For part of a problem, I need to show an open set $U \subset \mathbb{C}$ is a $\sigma$-compact locally compact hausdorff space. ($\sigma$-compact is countable union of compact sets)
I suspect its trivial – but I'm having trouble reasoning. Thoughts?
An open set in complex plane is the countable union of compact sets and locally compact hausdorff
compactnesscomplex-analysisgeneral-topology
Related Solutions
The argument is basically just fine, but the notation in that last line is a bit off: when you write $\bigcup_{n\in\mathbb{N}}$, you’re saying that you want to take the union of some sets indexed by natural numbers. What follows $\bigcup_{n\in\mathbb{N}}$ should be a typical one of these sets with index $n$. If you knew that every member of $\mathcal{B}$ was compact, for instance, you could write $X = \bigcup_{n\in\mathbb{N}}\operatorname{cl}B_n$. What you want here is simply $X = \bigcup \{\operatorname{cl}B:B \in \mathcal{B}\text{ and}\operatorname{cl}B\text{ is compact}\}$.
If fact, you never needed to index $\mathcal{B}$ in the first place. It’s perfectly fine to say this:
- Let $\mathcal{B}$ be a countable base. Since $X$ is locally compact, for each $x\in X$ there is an open nbhd $U_x$ of $x$ with compact closure, and since $\mathcal{B}$ is a base for $X$, there is some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq U_x$. Clearly each $B_x$ has compact closure, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$. There are only countably many distinct members of $\mathcal{B}$, so there are only countably many different sets $B_x$, and we have therefore expressed $X$ as the union of countably many compact subsets, as desired.
Or you could replace that last sentence with something like this:
- $\{B_x:x\in X\} \subseteq \{B \in \mathcal{B}:\operatorname{cl}B\text{ is compact}\} \subseteq \mathcal{B}$, so $\{B_x:x\in X\}$ is countable, and $X = \bigcup\limits_{x\in X}\operatorname{cl}B_x$ therefore expresses $X$ as a countable union of compact subsets.
There are many other perfectly good ways to do it; I’m just trying to give you some idea of what’s possible, since you’re working on the proof-writing as much as you are on the mathematics itself.
Suppose $X$ is a locally compact Hausdorff space and $Y \in \Gamma(X)$. I'll prove that $Y$ is a locally compact Hausdorff space with the subspace topology.
I
Suppose $x,y \in Y$ and $x \not= y$.
- Then $\exists V_i \in \Gamma(X),i \in V_i$ where $i=x,y$ and $V_x \cap V_y = \phi$. This follows from X being Hausdorff.
- For $i=x,y$, put $G_i = Y \cap V_i$. Then $G_i \in \Gamma(Y)$, $i \in G_i$ and $G_x \cap G_y=\phi$.
Hence $Y$ is Hausdorff.
II
Let $x \in Y$. We have:
- $\{x\} \subset X$. $\{x\}$ is $X-$compact.
- By theorem 2.7 of RCA Rudin:
$\;\;\;\;\exists V \in \Gamma(X), \{x\} \subset V \ \subset cl_X(V) \subset Y \ \text{ such that } cl_X(V)\text{ is $X-$compact.}$
$cl_X(V)$ is $Y-$compact as well. Also $V \in \Gamma(Y)$
As $cl_X(V) = cl_X(V) \cap Y$, $cl_X(V)$ is $Y-$closed.
By 2 and 4, $cl_Y(V) \subset cl_X(V)$.
As closed subsets of compact sets are compact,$cl_Y(V)$ is $Y-$compact.
Now we have all the ingredients ready. We have:
- $V \in \Gamma(Y)$ such that $x \in V$.
- $cl_Y(V)$ is $Y-$compact. Therefore Y is locally compact.
Best Answer
Local compactness is trivial: for each $w\in U$, consider a closed ball centered at $z$ and contained in $U$. Being Hausdorff is also trivial: each metric space is Hausdorff.
In order to prove that $U$ is $\sigma$-compact, let$$K_n=\left\{z\in U\,\middle|\,\lvert z\rvert\leqslant n\wedge d(z,\delta U)\geqslant\frac1n\right\}.$$Then each $K_n$ is compact and $U=\bigcup_{n\in\mathbb N}K_n$.