Initial Topology on a Locally Compact Hausdorff Space If I consider the initial topology generated from the continuous functions out of a locally compact Hausdorff space into $\mathbb{R}$, is this initial topology the original topology? I know it's Hausdorff, how could I show that it is once again locally compact?
 A: If $(X, \mathcal{T})$ is Tychonoff ($T_{3{1 \over 2}}$) then the initial topology (say $\mathcal{T}'$) w.r.t. $C(X)$, the set of all continuous functions from $(X, \mathcal{T})$ to $\mathbb{R}$, is the original topology of $X$. And locally compact Hausdorff spaces are Tychonoff.
The Tychonoff fact is not too hard to see: clearly $\mathcal{T}' \subset \mathcal{T}$, as the functions in $C(X)$ are all continuous w.r.t. $\mathcal{T}$.
And if $x \in O$ and $O$ is in $\mathcal{T}$, then by $X$ being Tychonoff we have a continuous function $f \in C(X)$ such that $f(x) = 0$ and $f[X \setminus O] = \{1\}$. But then $O' = f^{-1}[(\leftarrow, \frac{1}{2})]$ is in $\mathcal{T'}$ and $x \in O'$ and $O' \subset O$ (if $x \in O'$, $f(x) < \frac{1}{2}$, so $x$ cannot be in $X\setminus O$, as all those map to $1$ under $f$, so $x \in O$). This shows that $\mathcal{T'}$ forms a base for $\mathcal{T}$, so $\mathcal{T} \subset \mathcal{T}'$ and we have equality.
One way to see that locally compact Hausdorff spaces $X$ are Tychonoff: the one point compactification $\alpha X$ is compact Hausdorff (so normal $T_1$, hence Tychonoff) and Tychonoff is a hereditary property, so $X$ is too.
A: Given a locally compact Hausdorff space $(X, \tau)$, I assume that the initial topology $(X, \tilde \tau)$ are the weak topology generated by ALL continuous functions from $X$ to $\mathbb R$. 
To show $\tau = \tilde \tau$, it suffices to show that all $U\in \tau$ is generated by elements in $\tilde \tau$. 
Let $x\in U$. As $X$ is locally compact, there is a compact $K$ so that $x\in K$ and $K \subset U$. By Urysohn's lemma, there is a continuous function (with respect to $\tau$) $f: X \to \mathbb R$ with compact support so that $0\leq f\leq 1$, $f\equiv 1$ on $K$ and $f\equiv 0$ outside $U$. As a result, the set
$$V_x = f^{-1} (1/2, 3/2)$$
is in $\tilde \tau$ and $V_x \subset U$. Thus 
$$U = \bigcup_{x\in U} V_x$$
and so $\tau \subset \tilde \tau$. 
