite{unitball} does not admit any averaging operator.
\end{prop}

Proof: Let $r_n$ a sequence of positive real numbers that defines a continuous surjection $2^\mathbb{N}\longrightarrow [0,1]$ by $(t_n)\mapsto \sum r_n t_n$. For example $r_n = 2^{-n-1}$. Let $Z = \{z\in\mathbb{N}^\mathbb{N} : \sum_{n=0}^\infty \frac{z_n}{r_n} \leq 1\}$ and $L = Z\times \prod_{n\in\mathbb{N}}\sigma_n(\kappa)^\mathbb{N}$. Denote elements of $L$ as $(z,x)$, where $x = (x[m,n]\in\sigma_n(\kappa) : n,m\in\mathbb{N}$). A continuous surjection $g:L \longrightarrow B^+(\kappa)$ is defined as $g(z,x)_i = \sum_{m\in\mathbb{N}}\frac{1}{r_m} x[m,z_m]_i$. The surjection $f$ from \cite{unitball} is the composition of $g$ with a continuous surjection $g:A(\kappa)^\mathbb{N}\longrightarrow L$. It is enough to see that $g$ has no averaging operator. This can be seen using a result of Ditor. Define a derivation procedure on $B^+(\kappa)$ in the following way: given a set $X\subset B^+(\kappa)$, its derived set $X'$ is the set of all $x\in X$ whose fiber $g^{-1}(x)$ contains two disjoint sets $S$ and $T$ that belong to the closure in the Vietoris topology of the set of fibers $\{g^{-1}(y) : y\in X\}$. Inductively, $X^{(n)} = (X^{(n-1)})'$. According to \cite[Corollary 5.4]{Ditor}, if $B^+(\kappa)^{(n)} \neq\emptyset$ for all $n$, then $g$ does not admit any averaging operator. Let $A = \{y\in B^{+}(\kappa) : \sum_{i\in\kappa}|y_i|<1\}$. We prove that $A' = A$. Let $y\in A$, and $\varepsilon = 1-\sum_{i\in\kappa}|y_i|$. We can consider a sequence of elements $y_n\in A$ converging to $x$ in which $y_n$ coincides with $y$ on the support of $y$, and $y_n$ has a single extra nonzero coordinate with value $\frac{3}{4}\varepsilon$. It is possible to find a natural number $k$ big enough so that for every $(z,x)\in g^{-1}(y_n)$ we have $\sum_{j<k}z_j \geq 1-\varepsilon/2$. Now let $m$ another integer big enough so that $\frac{3}{4m}\varepsilon<\sum_{j\geq k}r_j$. We pick a new sequence $y'_n\in A$ converging to $y$, in which $y'_n$ coincides with $y$ on the support of $y$ and has exactly $m$ extra nonzero coordinates with value $\frac{3}{4m}\varepsilon$ on each of these coordinates. The elements $y'_n$ have the property that for every $(z,x)\in g^{-1}(y'_n)$, $\sum_{i<k}z_i\leq 1-\frac{3}{4}$. Two cluster points in the Vietoris topology $S$ and $T$ of the sequences $\{g^{-1}(y_n)\}$ and $\{g^{-1}(y'_n)\}$ respectively are two disjoint closed sets of $g^{-1}(y)$, since $S\subset\{(z,x) : \sum_{j<k}z_j \geq 1 -\varepsilon/2\}$, while $T\subset\{(z,x) : \sum_{j<k}z_j \leq 1-\frac{3}{4}\varepsilon\}$. This proves that $y\in A'$.\quad{$\blacksquare$}\medskip

\begin{problem}
Is there a continuous surjection $f:A(\kappa)^\mathbb{N}\longrightarrow B(\kappa)$ with an averaging operator?
\end{problem}

\section{Eberlein compacta, WCG spaces and their subspaces}

There are certain natural hierarchies of classes of Banach spaces. Two of them --the so called \emph{descriptive hierarchy} and \emph{differentiability hierarchy}-- are described for example in \cite[Chapter 1]{habil}. Many of these classes can be characterized using a topological property of the weak topology. For example we have:

\begin{thm} Let $X$ be a Banach space.
\begin{itemize}
	\item $X$ is separable if and only if $(X,w)$ is separable.
	\item $X$ is reflexive if and only if $(X,w)$ is $\sigma$-compact.
	\item $X$ is WCG if and only if $(X,w)$ contains a dense $\sigma$-compact subset.
	\item $X$ is weakly Lindel\"of determined (WLD) if and only if $(X,w)$ is primarily Lindel\"of.
	\item $X$ is Asplund if and only if each separable subset of $(X,w)$ can be covered by countably many relatively closed metrizable subsets.
\end{itemize}
\end{thm}

There exist a number of characterizations of subspaces of WCG spaces (cf.~\cite{FMZ} and \cite[Theorem 6.13]{biorthogonal}), but all of them involve the linear structure of the Banach space.

\begin{problem} Is it possible to characterize subspaces of WCG spaces in terms of the weak topology? More precisely, suppose that $X$ is a subspace of a WCG space, and $Y$ is Banach space weakly homeomorphic to $X$, is $Y$ also a subspace of a WCG space?
\end{problem}

This problem has is meaningful as subspaces of WCG spaces need not be WCG. There are two basic examples -- the first one, by Rosenthal \cite{rosenthal}, is a subspace of a large $L^1(\mu)$ space; the second one is due to Argyros \cite[Theorem 1.6.3]{Fabiansbook} and it is a natural subspace of a $C(K)$ space with $K$ uniform Eberlein compact. But there are many spaces which are hereditarily WCG, apart from separable or reflexive spaces, an example is $c_0(\Gamma)$ for an arbitrary set $\Gamma$. So, the following problem seems to be natural.

\begin{problem} Characterize those compact spaces $K$ for which $C(K)$ is hereditarily WCG.
\end{problem}

This question was posed to us by Marian Fabian some time ago. Scattered Eberlein compacta have this property. Indeed, if $K$ is scattered and Eberlein,
then $C(K)$ is simultaneously Asplund and WCG. Further, WCG Asplund spaces are hereditarily WCG by \cite[Corollary 6]{OriVal}, cf. also \cite[Proposition 8.3.2]{Fabiansbook}. Another class with the property that $C(K)$ is hereditarily WCG is that of Eberlein compacta of weight less than $\mathfrak b$ by \cite[Corollary 7]{RNcompact}. Cardinal $\mathfrak b$ is the \emph{bounding number}: the least cardinality of a subset of $\mathbb{N}^{\mathbb{N}}$ which is not contained in any $\sigma$-compact subset of $\mathbb{N}^{\mathbb{N}}$. Consistently $\mathfrak b >\aleph_1$. On the other hand, the above mentioned example $C(K)$ of Argyros can be taken so that $K$ is a uniform Eberlein compact space of weight $\mathfrak b$. Hence $C(B(\mathfrak b))$ and $C(A(\mathfrak b)^\mathbb{N})$ are not hereditarily WCG, since every uniform Eberlein compact space of weight $\kappa$ is a subspace of $B(\kappa)$ and a continuous image of a subspace of $A(\kappa)^\mathbb{N}$.

One more problem inspired by the fact that WCG is not a hereditary property is the following one:

\begin{problem} Let $K$ be a convex Eberlein compact set, i.e. a convex compact subset of a locally convex space which is Eberlein compact. Is it homeomorphic
to a convex compact subset of $(X,w)$ for a Banach space $X$?
\end{problem}

Recall that one of the equivalent definitions of an Eberlein compact space is that these are just compact subsets of $(X,w)$ for a Banach space $X$. If $X$ is a subspace of WCG, then $(B_{X^*},w^*)$ is convex Eberlein compact (the underlying locally convex space being $(X^*,w^*)$). But $(B_{X^*},w^*)$ is \emph{affinely} homeomorphic to a convex compact subset of $(Y,w)$ for a Banach space $Y$ if and only if $X$ is WCG.

Indeed, suppose first that $X$ is WCG. Let $K\subset (X,w)$ be a generating compact subset. We can embed $(B_{X^*},w^*)$ affinely homeomorphically onto a bounded subset of $C_p(K)$ using the canonical restriction map $x^*\mapsto x^*|K$. As bounded pointwise compact subsets of $C(K)$ are weakly compact, the first implication is proved.

Conversely, suppose that $(B_{X^*},w^*)$ is affinely homeomorphic to a convex compact set $K\subset (Y,w)$ for a Banach space $Y$. Note that without loss of generality we can suppose that the spaces $X$ and $Y$ are real. (If they are complex, we consider them over the field of reals. This preserves all the respective properties.) Consider the standard restriction map defined by $y^*\mapsto y^*|K$. This is a continuous map of $(B_{Y^*},w^*)$ into $C_p(K)$. Moreover, the image $L$ of $B_{Y^*}$ separates points of $K$. This is a standard way to prove that $C(K)$ is WCG. But in our case, we suppose that $K$ is convex and so, $L\subset A(K)$, the Banach space of all affine continuous functions on $K$. It is now easy to check that $L\cup\{1\}$ is a weak compact generating subset of $A(K)$. Indeed, every element $\phi\in A(K)^*$ is of the form
$$\phi(f)=a f(x) - b f(y),\quad f\in A(K)$$
for some $x,y\in K$ and $a,b\in\mathbb{R}$. (To see it extend $\phi$ to an element of $C(K)^*$. This extension is of the form $a\mu_1-b\mu_2$ for some $a,b\in\mathbb{R}$ and Radon probabilities $\mu_1$,$\mu_2$. Then set $x$ to be the barycenter of $\mu_1$ and $y$ to be the barycenter of $\mu_2$.)
So, $L\cup\{1\}$ is a weak compact subset of $A(K)$ which contains $1$ and separates points of $K$. It follows from the above described representation of the dual that the only $\phi\in A(K)^*$ which is zero on $L\cup\{1\}$ is the zero functional.
So, by Hahn-Banach theorem $L\cup\{1\}$ generates $A(K)$, hence $A(K)$ is WCG.
But $A(K)$ is isometric to $A(B_{X^*},w^*)$ which is isomorpic to $X\times\mathbb{R}$. Thus $X$ is WCG as a hyperplane of a WCG space.

So, there are examples of Banach spaces $X$ such that $(B_{X*},w*)$ is Eberlein but it cannot be affinely embedded into $(Y,w)$ for any Banach space $Y$. But it is not clear whether it must be homeomorphic to a convex subset of some $(Y,w)$.

\section{Valdivia compacta and their interactions with other structures}

Valdivia compacta form a rich and natural class of compact spaces which is useful in the investigation of nonseparable Banach spaces with the focus on decompositions to separable pieces. For a detailed study of this class we refer to \cite{survey}. A collection of natural examples of Valdivia compacta can be found in \cite{valexa}, and a corresponding study on a more general class in \cite{kubisjmaa}.
We recall the necessary definitions.

\let\S\Sigma  \let\G\Gamma
Let $\Gamma$ be an arbitrary set. We define
$$\S(\G)=\{ x\in\R^\G : \{\gamma\in\G : x(\gamma)\ne 0\}
\mbox{ is countable} \}.$$
Further, let $K$ be a compact space.
\begin{itemize}
\item We
say that $A\subset K$ is a {\it $\S$-subset of $K$} if there is a
homeomorphic injection $h$ of $K$ into some $\R^\G$ such that
$A=h^{-1}(\S(\G))$.
\item $K$ is called a {\it Valdivia compact space}
if $K$ has a dense $\S$-subset.
\end{itemize}

Moreover, $K$ is called a {\it Corson compact space}
if $K$ is a $\S$-subset of itself, i.e. if it is homeomorphic to a subset of $\S(\G)$.

We start with the following problem. A subset $A$ of a complex vector space is called absolutely convex if $\sum_{i=1}^n t_i x_i\in A$ whenever $x_1,\ldots,x_n\in A$ and $t_1,\ldots,t_n\in\mathbb{C}$ with $\sum_{i=1}^n|t_i|\leq 1$.

\begin{problem} Let $X$ be a complex Banach space. Suppose that $(B_{X^*},w^*)$ admits a convex symmetric dense $\S$-subset. Does it admit an absolutely convex one?
\end{problem}

The answer is positive for $C(K)$ spaces (see \cite[Theorem 3.8]{complex}) and in case $(B_{X^*},w^*)$ admits only one dense $\S$-subset (which is trivial).
A related problem is the following one.

\begin{problem} Let $K$ be a Valdivia compact space and let $G$ be a compact (or even finite) abelian group of automorphisms of $K$. Is there a dense $\Sigma$-subset of $K$ which is invariant for each element of $G$?
\end{problem}

Even the following simplest case seems to be open.

\begin{problem} Let $K$ be a Valdivia compact space and $h:K\to K$ a homeomorphism satisfying $h\circ h=\mbox{id}$. Is there an $h$-invariant dense $\Sigma$-subset of $K$?
\end{problem}

The following problem is inspired by results of \cite{nova}.

\begin{problem} Is there a Radon-Nikod\'ym Valdivia compact space which contains a copy of $[0,\omega_2]$? Or even a scattered one?
\end{problem}

We recall that a Valdivia compact space which does not contain a copy of $[0,\omega_1]$ is Corson \cite{survey} and that a Corson Radon-Nikod\'ym compact space is Eberlein \cite{OriSchVal}. A positive answer would yield essentially new examples of Valdivia compacta, a negative one would yield that the dual of an Asplund space with countably $1$-norming Markushevich basis also contains such a basis (see \cite[Section 6]{nova}).

Another question from \cite{nova} is the following.

\begin{problem} Let $X$ be an Asplund space. Is there an equivalent norm on $X$ such that the respective bidual unit ball is Valdivia compact (in the weak* topology)?
Or even such that the dual admits a countably $1$-norming Markushevich basis?
\end{problem}

If $X$ is Asplund and has density $\aleph_1$, then $(B_{X^{**}},w^*)$ is Valdivia compact by \cite[Corollary 4.5]{nova}. In this case no renorming is needed.
On the other hand, there is an Asplund space of density $\aleph_2$ such that the respective bidual unit ball is not Valdivia compact\cite[Example 4.10(b)]{nova}.
This space is a renorming of $C[0,\omega_2]$. However, if we consider this space
with the canonical norm, the bidual unit ball is Valdivia compact (as it is for each $C(K)$ space, see e.g. \cite[Example 4.10(a)]{nova}). In this context it is worth to mention the result of \cite[Proposition 26]{kubisjmaa}, which implies that the dual to an Asplund space admits a $1$-projectional skeleton, hence the bidual unit ball admits a retractional skeleton. The definitions of these notions can be found in \cite{kubisjmaa}. We only stress that this is a kind of a noncommutative analogue of Valdivia compacta (Valdivia compact spaces are exactly those admitting a commutative retractional skeleton by \cite{kubmich}).
So, the question is whether we can by a renorming ensure commutativity.

The next problem is from \cite{valexa} and \cite{valdline}.

\begin{problem}\label{orderpreservingimage} Let $K$ be a linearly ordered compact space which is a continuous image of a Valdivia compactum. Is $K$ an order-preserving continuous
image of a linearly ordered Valdivia compactum?
\end{problem}

Linearly ordered Valdivia compacta have a quite nice structure by \cite[Section 3]{valexa}. A characterization of such spaces is the content of \cite{valdline}.
The structure of linearly ordered compact spaces which are continuous images of Valdivia compacta is less clear. A criterion for recognizing whether $K$ is an order-preserving continuous image of a linearly ordered Valdivia compactum is contained in \cite[Theorem 6.1]{valdline}. In view of this the following question
is also interesting.

\begin{problem}\label{uncountablecharacter} Let $K$ be a linearly ordered compact space which is a continuous image of a Valdivia compactum. Is the cardinality of the set of all points of uncountable character at most $\aleph_1$?
\end{problem}

A positive answer to Problem~\ref{orderpreservingimage} would imply a positive answer to Problem~\ref{uncountablecharacter}, due to \cite[Proposition 5.5]{K2006a} and \cite[Theorem 6.1]{valdline}.
However, the opposite direction is not clear due to \cite[Example 3.5]{valdline}.

The last problem is from \cite[Section 7]{valexa}.

\begin{problem} Let $A$ be a $C^*$ algebra. Is the bidual unit ball $(B_{A^{**}},w^*)$ Valdivia compact? Is it true for $A=B(\ell_2)$, the algebra of all bounded linear operators on $\ell_2$?
\end{problem}

For commutative $C^*$ algebras, i.e. for spaces $C_0(T)$ with $T$ locally compact, the answer is positive (see e.g. \cite[Theorem 5.5]{valexa}).
More generally, the answer is positive for type I $C^*$ algebras, i.e. for those $C^*$ algebras for which the bidual is a von Neumann algebra of type I (which is in this case equivalent to semifiniteness). This follows from \cite[Theorem 7.1]{valexa}.

\section{Radon-Nikod\'{y}m compact spaces}

Radon-Nikod\'{y}m (RN) compact spaces are defined as weak$^\ast$ compact subsets of duals of Asplund spaces. Ever since it was asked by Namioka \cite{Namioka}, the main open question on this subject has been whether the continuous image of an RN compactum is RN compact.

\begin{problem}
Is the continuous image of an RN compact space also RN compact?
\end{problem}

There has been a number of papers addressing this problem \cite{OriSchVal,StegallRN,Arvanitakis, FabHeiMat,MatSte,Namiokanote,RNcompact,RNorder,ArvAvi}, including a survey article \cite{Fabiansurvey}. We collect in this section a list of problems scattered in the litterature and some remarks, all of them related to the continuous image problem. First, we recall the intrinsic topological characterization of RN compacta. Let $d:K\times K\longrightarrow [0,+\infty)$ be a function with the property that $d(x,y)=0$ if and only if $x=y$. The function $d$ is said to \emph{fragment} $K$ if for every nonempty $L\subset K$ and every $\varepsilon>0$ there exists $U\subset K$ open such that $U\cap L\neq\emptyset$ and $\sup\{d(x,y) : x,y\in U\cap L\}<\varepsilon$. A compact space is $RN$ if and only if there is a lower semicontinuous metric $d$ on $K$ that fragments $K$. If one does not assume $d$ to satisfy the triangle inequality one obtains the concept of \emph{quasi $RN$ compact}. The class of quasi RN compacta is closed under continuous images \cite{Arvanitakis} but the following is unknown:

\begin{problem}
Is every quasi RN compact space an RN compactum? Is every quasi RN compact space a continuous image of an RN compactum?
\end{problem}

The class of quasi RN compacta was discovered independently by different authors \cite{Arvanitakis, FabHeiMat}, and it admits several nice characterizations \cite{RNcompact, Fabiansurvey}. If $K$ is quasi $RN$ compact, then $K$ is $RN$ compact if any of the following two conditions hold: (1) the weight of $K$ is less than $\mathfrak b$ \cite{RNcompact}, or (2) $K$ is \emph{almost totally disconnected} \cite{Arvanitakis}, i.e., $K\subset [0,1]^I$ and for every $x\in K$, $|\{i\in I : x_i\not\in\{0,1\}\}|\leq\aleph_0$ (that includes $K$ being Corson, Eberlein or zero-dimensional).

All classes of quasi RN, RN and their continuous images are closed under the following operations: passing to subspaces, making countable products, taking spaces of probability measures $P(K)$, and taking spaces of Borel measures of variation at most $1$, $M_1(K) = (B_{C(K)^\ast},w^\ast)$. The closed convex hull of compact space $K$ is a continuous image of the space $P(K)$, so the following problem by Namioka is a particular case of the continuous image problem:

\begin{problem}
Let $K$ be a compact subset of a locally convex space. Assume that $K$ is RN compact. Is the closed convex hull of $K$ also RN compact?
\end{problem}

Arvanitakis \cite{Arvanitakis} proved the following:

\begin{thm}\label{0dimRN} For a zero-dimensional compact space, the following are equivalent
\begin{enumerate}
\item $K$ is RN
\item $K$ is quasi RN
\item $K$ embeds in a countable product of scattered compact spaces.
\end{enumerate}
\end{thm}

In this direction, Argyros asked whether the whole class of RN compacta (and not only zero-dimensional) can be generated in a sense from the class of scattered compacta. Namely,

\begin{problem}
Let $K$ be an RN compact space. Is $K$ homeomorphic to a subspace of $P(S)$ for $S$ scattered compact? Is it a subspace of $M_1(S)$? Is it a continuous image of a subspace of $M_1(S)$?
\end{problem}

\begin{problem}
Let $K$ be a RN compact (or quasi RN compact) space. Is $K$ the continuous image of a zero-dimensional RN compact space $L$? Can a continuous surjection $f:L\longrightarrow K$ be obtained with an averaging operator?
\end{problem}

The interest of getting the averaging operator is the relation with the continuous image problem. If $f:L\longrightarrow L$ has an averaging operator, then $C(L)$ is a complemented subspace of $C(K)$. A compact space is RN compact if and only if $C(K)$ is an Asplund generated space \cite[Theorem 1.5.4]{Fabiansbook}, and this class is closed under complemented subspaces. The fact that these classes of compact spaces can be characterized through their $C(K)$ Banach spaces, allows to establish a connection with the problem of getting $C(K)$ isomorphic to $C(L)$ with $L$ zero-dimensional, mentioned in Section~\ref{sectionCK}.

\begin{prop}
Suppose that there exists a compact space $K$ that is quasi RN compact but not RN compact. Then $C(K)$ is not isomorphic to $C(L)$ for any zero-dimensional compact space $L$.
\end{prop}

Proof: Suppose that $C(K)$ is isomorphic $C(L)$. Since $K$ is quasi RN compact space, $C(K)$ is a $\sigma$-Asplund generated Banach space \cite{sigmaasplund}. Hence so is $C(L)$, so $L$ is quasi RN compact \cite{sigmaasplund}. But $L$ being zero-dimensional, Arvanitakis' Theorem \ref{0dimRN} implies that $L$ is RN. Hence $C(L)$ is isomorphic to $C(K)$ is Asplund generated, and  $K$ is RN \cite[Theorem 1.5.5]{Fabiansbook}. \quad{$\blacksquare$}\medskip

The approach from \cite{FabHeiMat} to quasi RN compact spaces and the recent characterization of continuous images of RN compacta in \cite{ArvAvi} suggest a relevant role of the combinatorics of $\mathbb{N}^\mathbb{N}$ in the problem of distinguishing quasi RN, RN compacta and their continuous images. We have also the fact that the three classes coincide for compact spaces of weight less than $\mathfrak b$ \cite{RNcompact}. Nevertheless, we do not know the answer to the following

\begin{problem}
Assume that every continuous image of an RN compactum $K$ of weight $\mathfrak c$ is RN compact. Does this imply that every continuous image of an RN compact space is RN compact? Suppose that all quasi RN compacta of weight $\mathfrak c$ are continuous images of RN compacta. Does this imply that this holds for all quasi RN compacta?
\end{problem}

A natural way of addressing this problem is through this other one:

\begin{problem}
Let $K$ be a  continuous image of an RN compact space. Suppose that all continuous images of $K$ of weight at most $\mathfrak c$ are RN compact. Is $K$ RN compact? Suppose that $K$ is qRN and all its continuous images of weight $\mathfrak c$ are continuous images of RN compacta. Is $K$ a continuous image of an RN compactum?
\end{problem}

Of course, we may ask by the way a similar question like:

\begin{problem}
Let $K$ be a  continuous image of an RN compact space. Suppose that all closed subspaces of $K$ of weight $\leq \mathfrak c$ are RN compact. Is $K$ RN compact? Similarly for quasi RN compacta and continuous images of RN compacta.
\end{problem}

After the results from \cite{RNcompact}, it would be natural to ask the same problems for cardinal $\mathfrak b$ instead of $\mathfrak c$, because quasi RN compact spaces of weight less than $\mathfrak b$ are RN, and it is for weight $\mathfrak b$ where first essential difficulties in the continuous image problem appear.

In \cite{ArvAvi}, Arvanitakis and the first author developped a method of construction of continuous images of RN compacta that are natural candidates not to be RN compact. We mention here a concrete one: the space of almost increasing functions. Let $(\prec)$ be a well order on $\mathbb{N}^\mathbb{N}$, and $$AIF := \left\{x\in [0,1]^{\mathbb{N}^\mathbb{N}} : x_\sigma \leq x_\tau + \frac{1}{2^{min\{n : \sigma_n\neq\tau_n\}}}\text{ whenever }\sigma\prec\tau\right\}.$$
The space $AIF$ is a continuous image of an RN compactum \cite{ArvAvi}.

\begin{problem}
Is $AIF$ an RN compact space?
\end{problem}

Finally we mention another particular instance of the problem of continuous images studied by Matou\v{s}kov\'{a} and Stegall \cite{MatSte}:

\begin{problem}
Let $K$ be a compact space that can be written as the union of two RN compact spaces $K_1$ and $K_2$. Is $K$ RN compact?
\end{problem}

Notice that the union of two compact spaces is the continuous image of the disjoint union, and clearly the dijoint union of two RN compact spaces is RN compact. Matou\v{s}kov\'{a} and Stegall \cite{MatSte} prove that the answer to this question is positive in some particular cases: when $K_1\cap K_2$ is either metrizable, scattered or a $G_\delta$ set, or when $K_1$ is a retract of $K$ or when $K\setminus K_1$ is scattered. Fabian \cite{Fabiansurvey} asks whether one could prove the result for $K_1\cap K_2$ Eberlein compact.

\begin{acknow}
The first author was supported by MEC and FEDER (Project MTM2008-05396), Fundaci\'{o}n S\'{e}neca (Project 08848/PI/08) and Ramon y Cajal contract (RYC-2008-02051).

The second author was supported in part by the grant GAAV IAA 100190901 and in part by the
Research Project MSM 0021620839 from the Czech Ministry of Education.
\end{acknow}

redpajama_set_name: RedPajamaArXiv


text:
It Was Like a Country Boy in the Big City
Charles Bell considered selling to be the most important aspect of his business. In fact, most people associated with the business considered it his passion. Bell's love for selling grew as he went door-to-door offering his wares, but, as his skills developed, he came to believe that this mode of operation had its limitations. If Le-Al-Co was going to grow he needed to move away from households and toward stores and suppliers.
Bell's chance to break into the world of wholesale and retail came from a friend who mentioned Tennessee Storm Window, owned by Lefty and Margaret Hardcastle. After discovering that it was Nashville's largest supplier of storm products, Bell borrowed Thompson's car for the drive to the Nolensville Road location. Shelah Thompson's car was nicer than his own, and he wanted to make a good impression.
The Hardcastle's did not buy anything on that first visit, but Bell visited them once a week for the next three months. Finally, his persistence paid off when they placed an order for 120 windows. The size of the order almost panicked Thompson, and he said the task was impossible. However, as he would do time and again when faced with such tasks, he completed the order. As a result, the Le-Al-Co and Tennessee Storm Window relationship lasted for many years, and Hardcastle always considered himself Bell's benefactor. Regrettably, the Hardcastle's felt resentment instead of pride when Le-Al-Co became too large to sell to them.
The growth of Le-Al-Co in the 1960s resulted from the hard work of Bell and a staff of dedicated salesmen. Sam McReynolds worked for Sears before making the transition to Le-Al-Co. Jack Robertson sold windows and doors for six years and has been credited with making great contributions to the company's sales philosophy. Henry Harris managed the account of Hendersonville's Jones Homes, Le-Al-Co's second largest customer. Tragically, Harris suffered critical injuries in an automobile accident. He recovered and returned to the company years later to oversee the parts department.
Tennessee Storm Window and Jones Homes, the two largest accounts, provided a strong foundation for a growing group of customers that included Missouri-based Riggs Cash and Carry. Indiana was the home of Lensing Wholesale and Lumberman's Wholesale. The sales territory stretched into Georgia with Atlanta's Addison Millwork. Tennessee customers included Dealers Warehouse in Knoxville; Cole Manufacturing of Memphis; and Madison Millwork in Jackson. However, Louisville's Jacob Levy and Sons was the most valued customer of this era.
In 1966, Bell met with their buyer, Harold Skaggs, and began a special relationship. Skaggs respected Bell as a salesman and a person and went to great lengths to help the young entrepreneur. He provided Bell with the quotes of competitors and covered any delivery problems with his superiors. This began a business arrangement that remained throughout the existence of Le-Al-Co.
The 1970s brought changes to the industry and to Le-Al-Co. "Do-It-Yourself" stores began to appear on the scene and looked to be the future of the market. Historically, locally owned stores sold products to builders, but DIY's were chains that catered to homeowners wanting to improve their residences. With an eye on an altering future, Bell decided to change the ideology of his company.
In 1972, he raided a major supplier, Alumax Extrusions, for a young salesman, Mike Dinwiddie. After a series of negotiations, Dinwiddie became the vice-president of Le-Al-Co and changed its direction immediately. Instead of hiring salespeople directly, Le-Al-Co used sales representation firms throughout the nation. Among these firms was SJS Sales in Brooklyn, New York. Jeff Saul, owner of the business, fascinated Bell and Dinwiddie. He impressed them with limousine tours around Chicago and with stories of his days as a champion swimmer. Unfortunately, Saul's body was found floating by his boat. Apparently, he had dealings with organized crime that ended badly.
Despite the detailed sales strategy and the interesting characters that it introduced, the sales representative idea never took flight. With representatives throughout the nation, Bell realized that he and Dinwiddie were making all of the sales. Eventually, contracts with all of the firms ended. A sales force that once consisted of over twenty people and grew to a force of representatives around the nation shrank to a total of two.
Soon, Bell discovered that selling to large chain stores did not differ from selling to the "Mom and Pop" stores of the 1960s. Simply, the salesman sold himself, and the products would eventually be bought. Everyone associated with Le-Al-Co agreed that Bell could employ his skill and personality better than anyone, as most in the industry respected and liked him.
A perfect example of his abilities took place while making a sales call to Atlanta-based Williams Lumber in 1978. As Bell waited to meet with their buyer, he began a conversation with another salesman who mentioned a new store in the area. After the meeting, Bell had some spare time before catching his flight.
Sensing an opportunity, he visited the offices of the new business and found a card table and a telephone. No chairs could be found. He soon learned that they wanted to open a store, but no one would sell to them because of bad credit. Bell gambled and sold them an order before leaving. However, the deal had one stipulation. Le-Al-Co would deliver half a shipment and send the rest after payment. With the agreement in place, the first Home Depot opened and, through the years, would purchase over $50 million worth of products.
More chain stores followed as Bell sold products to Angels in California; Cashway in Texas; Marvin's in Alabama; and Scotty's in Florida. There was also Handy City, Handy Dan, K-Mart and J.C. Penney. Through these stores and others Le-Al-Co products could be found in 49 states and in Great Britain.
In 1984, Dinwiddie left Le-Al-Co and was replaced by Tom Ruban as vice-president of sales. Over the next ten years annual sales increased from $12 million to $25 million. This can be attributed to Bell's abilities as a salesman and Ruban's abilities as an organizer.
He hired Tim Burroughs to oversee customer service, the department that took daily orders and worked closely with store employees. Ironically, Burroughs had grown up close to the Bell household as a friend of their oldest son, Jack. He had never experienced the business side of the family but soon realized that it was not too different from the personal side. Burroughs remembered traveling to his first convention with Bell. As he stated, "We rode around Chicago looking at the buildings. It was like a country boy in the big city." Burroughs had to learn the window and door business quickly and was mostly helped by the women who worked for him. Two of the most helpful were Pam Helm and Shirley Dematteo, Elaine Bell's aunt.
Bell and Ruban could not be the only salesmen of a rapidly growing business; therefore more people were brought in. Alabamian Nick Compton was considered by many to be the best of these. However, Walt Costello was undoubtedly the most colorful. He received word that some windows had been installed but were not working correctly. As the salesman, he had to visit the site to determine a solution to the problem. Upon arrival, Costello discovered that he was at a nudist colony. Shockingly, it took all day to replace the windows. Later, he learned that his aunt was one of the residents.
Obviously, many changes in the sales department of Le-Al-Co took place over 35 years. Sales philosophies, salesmen and customers came and went. However, Charles Bell was always there. Sales were his passion, and he believed that they were the driving force of success. In the early days, he drove through the night and sold during the day. Eventually, he flew to the customers. However, through the decades and the changes Bell always remembered one thing – it does not matter how many windows and doors can be made if they can not be sold.
Tags: Business, Community, Do It Yourself, History, Home Depot., JC Penney, K Mart, Le-Al-Co, Sales, Storm Doors, Storm Windows
Categories Community, History
← Give Me My Money Back, and I Don't Care What You Do
He Could Have Built a Battleship – Under Cost →
2 Responses to "It Was Like a Country Boy in the Big City"
DyingNote March 8, 2013 at 05:13 #
I take it this is the second part of the story you had promised to publish on this blog.
Rick March 8, 2013 at 15:59 #
It is. The next several posts will continue the story.

redpajama_set_name: RedPajamaCommonCrawl


text:
ACM SIGUCCS
Share SIGUCCS
Conference Attendance Grants
Conference Exhibitors
SIGUCCS Board Update – October 2022
October 27, 2022 Chair's Corner, SIGUCCS BoardSIGUCCS Editor
In late summer, the executive committee asked the ACM SGB (SIG Governing Board) and the SIGUCCS Community to allow the current board to extend their positions for another term of three years. This passed approvals and the executive committee is happy to continue our journey through June 2026. This will allow this group of individuals to host the 2023 conference, begin planning a 2024 conference, and work through the SIG Viability review that will happen in December 2023 without an interruption or disruption to the team. It will also allow us to begin to put into place some plans now for expanding appointed positions on the board to assist with all of these future plans.
SIGUCCS Executive Committee • Chicago • 2022
In looking at a 2024 conference, the board is beginning to solicit proposals for a location for another in-person conference in the spring of 2024 and hopes to have the location and dates available at our 2023 conference.
And speaking of appointed positions, the 2023 awards cycle will be the last term for Kathy Fletcher and her management of the Communication Awards. Becky Johnson has agreed to take on this role for the next award cycle. Becky has been an active Communication Award nominee and winner so she is perfect for taking over. We also want to thank Kathy for six years of service in this position. She has done an excellent job!
Lisa Brown
Chair, SIGUCCS Executive Committee
SIGUCCS Board Update – July 2022
July 30, 2022 2023 Conference, SIGUCCS BoardSIGUCCS Editor
As many of you are aware, SIGUCCS is hosting the 2023 conference in Chicago as an in-person event from March 26-29, 2023. In order to facilitate the shortened planning timeline, the Executive Committee took on the roles of the core committee. I am proud to announce the following roles:
Conference Chair: Laurie Fox
Program Co-Chairs: Miranda Carney-Morris and Shawn Plummer
Treasurer: Parrish Nnambi
Continue reading SIGUCCS Board Update – July 2022 →
January 2022 – Updates from the Chair
January 13, 2022 2022 Conference, Chair's Corner, ConferenceSIGUCCS Editor
Happy New Year to all! The tenure of this executive committee is about halfway over and yet it feels like we have just begun. All of our energies have been focused on finding ways to keep the community engaged during a pandemic. And it has not been easy.
It was with great sadness that we decided to move the 2022 conference in San Diego to a virtual event. With the pandemic continuing, and travel budgets at institutions in a state of flux, we decided that while some might be able to attend in person, that it would be better for the overall health of the community to be inclusive and do another virtual event. Additionally, we wanted to be respectful of the conference planning committee and make sure that they had enough time to plan a well-run virtual conference. We are taking what we learned from the 2021 conference and doing that work now! Watch for announcements in the coming weeks on schedule, sessions, and other activities.
Continue reading January 2022 – Updates from the Chair →
August 2021 – Updates from the Chair
August 20, 2021 Chair's Corner, SIGUCCS BoardSIGUCCS Editor
Update from the Board – August 2021
ACM Senior Member Award
It is with great pleasure that we announce that Laurie Fox, past chair of SIGUCCS, was honored as an ACM Senior Member this past month. This award recognizes Laurie's dedication to both SIGUCCS and ACM over the years. For those not aware, in addition to serving on the SIGUCCS executive committee, Laurie has been on the SIG Governing Board of ACM and holds the position of New SIG Advisor.
Continue reading August 2021 – Updates from the Chair →
Laurie Fox Recognized as ACM Senior Member
July 30, 2021 SIGUCCS Commuity, SIGUCCS PeopleSIGUCCS Editor
Laurie Fox was recently recognized by ACM as an ACM Senior Member. The Senior Members Grade recognizes those ACM members with at least 10 years of professional experience and at least 5 years of Professional Membership in the last 10 years, who have demonstrated performance through technical leadership, and technical or professional contributions.
Laurie is a frequent presenter at SIGUCCS Conferences, Past Chair of the SIGUCCS Executive Committee (EC), and New SIG Advisor on the ACM SIG Governing Board. She is currently serving as Past Chair on the EC, Chair of the Marketing Committee, and liaison to the Awards Committee.
Kathy Fletcher and Tom Gerace are also ACM Senior Members. We are proud of their contributions!
Stay Up-to-Date on SIGUCCS News
July 16, 2021 SIGUCCS CommuitySIGUCCS Editor
The SIGUCCS Marketing and Conference Planning Committees work closely together to plan communications to our community through email and social media channels. The marketing committee also helps other volunteers with publicity and communication.
We have planned theme days:
Wednesday: Conference Updates (all channels)
Wednesday: Weekly Challenge Question (Slack)
Thursday: Throw Back Thursday pictures (Facebook Community group)
Monthly on Fridays: Board Updates (Newsletter) and Monthly Recaps (email)
Here are all of the SIGUCCS Communication channels:
email list (SIGUCCS-L)
Conference News Blog (https://siguccs.org/Conference/2022/blog/)
Twitter (@siguccs)
Facebook community (https://www.facebook.com/groups/siguccsfolks/)
SIGUCCS Slack channel (https://siguccs.org/wp/?p=153)
LinkedIn group (https://www.linkedin.com/groups/1675517)
If there is content you'd like to see from SIGUCCS, or you have something you would like to contribute, I'd be happy to talk. Contact me at fox@geneseo.edu
June 2021 – Updates from the Chair
June 25, 2021 Chair's Corner, SIGUCCS BoardSIGUCCS Editor
As we finish our first year as a new board, and what a year it has been, we are excited to share the following updates.
Continue reading June 2021 – Updates from the Chair →
SIGUCCS Hosts Hangout: Past, Present, and Future
April 27, 2021 SIGUCCS PeopleSIGUCCS Editor
While gathering in person wasn't in the cards this year, having a virtual conference came with its unexpected benefits – allowing members past and present from all over the world to connect.
Continue reading SIGUCCS Hosts Hangout: Past, Present, and Future →
SIGUCCS 2021 Conference Takeaway – Accessing Resources
April 21, 2021 2021 Conference, TakeawaySIGUCCS Editor
The conference session on March 14 featured two presentations:
Check This Out: Loaning Laptops and Other Technology with Ashley Koebel from University of Michigan
Revamping a Neglected Technology Space for Success: A Team-Based Approach to Developing Training, Documentation, and Communication for a Virtual Reality Space with Doralyn Rossmann from Montana State University
Continue reading SIGUCCS 2021 Conference Takeaway – Accessing Resources →
SIGUCCS 2021 Conference Takeaway – Lightning Talks
April 9, 2021 2021 Conference, TakeawaySIGUCCS Editor
Fist with blue lightning illustration in vector
Lightning Talks are short seven minute presentations. The presenter provides a concise summary of a single topic with useful information in a short time. This year, we had a nice grouping of lightning talks!
Continue reading SIGUCCS 2021 Conference Takeaway – Lightning Talks →
Newsletter Categories
Newsletter Categories Select CategoryAwardsConference 2014 Conference 2015 Conference 2016 Conference 2017 Conference 2018 Conference 2019 Conference 2020 Conference 2021 Conference 2022 Conference 2023 Conference TakeawayNew PostProfessional DevelopmentSIGUCCS Board Chair's CornerSIGUCCS CommuitySIGUCCS People


text:
Charge Tracker Documentation {#mainpage}
=========
This is the documentation for the [ChargeTracker](https://github.com/ritstudentgovernment/chargeflask) backend, the front-end documentation is located in the [chargevue](https://github.com/ritstudentgovernment/chargevue) repository.

[TOC]

# Getting Started {#gettingstarted}

This guide contains step-by-step instructions on how to install the server, how to get data to the backend and how to get data from it.

## Server Installation {#serverinstall}
Prerequisites: Flask and PostgreSQL
1. Pull project from [this repo](https://github.com/ritstudentgovernment/chargeflask).
2. Install and create [virtual environment](http://python-guide-pt-br.readthedocs.io/en/latest/dev/virtualenvs/).
3. Create a database for the project.
4. Specify the ``SQLALCHEMY_DATABASE_URI`` in the ``config.py`` file with the database path.
5. Activate virtual environment.
6. Execute `pip install -r requirements.txt`
7. Complete the `secrets.py` file (Contact System Administrator for relevant data).
8. Run ``python run.py``

## Testing Utilities {#testutil}
In some cases, you'd like to test if your backend code is properly working. But, because you don't have access to a frontend yet, it might seem quite hard to test your code while developing it. For this, it's recommended to install in your machine a tool called [Socket.io tester](https://electronjs.org/apps/socket-io-tester), which let's so send and retrieve data from the backend without the need of a developed client.

# Data Manipulation {#datamani}
ChargeTracker uses a technology called [SocketIO](https://socket.io/), which basically provides real-time communication between two different machines through WebSockets. SocketIO allows the easy implementation of WebSockets cross-platform and cross-language. For example, the [chargevue](https://github.com/ritstudentgovernment/chargevue) client for this application, uses the native Javascript SocketIO library, meanwhile this backend utilizes the Flask library [Flask-SocketIO](https://flask-socketio.readthedocs.io/en/latest/). Bellow, there are a few detailed examples on how to request or submit data to and from the ChargeTracker backend.

## Submitting Data {#submitData}
Submitting data to ChargeTracker is very simple. Basically, inside each the controllers file of each module there are certain functions that _listen_ for data in a specific event name. For example, inside the controllers of the Committees module, we can see the `get_commitees` method:

~~~~~~~~~~~~~python
##
##  Gets list of all committees.
##
##  	@param      broadcast  Flag to broadcast list of committees to all users.
##
## @emit       Emits a list of committees.
##
@socketio.on('get_committees')
def get_committees(broadcast = False):
    committees = Committees.query.filter_by(enabled = True).all()
    comm_ser = [{"id": c.id, "title": c.title} for c in committees]
    emit("get_committees", comm_ser, broadcast= broadcast)
~~~~~~~~~~~~~

We can see that every method inside the controllers start with the line `@socketio.on('<event_name>')`, this basically means that once the server listens to a request in that specific event, the function under it will be executed. In this case, when a person sends a request with the event `get_committees` to the server, the function get_committees will be executed.

To send this request to the server, the request can be written in the following way:

~~~~~~~~~~~~~{.js}
// Javascript Socket-IO request.
var socket = io.connect('http://localhost:5000');
socket.emit('get_committees');
~~~~~~~~~~~~~

We can also notice that the `get_committees` method gets the parameter broadcast, which basically broadcasts the list of committees to all the users listening, to send data to the server, you can send a JSON object or a String.

For example, if I want my request of committees to be broadcasted to everyone that is listening to `get_committees`, I can do:

```javascript
// Javascript Socket-IO request with params.
var socket = io.connect('http://localhost:5000');
socket.emit('get_committees', true);
```

## Retrieving Data {#retrieveData}

**IMPORTANT**: Also take into account that to receive this data, you have to be listening to the events with the same event_name. So, if I'd like to receive all the committees, you have to listen to the `get_committees` event name:

```javascript
// Javascript Socket-IO listener.
var socket = io.connect('http://localhost:5000');
socket.on('get_committees', function(data){
  	console.log(data);
});
```

These channels are always open, so you only have to listen to data once and you'll keep receiving new data as it becomes available, without having to create new requests!

# Module Structure {#modulestruct}

To keep the source code organized, each functionality of [ChargeTracker](https://github.com/ritstudentgovernment/chargeflask) is divided in different modules. For example, the Users module contains everything relevant to user authentication, meanwhile the Committees module has everything relevant to manipulating committees.

Every module in ChargeTracker contains the following files:

- `__init__.py` (required): This file is usually empty, its necessary to create a Python module.
- `models.py` (required): This file contains the model declaration of the module itself. This usually contains the SQLAlchemy model declaration.
- `controllers.py` (required): This file contains all the relevant controllers to manipulate data on the server, please refer to the examples on the [Data Manipulation](#datamani) section.
- `test_<module_name>.py` (required): This file contains all the test cases for this specific module, please refer to the examples on the Testing Modules section.
- `<module_name>_response.py`: Instead of hard-coding response strings inside the controllers class, all the response strings specific to this module are located inside this class. In the future, this will allow the easy implementation of [ChargeTracker](https://github.com/ritstudentgovernment/chargeflask) in a different language.

# Unit Tests and Code Coverage {#unittests}

Every module in this project is unit tested to check if the code implemented is fully functional. For unit testing, ChargeTracker utilizes the Python testing utility, [pytest](https://docs.pytest.org). To execute your unit tests locally, you can simply run the command: `py.test --cov=./app`, if you want to generate a html coverage report, the same command can be executed with the flag `--cov-report=html`. For code coverage, this project uses [TravisCI](https://travis-ci.org/) and [Codecov](https://codecov.io/gh), the TravisCI .yml file is located under `.travis.yml`, which basically contains the configuration that Travis needs to execute your code. Also, take into account that TravisCI needs a copy of your `secrets.py` to execute your code, this is done by encrypting your secrets file with TravisCI.

redpajama_set_name: RedPajamaGithub


(CNN) -- Hundreds of pages of search warrant documents for Michael Cohen released Tuesday give the public a new understanding of how special counsel Robert Mueller has tackled his broad mandate and provide fresh details of the investigation into President Donald Trump's former lawyer and fixer.
The materials show that Cohen was an early target of Mueller's investigation into possible collusion between Trump's campaign and Russia and obstruction of justice, and came under intense scrutiny from New York federal prosecutors, with investigators seeking everything from Cohen's email accounts to phone records to historical data of his cell phones' locations the month before the 2016 election.
The Cohen-related material that has posed the greatest potential danger to the President comes from the Southern District of New York's investigation into the hush-money payments to women alleging affairs with Trump. The search warrant materials related to that matter, however, were redacted in Tuesday's disclosures.
It didn't take Mueller's team long to start ramping up its probe into Cohen, search warrants unsealed Tuesday show, and the special counsel's office likely took a similar tack with its other targets.
Prosecutors and the FBI were given approval to search Cohen's various email accounts in July, August and September 2017, as the investigation into Trump's then-personal lawyer played out quietly behind the scenes before any charges had been filed by the special counsel's office.
It's been a familiar pattern for Mueller's investigation where the public only learns about investigations into key figures months later -- often through court filings -- like when former Trump campaign adviser George Papadopoulos' guilty plea was revealed in October 2017 on the day that Mueller's first charges were unsealed.
Republicans have criticized Mueller's investigation for stretching on for nearly two years, and many have urged him to wrap it up already. But the search warrants released Tuesday show that Mueller's team has moved at a swift pace throughout its 22-month probe, with months of investigative work preceding the actual public filing of charges.
If Mueller is in fact nearly done, the length of his investigation could be dwarfed by similar probes, like special prosecutor Ken Starr's investigation into President Bill Clinton.
Cohen's campaign finance crimes -- in which he made or orchestrated payments to silence women during the 2016 presidential election who claimed sexual encounters with then-candidate Trump -- were arguably the most significant charges against him because he implicated the President when he pleaded guilty, but the documents released Tuesday contain only redacted material related to those payments.
That's because the campaign finance violations are related to an ongoing Manhattan US Attorney's investigation into whether any executives at the Trump Organization committed crimes connected to the company's efforts to reimburse Cohen for the hush money he paid to one of the women, adult-film actress Stormy Daniels, CNN has previously reported.
After media organizations, including CNN, sought to have the search warrant materials redacted, Manhattan federal prosecutors argued that they should be able to redact information related to ongoing probes, and US District Court Judge William Pauley agreed.
"The Government represents that aspects of its investigation remain ongoing, including those pertaining to or arising from Cohen's campaign finance crimes," Pauley wrote in his order. "Indeed the search warrant applications and affidavits catalogue an assortment of uncharged individuals and detail their involvement in communications and transactions connected to the campaign finance charges to which Cohen pled guilty."
The inquiry regarding the Trump Organization has been active since Cohen pleaded guilty in August 2018, and last month CNN reported that federal prosecutors in New York had requested interviews with executives at the company.
The search warrant materials document prosecutors' ability to examine multiple email accounts held by Cohen, but one matter left unexplained by the materials is why prosecutors appear to have been unable to search the account he used while working for Trump's family business, the Trump Organization.
Prosecutors sought to examine that account, according to the documents, but weren't able to do so. "Indeed from my involvement in this investigation, I know that Cohen had an email account with the Trump Organization, but the [US Attorney's Office] and FBI have not been able to obtain the contents of that account to date," the unnamed FBI agent who filed the affidavit for the search warrant application wrote.
It's unclear from the document whether prosecutors sought and were denied a search warrant for that account or whether they were unable to access it for some other reason. It's also not clear whether prosecutors were ultimately able to examine that account at some point after the April 9, 2018, searches of Cohen's home, hotel room, office and safety deposit box.
But since material seized from Cohen had to go through a process wherein a third party determined what was subject to attorney-client privilege -- and it doesn't appear that any material seized after the April 9 searches went through such a process -- it is unlikely that prosecutors were ultimately able to review Cohen's Trump Organization email account, at least not via a search warrant.
When Cohen was sentenced to three years in prison in December, Judge William Pauley said that Cohen had pleaded guilty to "a veritable smorgasbord of fraudulent conduct."
He was actually investigated for even more crimes that he wasn't ultimately charged with.
Cohen pleaded guilty to tax crimes, campaign finance violations tied to the hush money payments, false statements to a bank and lying to Congress about the length of discussions surrounding the Trump Tower Moscow project. The search warrant documents released Tuesday show federal prosecutors also suspected that Cohen could have violated foreign lobbying laws and committed money laundering.
Federal investigators asked for access to records related to Cohen's consulting business, Essential Consultants LLC, which was also the shell company used to make the hush-money payments to women.
It's not clear exactly why Cohen wasn't charged with those crimes. One possibility is simply that investigators didn't have enough evidence to charge him with those crimes. Another is that when Cohen agreed to plead guilty, prosecutors agreed to drop other potential charges.
Violations of the foreign lobbying law, the Foreign Agents Registration Act, have rarely been prosecuted, although Mueller's investigation -- including charges against former Trump campaign chairman Paul Manafort and his deputy Rick Gates — has shown that sentiment could be on the way out.
Cohen has company as someone who was questioned over foreign lobbying work but not ultimately charged: former Trump national security adviser Michael Flynn was scrutinized for his Turkish business, but he was charged by Mueller only with lying to the FBI.
Mueller has not spoken publicly since he was appointed special counsel in May 2017. Instead, the special counsel's office has let its court filings do the talking to lay out its cases against senior Trump officials and Russian intelligence officers it says hacked Democrats during the 2016 campaign.
But the search warrant documents show just how much Mueller's team has uncovered in its 22-month investigation, which has led to charges against 37 individuals and entities.
Mueller's team was allowed to review years of Cohen's emails and other online data during his time working for Trump. Prosecutors and the FBI sought search warrants for Cohen's various email accounts beginning in July 2017, and received approvals in November 2017 and January 2018 to track Cohen's incoming and outgoing calls. The search warrants sought Cohen's emails related to his business dealings under a shell company used to cut hush money deals with women accusing then-candidate Trump of extramarital affairs, which Trump has denied.
It's still unknown what Mueller will say when he submits a report to Attorney General William Barr concluding his investigation, but the material offered in the search warrant documents provides a small glimpse into the depth of Mueller's information-gathering.
When Democratic investigators like House Judiciary Chairman Jerry Nadler and House Intelligence Chairman Adam Schiff say they want to obtain Mueller's underlying evidence, this is what they're after.
CORRECTION: This story has been updated to correct the length of the Mueller probe compared with similar investigations.


text:
Up2.1. Introduction2.2. Scientific journals2.3. Professional journals2.4. Conference papers2.5. Books and book chapters2.6. Textbooks2.7. Technical reports2.8. The world wide web2.9. Encyclopedias2.10. Library catalogues2.11. Bibliographic databases Exercise2.12. Exercise
Information literacy course
Selecting the right information sources
See 2. Selecting the right information sources
2.2. Scientific journals
A scientific journal is a collection of articles written by scholars in an academic field. An editorial board reviews articles to decide whether they should be accepted. Articles in scientific journals usually cover very specific topics or narrow fields of research. Since journals are published on a regular or periodic basis they are grouped in the category called "periodicals." Electronic journals, called e-journals, are published on the Web by scholarly organizations and publishers and are made available to you from the ITC library web pages.
A scientific journal article is an original contribution that appears in a published scientific journal.
We distinguish two types of scientific journals:
ISI journals
The Institute for Scientific Information (ISI) is part of Thomson Scientific, a commercial scientific information provider, which includes the search engine and citation index Web of Science and the Journal Citation Reports (JCR) available via the ITC library website.
The ISI selects among all the journals a list of so-called ISI journals. Many institutions and organizations, including ITC, only give credit to researchers for papers published in thee journals.
There are currently about 9 000 ISI journals in all fields, about 3 700 of them in the sciences. These are selected from a much larger group of candidate journals by expert editors by a holistic process.
The selected journals are referred to as ISI journals, and are listed in the master journal list and in the Journal Citation Reports (JCR).
Papers in these journals are included in three Citation Indices, including the Science Citation Index, also from Thompson. These are the primary searchable databases to find references.
The practical point for the beginning researcher (e.g. in an MSc project) is that ISI journals have been evaluated as:
reliable (especially their quality control and editorial policies), and
important (i.e. their papers are cited by other researchers)
so articles in ISI journals should be used in preference to those in non-ISI journals if possible
Examples of ISI Journals:
Canadian Journal of Remote Sensing (Canadian Aeronautics & Space Institute)
GIScience & Remote Sensing (Bellwether Publishing, USA)
IEEE Geoscience and Remote Sensing Letters (IEEE)
IEEE Journal of Selected Topics in Applied Earth Observations and Remote Sensing
IEEE Transactions on Geoscience and Remote Sensing (IEEE)
International Journal of Remote Sensing (Taylor & Francis)
ISPRS Journal of Photogrammetry and Remote Sensing (Elsevier Science for ISPRS)
Photogrammetric Engineering and Remote Sensing (American Society of Photogrammetry and Remote Sensing)
Non-ISI journals
Recalling the standard to qualify as an ISI journal (selectivity, review process, reputation of its editorial board, international diversity of its authorship; importance of its papers) there are journals that fail some of these (e.g. not diverse, not important) yet still have good quality control.
Thus there may be useful information in their papers. For example, as of 28-April-2009 there were 2 082 journals published by Springer, of which 794 were listed in JCR; this is a ratio of about 38%. So only a bit over 1/3 of the journals from this well-regarded publisher are ISI.
This is especially true for non-ISI journals published by regional or national societies whose main reason for omission from the ISI list is their narrow geographic focus. Papers in these journals may include valuable information, especially about local conditions, for example in a study area.
You may use a Scientific journal
when doing scholarly research.
to find out what has been studied on your topic.
to find reference lists that point to other relevant research.
Examples of scientific journals
Remote Sensing of Environment
IEEE Transactions on Geoscience and Remote Sensing
International journal of geographical information science IJGIS
Further reading: Peer review: a guide for researchers


I've been participating in the dwicontest the last couple of weeks, and here's the six icons I've made for two challenges, plus four outtakes that weren't submitted.
Sally Sparrow! I loved her. These are really pretty. Congratulations on winning the award!
Thanks! Yeah, I really liked Blink--well, in a terrifying way. ;) But I like Sally Sparrow, too--they should bring her back.


FinTech startup Koho recently launched a new feature and website to help Canadians find hidden bank fees.
The initiative, entitled What The Fee, allows people to connect their bank account to the site and scan bank statements to see how much they are paying when it comes to additional fees.
Toronto-based Koho, which looks to provide an alternative to traditional banking, and noted that the average Canadian pays around $159 in bank fees per year. It also pointed out that is some of the highest fees in the world, and more than most savings accounts (with a $12,000 balance) would earn in annual interest.
To create and launch its new tool, Koho teamed up with Flinks, a Montreal-based financial data company that works with businesses to connect users' bank accounts to gain financial insights into their customers and provide better risk products.
Flinks itself closed a $1.75 million seed funding round in July, using the funds to help it expand its product platform. The funding was led by Luge Capital and National Bank, with participation from Innostart Capital, Panache Ventures, iNovia Capital, and Conconi Growth Partners.
The Montreal company has previously partnered with UK-based Transferwise, and online lender, Merchant Advance Capital.
Earlier this year, Koho launched joint account feature, allowing users to share finances with their partner, family member, or friend using a joint prepaid Visa card.
In 2017, the company raised $8 million from Portag3 Ventures, after securing funding from Power Financial Corporation and other angel investors a year before. In 2015, it also raised $1 million seed round and earlier that year partnered with Visa, Peoples Trust Company, and Galileo Processing in preparation for launching its banking service.
Regarding What The Fee's launch, Paul Desmarais III, the executive chairman of Portag3 Ventures noted that Canada has the opportunity to join other countries and become a North American innovator in open banking.
"Canadians have the right to transparency in their interactions with their bank account and people living pay cheque to pay cheque are paying very high fees relative to their means. They deserve more access and innovation at the lowest possible cost," he stated.


text:
Q: Comparing statsmodel predictions with actual y-values (indexing issue) Objective: I would like to calculate the test error for predictions made by a fitted multivariate linear regression model.
Problem: Here is my code. It is intended to fit a linear regression model to training data, and then predict y-values (price) based on X_test variables:
X.insert(0, 'constant', 1)   # insert constant column
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2)
lm_sm = sm.OLS(y_train, X_train).fit()

y_pred = pd.DataFrame()   # dataframe for predictions vs actual y-values
y_pred['predictions'] = lm_sm.predict(X_test)

print y_test.sort_index().head()
print y_pred.sort_index().head()

The code outputs the following:
       price
6   257500.0
17  485000.0
23  252700.0
25  233000.0
26  937000.0
     predictions
0  509428.615367
1  324403.584917
2  477385.431339
3  484962.235105
4  827039.820936

Comparing predicted and actual price, this is obviously not right. The predict() method does not retain the indexes from my train/test split. So when I compare the predicted prices to actual prices, I can't be sure I'm comparing the right values.
The only solution I've thought of (and I'm not sure if this is right thinking) is to sort X_test when making the prediction, i.e. y_pred['predictions'] = lm_sm.predict(X_test.sort_index()). The predictions look a little more aligned with my expectations (note this is a first linear regression / benchmark so no feature engineering has been applied yet):
       price
6   257500.0
12  310000.0
18  189000.0
25  233000.0
29  719000.0
     predictions
0  259985.788272
1  590648.478023
2  339621.126287
3  316402.199424
4  635513.611634

I would then perform the test error calculation based on these sorted dataframes. Is this right? Is there a cleaner way to do this? A method I don't know about? Any help/thoughts would be appreciated, thank you!

A: I actually don't think there's anything wrong with the ordering. The mixed-up indices of y_test, when compared to the clean indices of y_pred, are the source of confusion.
When you use train_test_split on your dataset (X, y), it's obviously going to shuffle things around, which is why y_test has a shuffled subset of indices.
When you do lm_sm.predict(X_test), the output is a plain numpy array. It's not a pandas object with an index, so that index info is already lost. Plus, in your code, you store the results in a new y_pred dataframe. And in that fresh y_pred, the indices will be a fresh auto-increment: 0, 1, 2 ...
So just don't use sort_index(), and you can be sure that y_pred as you have it will line up with X_test and y_test.
EDIT. Hopefully this illustrates things better.
import numpy as np
import pandas as pd
from sklearn.model_selection import train_test_split
from statsmodels.regression.linear_model import OLS

X = pd.DataFrame(np.random.random((60, 3)))
y = pd.DataFrame(np.random.random((60, 1)))

X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2)

print(X_test)

The test set's features:
           0         1         2
2   0.547993  0.479149  0.495539
48  0.332964  0.857635  0.501391
23  0.380500  0.377257  0.088766
35  0.045725  0.432096  0.239523
52  0.254861  0.207215  0.985722
37  0.099525  0.205250  0.054000
22  0.426227  0.253524  0.336110
43  0.716443  0.006443  0.423447
49  0.146820  0.803366  0.390921
6   0.127666  0.848561  0.936604
46  0.303034  0.548064  0.852688
33  0.516726  0.977396  0.829725

And the test set's targets with print(y_test):
           0
2   0.123253
48  0.494307
23  0.312021
35  0.939558
52  0.958955
37  0.681215
22  0.181427
43  0.907552
49  0.589316
6   0.613305
46  0.947220
33  0.696609

Indices are shuffled, but they agree. Now do:
lm_sm = OLS(y_train, X_train).fit()
y_pred = pd.DataFrame()   # dataframe for predictions vs actual y-values
y_pred['predictions'] = lm_sm.predict(X_test)

# Print this directly
print(lm_sm.predict(X_test))

Last line is just a plain numpy array:
[ 0.44549342  0.44973765  0.24465328  0.17840542  0.42329909  0.09567253
  0.30675321  0.38496281  0.33836597  0.49959203  0.47488055  0.63751567]

And when you look at the new y_pred dataframe, the indices are a fresh 0, 1, 2 ... print(y_pred)
    predictions
0      0.445493
1      0.449738
2      0.244653
3      0.178405
4      0.423299
5      0.095673
6      0.306753
7      0.384963
8      0.338366
9      0.499592
10     0.474881
11     0.637516

You might have been surprised that these indices don't match those of y_test, but as I showed, the predict() function returns a plain numpy array, and there's nothing connecting the resulting y_pred to the original indices anymore. Nevertheless, you can be sure that everything is aligned.

A: Note that train_test_split now has optional parameter shuffle=False to avoid shuffling.  That should solve your problem with alignments in the df.

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