# How is $[0,1]$ locally compact if 0 and 1 do not have a neighborhood in $[0,1]$?

I am a physics student and I am trying to learn the concept of compactness as I need it to understand some group theory issues. I am having trouble understanding the statement that every compact space is locally compact. I understand the open cover definition of compactness and could prove that $[0,1]$ is compact using the supremum method. Now, according to the definition on Wikipedia of local compactness, a topological space $X$ is locally compact if every point in $X$ has a compact neighborhood. My understanding is that neighborhood of a point $p$ in $X$ should contain an open set containing $p$ itself. It seems to me though that $0$ and $1$ in $[0,1]$ do not have open sets/intervals in $[0,1]$ which contain $0$ and $1$. Am I making a conceptual mistake here ? I am not an expert, so please forgive my stupidity here. Thank you.

• You must distinguish between a subset of $\Bbb R$ and a subspace of $\Bbb R.$ – DanielWainfleet Jan 11 '18 at 5:42
• For this reason, officially, on math exams in high schools in Norway, a function on a domain with end points (say $f:[0,1]\to \Bbb R$ with $f(x) = x$) is not considered to have local max and min at those end points. They may be global extreme points, if the function value there is larger or smaller than any other place in the domain. But since "$f$ isn't defined on a neighbourhood" (paraphrased), they can't be local maxima or minima. So don't worry, you're not the only one who's confused about this. (Also, it infuriates me to no end that we are teaching this misconception to students.) – Arthur Jan 11 '18 at 8:40

The open sets of $[0, 1]$ (in the topology we usually think of*) are the intersections of open sets in $\Bbb R$ with $[0, 1]$. So for instance, because $(-0.3, 0.2)$ is open in $\Bbb R$, we know that $[0, 0.2)$ is open in $[0, 1]$.

* (This is called the "subspace topology", by the way.)

Does that help?

• Thank you for your time, John. I think now I see what I was missing. – singularity Jan 11 '18 at 2:37

It's important to bear in mind the topology on $[0,1]$, which is the subspace topology inherited from $\mathbb{R}$. Conferring with the definition of the subspace topology, you'll find that $[0,a)$ is an open set in $[0,1]$ for $0<a<1$, though indeed such a set is not open in $\mathbb{R}$.

As an aside, there is a theorem regarding local compactness for which this scenario arises as a special case:

Theorem: Let $X$ be a locally compact Hausdorff space. If $U$ is either an open or a closed subset of $X$, then $U$ is itself locally compact.

Proof: We want to show that any point $x \in U$ has a compact neighborhood $L_x \subset U$. To do this, we begin by noting that the local compactness of $X$ guarantees that $x$ has a compact neighborhood $K \subset X$. This will serve as a starting point.

The intersection of a closed set and a compact set is itself compact, so our strategy going forward will be to find a (finite) collection of closed neighborhoods $\{ C_i \}$ of $x$ such that $\displaystyle K \cap \left( \bigcap_i C_i \right)$ is a proper subset of $U$ (necessarily compact). A good step in the right direction is to take $K \cap \overline{U}$, which prunes away many points in $K \setminus U$.

To get a third closed set $S$ so that $S \cap K \cap \overline{U} \subset U$, we can do the following: note that $\partial U \cap K$ is compact because $\partial U$ is closed (where $\partial U$ denotes the boundary of $U$), and since $X$ is Hausdorff, we can cover $\partial U\cap K$ with open sets that are "far" from $x$. Let $T$ be the union of the open sets contained in the finite subcover this will necessarily admit, and let $S = X \setminus T$ (which is a closed neighborhood of $x$).

$L_x = S \cap K \cap \overline{U}$ is a compact neighborhood of $x$ contained in $U$.

In this scenario, of course, $X = \mathbb{R}$ (a locally compact Hausdorff space), and $U = [0,1]$.