I was wondering if someone could review my proof for correctness. Thanks in advance!
Let X = [0,1]$^{\omega}$, the infinite dimensional space as indicated by $\omega$. And let X have the uniform topology.
Hence X is a metric space under the metric $\bar{p}$(x,y) = sup{$\bar{d}$(x$_\alpha$, y$_\alpha$) | $\alpha$ $\in$ $J$ }, where $\bar{d}$(x, y) = min{|x - y|, $1$}
Then X is also a Hausdorff space since the product of Hausdorff spaces is a Hausdorff space.
Now suppose X is locally compact. Then take x = (${\frac12}$,${\frac12}$, ${\frac12}$, ... ). Then since X is locally compact Hausdorff there must be a neighborhood U of x such that there is a neighborhood V of x where $\bar{V}$ is compact and $\bar{V}$ $\subset$ U.
In the neighborhood V take the open ball, $B$ = B$_\bar{p}$(x,$\epsilon$) $\subset$ $\bar{V}$. $\bar{B}$ is closed in a compact space $\bar{V}$ since $\bar{B}$ is closed in X and fully contained in $\bar{V}$, so $\bar{B}$ must itself be compact as it is a closed set in a compact space.
Then consider the point:
$z$ = (${\frac12}$, ${\frac12}$ + $\frac\epsilon2$, ${\frac12}$ + ($\epsilon$ - $\frac\epsilon3$), ${\frac12}$ + ($\epsilon$ - $\frac\epsilon4$), ${\frac12}$ + ($\epsilon$ - $\frac\epsilon5$) , $...$ )
Then $\bar{p}$(x,z) = $\epsilon$ since the difference between each component of x (the constant $\frac12$ tuple) and z is an increasing sequence along the indices with a supremum of $\epsilon$. Hence $z$ $\notin$ B. But $z$ is also not in $\bar{B}$ since one can take extremely small epsilon neighborhoods around each component of $z$ and the points in those neighborhoods will also create a similar sequence that converges to ${\frac12}$ + $\epsilon$. [ Note: I say sequence as in it's clear that when the tuple is looked at as a sequence it converges to ${\frac12}$ + $\epsilon$. ]
Now consider the sequence made up of points of the compact space $\bar{B}$, where $z_i$ = $i^{th}$ component of $z$. Set t = x + $\epsilon$.
{x$_n$} =
x$_1$ = ($z_1$, $t$, $t$, $t$,$...$)
x$_2$ = ($z_1$, $z_2$, $t$, $t$, $t$, $...$)
x$_n$ = ($z_1$, $z_2$, $z_3$, ... , $z_n$, $t$, $t$, $t$, $t$ ,$...$ )
Then the above sequence converges to $z$. Since given any $\epsilon$$_0$ $\gt$ $0$ we can find an $N$ such that for n $\ge$ $N$ we have $\bar{p}$(x$_n$, $z$) $\le$ $\epsilon$$_0$.
Since the sequence {z$_i$} converges to $t$ we can find an $I$ such that for $i$ $\ge$ $I$ we have: (z$_i$ - t) $\le$ $\epsilon$.
Then take $N$ = $I$ and we have for any n $\ge$ $N$, $\bar{p}$(x$_n$, z) < $\epsilon$, since for all the components in x$_n$ up to $n$ the difference between $z_i$ and $x_i$ will be $0$. For the components greater than $n$ we have the difference in all cases equal to: t - z$_i$ where z$_i$ has already been guarenteed to be within $\epsilon$ of t.
So, we have constructed a sequence of points of $\bar{B}$ that converges to a point $z$ $\notin$ $\bar{B}$. But $\bar{B}$ is a compact metric space. Now every sequence in a compact metric space must have a convergent subsequence, but we have a sequence that converges to a point outside $\bar{B}$. Hence no subsequence of {x$_i$} can converge since otherwise the {x$_i$} as a sequence of points in the total space X would converge to $z$ $\notin$ $B$ and have a subsequence converge to a different point which would be in $\bar{B}$, which is impossible. Hence $\bar{B}$ cannot be compact, a contradiction.
Therefore $X$ cannot be locally compact.