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.
[Math] How is $[0,1]$ locally compact if 0 and 1 do not have a neighborhood in $[0,1]$
compactnessgeneral-topology
Best Answer
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?