What is the meaning of $A =\prod_{jI am trying to parse this proof in Hungerford's book:



When he gets into computing this:
$$A =\prod_{\substack{j<k\\ j,k \neq c,d}}^{}(i_j-i_k)$$
What does this means?

*

*Does it means that $j<k$ and  $j\neq c$ and $k\neq d$; or

*Does it means that $j<k$ and  $j\neq c$ and $k\neq d$ and $h\neq d$ and $k\neq c$?

I computed the indices I'd get in both cases in a simple example where $c=3$, $d=5$ and $j=9$ and it for the first case, I'd get the following indices:
$$\begin{array}{cccccccc}
\begin{array}{cc}
 1 & 2 \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 3 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 3 \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 4 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 4 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 6 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 6 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 6 \\
\end{array}
 & 
\begin{array}{cc}
 5 & 6 \\
\end{array}
 & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 7 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 7 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 7 \\
\end{array}
 & 
\begin{array}{cc}
 5 & 7 \\
\end{array}
 & 
\begin{array}{cc}
 6 & 7 \\
\end{array}
 & \text{} & \text{} \\
\begin{array}{cc}
 1 & 8 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 8 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 8 \\
\end{array}
 & 
\begin{array}{cc}
 5 & 8 \\
\end{array}
 & 
\begin{array}{cc}
 6 & 8 \\
\end{array}
 & 
\begin{array}{cc}
 7 & 8 \\
\end{array}
 & \text{} \\
\begin{array}{cc}
 1 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 9 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 5 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 6 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 7 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 8 & 9 \\
\end{array}
 \\
\end{array}$$
For the second case, I'd get the following indices:
$$\begin{array}{cccccccc}
\begin{array}{cc}
 1 & 2 \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
 \diamond \;\diamond  \\
\end{array}
 & 
\begin{array}{c}
\diamond \;\diamond  \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 4 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 4 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 6 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 6 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 6 \\
\end{array}
 & 
\begin{array}{c}
 \diamond \;\diamond  \\
\end{array}
 & \text{} & \text{} & \text{} \\
\begin{array}{cc}
 1 & 7 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 7 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 7 \\
\end{array}
 & 
\begin{array}{c}
 \diamond \;\diamond \\
\end{array}
 & 
\begin{array}{cc}
 6 & 7 \\
\end{array}
 & \text{} & \text{} \\
\begin{array}{cc}
 1 & 8 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 8 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 8 \\
\end{array}
 & 
\begin{array}{c}
 \diamond \;\diamond  \\
\end{array}
 & 
\begin{array}{cc}
 6 & 8 \\
\end{array}
 & 
\begin{array}{cc}
 7 & 8 \\
\end{array}
 & \text{} \\
\begin{array}{cc}
 1 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 2 & 9 \\
\end{array}
 & 
\begin{array}{c}
 \text{$\bullet$ $\bullet$} \\
\end{array}
 & 
\begin{array}{cc}
 4 & 9 \\
\end{array}
 & 
\begin{array}{c}
 \diamond \;\diamond \\
\end{array}
 & 
\begin{array}{cc}
 6 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 7 & 9 \\
\end{array}
 & 
\begin{array}{cc}
 8 & 9 \\
\end{array}
 \\
\end{array}$$
Where $\bullet \;\bullet$ and $\diamond \;\diamond$ are the deleted pairs of indices. My guess is that it means the second case because it seems only this way would allow us to have $\sigma(A)=A$, since no $(i_j - i_k)$ would be changed. Is my reasoning correct?

I feel like Hungerford may have painted himself into a corner with that notation, so let me offer you a different way of doing exactly what the proof is doing.
For a fixed $n$, consider commuting variables $x_1,\ldots,x_n$ and form the Vandermonde matrix $V=V(x_1,\ldots,x_n)$ with $V(x)_{ij} = x_i^{j-1}$. It is a pleasant exercise to verify that $
\det V$ is equal to $\prod_{i<j}(x_i-x_j)$ and, in particular, it evaluates to a non-zero value whenever we pick distinct values for the variables, as Hungerford stated. Let us call this polynomial $\Delta$.
The symmetric group $S_n$ acts on the variables $x_i$ by its action on the indices, so in particular it acts on each polynomial in the variables $x_1,\ldots,x_n$ and hence on $\Delta $.
The first observation we make is that for a transposition the determinant changes by $-1$,  since in the way we have written it, it is clear two rows are swapped. This means that if $\sigma$ is written as a product of $N$ transpositions, then $\sigma \Delta  = (-1)^N\Delta$. Assume that $\sigma$ is also written as a product of $M$ traspositions.
To conclude that $N=M$ modulo $2$, it suffices we evaluate $\Delta$ at a the point say $p=(1,2,\ldots,n)$ as suggested in your post, in which case $\Delta(p)\neq 0$ and we obtain that $(-1)^N = (-1)^M$.
