Is being locally compact a topological property?

Question

I want to make sure that locally compact is a topological property. If it is right, then the rational number are not homeomorphic to positive integers numbers.

Answer 1

There are several slightly different definitions of local compactness; they are equivalent for Hausdorff spaces but not necessarily for non-Hausdorff spaces. All of them define topological properties, however. I’ll prove this for one of the most common definitions: $X$ is locally compact if each point of $X$ has a local base of compact nbhds.

Suppose that $X$ is locally compact in this sense and that $f:X\to Y$ is a homeomorphism onto $Y$. Let $y\in Y$; there is an $x\in X$ such that $f(x)=y$. Let $\mathscr{B}$ be a local base of compact nbhds of $x$; continuous maps preserve compactness, so $\{f[B]:B\in\mathscr{B}\}$ is a family of compact sets, each of which contains $y$. Let $\mathscr{C}=\{f[B]:B\in\mathscr{B}\}$; we want to show that $\mathscr{C}$ is a local base at $y$.

Suppose that $U$ is any open nbhd of $y$. Then $f^{-1}[U]$ is an open nbhd of $x$, so there is a $B\in\mathscr{B}$ such that $x\in\operatorname{int}_XB\subseteq B\subseteq f^{-1}[U]$; clearly $y\in f[\operatorname{int}_XB]\subseteq f[B]\subseteq U$. Moreover, $f$ is a homeomorphism, so $f[\operatorname{int}_XB]$ is open in $Y$, and therefore $y\in\operatorname{int}_Yf[B]$. Thus, $f[B]$ is a compact nbhd of $y$ that is contained in $U$, and $\mathscr{C}$ is indeed a local base of compact nbhds of $y$.

However, there are easier ways to show that $\Bbb Z^+$ is not homeomorphic to $\Bbb Q$. The simplest is to note that $\{1\}$ is an open set in $\Bbb Z^+$. If $f:\Bbb Z^+\to\Bbb Q$ were a surjective homeomorphism, $f[\{1\}]=\{f(1)\}$ would be an open set in $\Bbb Q$, which it is not: $\Bbb Q$ has no isolated points.

Answer 2

Let $X,Y$ be topological spaces and $f$ be a homeomorphism between them, X - locally connected. Take a point $y\in Y$. There is a point $x\in X, f(x)=y$ and open, homeomorphic surroundings $x\in U_x\subset X$,$y\in U_y\subset Y$ such that $U_x$ is connected, but this is topological invariant, so $U_y$ is also connected. This is connected surrounding of a arbitrary point of $Y$.

There you go, I think it's ok.

You don't have to use local property here, there's no bijection between $\mathbb{R}$ and $\mathbb{Z_+}$, it's enough.

Answer 3

Yes, local compactness is a topological property, for it is defined purely in terms of set theory and open sets.

That homeomorphisms preserve this property is the definition of a topological property. The last question might depend on how you define "locally P" for a property P. If you say X is locally P iff every point has a neighbourhood (in the general sense) that has P, and P is itself a topological property, then yes, so is locally P, obviously. You need that a homeomorphisms preserves open sets and hence neighourhoods and is surjective, so that all points in the image are reached.