Question: > 
> Fix a set $X$ with at least 2 elements and $p\in X$. Let $\mathcal{T}\_p = \{ U \subset X \mid p \notin U \} \cup \{ X \}$.
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> 1. Show that $\mathcal{T}\_p$ is a topology and denote this topological space by $X$.
> 2. Is $X$ connected?
> 3. Is $X$ Hausdorff?
> 4. Let $(x\_n)$ be the sequence with $x\_n = a$ for all $n$ for some $a \in X$. For what values of $a$ does $(x\_n)$ converge? If $(x\_n)$ converges, where does it converge to?
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My work.

1. $X, \emptyset \in \mathcal{T}\_p$.

Suppose $U\_1 \subset X$ and $U\_2 \subset X$ and $p \notin U\_1$ and $p \notin U\_2$ then $p \notin U\_1 \cup U\_2 \subset X$ so $U\_1 \cup U\_2 \in \mathcal{T}\_p$ Hence unions of elements of $\mathcal{T}\_p$ are in $\mathcal{T}\_p$.

Suppose $U\_1 \subset X$ and $U\_2 \subset X$ and $p \notin U\_1$ and $p \notin U\_2$ then $p \notin U\_1 \cap U\_2 \subset X$ so $U\_1 \cap U\_2 \in \mathcal{T}\_p$ Hence intersections of elements of $\mathcal{T}\_p$ are in $\mathcal{T}\_p$.
2. $X$ is connected since the only set in $\mathcal{T}\_p$ containing $p$ is the entire set $X$, and hence there cannot exist a separation.
3. $X$ is not Hausdorff since it is connected

I think (1) and (2) are right, I think (3) is wrong and I don't know what to do for (4).

Answer: 
1. This is almost correct, but for the union part write it up for arbitrary unions, not just finite ones. Make explicit why $\emptyset \in \mathcal{T}\_p$.
2. The reason is essentially correct, make the write-up more formal: suppose $X = A \cup B$ is a separation of $X$. One of the sets contains $p$ etc...
3. This has a bogus argument: the reals are connected and very Hausdorff. Think of the same reason why $X$ is connected (neighbourhoods of $p$) and use that $X$ has another element besides $p$...
4. Just use the definition of convergence. And consider for limits all $x \neq p$ and $p$ separately. What is a typical open neighbourhood of $x \neq p$ and $p$? Does the sequence stay inside the neighbourhood or not?
An open neighbourhood of a point $x\neq p\in X$ is $X$ or any subset of $X$ which contains $x$, but not $p$. In particular $\{x\}$ is an open neighbourhood of $x$, hence two distinct points $x,y\neq p$ have disjoint neighbourhoods and $X$ would be Hausdorff, except… the only neighbourhood of $p$ is $X$, so that the T2 axiom (Hausdorff space) is not true for $x$ and $p$.

However, $\mathscr T\_p$ satisfies axiom T0 (Kolmogorov space): for any points $x\neq y$, there exists a neighbourhood of one of them that doesn't contain the other.