0
$\begingroup$

In Big Rudin, theorem 2.7, Rudin states that if there is a compact set $K\subset U$ where $U$ is an open set in a locally compact Hausdorff space $X$, then there is an open set $V$ with compact closure such that:

$K\subset V\subset \overline{V}\subset U$.

In theorem 2.13, he states that if we have $K\subset V_1\cup\ldots\cup V_n$ where $V_1,\ldots,V_n$ are open, then by theorem 2.7, every $x\in K$ has a neighborhood $W_x$ with compact closure $\overline{W_x}\subset V_i$ for some $i$ depending on $X$.

I don't see why this is true to be honest. All that theorem 2.7 tells us is that $\overline{W_x}\subset V_1\cup\ldots\cup V_n$.

What did I miss?

$\endgroup$
1
  • 1
    $\begingroup$ Since $K\subset V_1\cup\dots\cup V_n$, for every $x\in K$ there is some $i$ such that $x\in V_i$. Since $V_i$ is open, there is an open set $U$ such that $x\in U\subset V_i$. Notably, $\{x\}\subset U$, and $\{x\}$ is compact. Now apply Theorem 2.7. (edit: I guess you can just take $U = V_i$...) $\endgroup$
    – Joey Zou
    Commented Dec 20, 2015 at 17:40

1 Answer 1

Reset to default
2
$\begingroup$

Since $K\subset V_1\cup\dots\cup V_n$, for every $x\in K$ there is some $i$ such that $x\in V_i$. Notice that $\{x\}\subset V_i$, $\{x\}$ is compact, and $V_i$ is open. Now apply Theorem 2.7.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .