# Determining whether a subspace of the plane (with Euclidean topology) is locally compact

There is a theorem that states that a Hausdorff space is locally compact if and only if every point has a compact neighborhood. Checking whether every point has such a neighborhood can be tedious and time-consuming for more complex spaces, so is it possible to determine whether a subspace of a plane is locally compact by a simpler means; e. g. are there any sufficient conditions for local compactness in such spaces that might be simpler to prove than using the above theorem, using only general topology? I appreciate your help.

Your question is bit vague. In general there is no way to avoid checking whether every point has a compact neighborhood, even it the space is a subset of a (Euclidean) plane. There are some general theorems that might help you, for example the following:

A subspace $X$ of a locally compact Hausdorff space $L$ is locally compact if and only if it can be written as $X = A \cap U$, where $A$ is closed and $U$ is open in $L$. This applies to subsets $X$ of any finite-dimensional Euclidean space.

Moreover, your space $X$ must be defined somehow, and in some cases it is fairly obvious from the definition that it is locally compact. For example, if you have a continuous map $f : L \to Y$ defined on a locally compact Hausdorff space $L$, then all preimages of subsets $M \subset Y$ having the form $M = B \cap V$, where $B$ is closed and $V$ is open in $Y$, are locally compact.

Maybe this is not quite what you were expecting, but here's an interesting way to get a large class of locally compact subspaces of $\mathbb{R}^2$.

The Heine-Borel theorem states that in $\mathbb{R}^n$ any subset $A \subseteq \mathbb{R}^n$ is compact if and only if it is closed and bounded.

Now take any closed and bounded subset $C$ of $\mathbb{R}^n$ , by the Heine-Borel theorem $C$ is compact and since any compact topological space is locally compact it follows that $C$ is locally compact.

So any closed and bounded subset of $\mathbb{R}^n$ is locally compact.

In the case when $n=2$, we can easily generate a large number of examples of locally compact spaces from our above deductions.

Examples:

• The circle $\mathbb{S}^1$ is compact, and thus locally compact.

• Any finite set of points is locally compact.

• Any finite union of closed balls are closed, and thus locally compact.

• $[a,b] \times [c,d]$ is closed and bounded and thus locally compact for any $a,b,c,d \in \mathbb{R}$