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I am trying to show that every manifold (a Hausdorff and second countable topological space) is locally compact (has a precompact basis).

Consider an open set of a manifold, $M$. I must show that for each $x\in U$ we can find an open set in $M$ such that its closure is compact.

So let $x\in U$ where $U$ is an open set in $M$. Because $M$ is a manifold, we can find an open set contained in $U$ containing $x$ homeomorphic to an open Euclidean ball. Call this set $V$. Now since $V$ is homeomorpic to a Euclidean ball, we can find another open set, $G$ contained in $V$, containing $x$ such that the closure of $G$ in $V$ is compact.

To summarise, we have an open set $V$ in $M$ containing an open set $G$ and $\text{cl}_V{(G)}\subseteq V$ is compact. To get the result I need, I want to say that this mean the closure of $G$ in $M$ is compact. However I get stuck here.

My question is: how can I show that the closure of $G$ in $M$ is also compact? Am I in the right direction?

EDIT: I was thinking of arguing that the closure of $G$ in $M$ is contained in $V$ and then the result follows immediately since then the two closures are equal. Is this a true statement?

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  • $\begingroup$ You have a map that is a homeomorphism $\endgroup$ – H. Gutsche Jun 18 '18 at 21:53
  • $\begingroup$ I'm not sure I am getting your suggestions. Could you elaborate please? $\endgroup$ – fosho Jun 18 '18 at 22:03
  • $\begingroup$ I was coming from the coordinates, but you already know that the closure of G is compact in V. V carries the trace topology from M. Look at the open covers in V and how they relate to open covers (that cover V) in M and you are nearly there. That argument depends on how compact was defined for you. In general I use the Bourbaki versions, because I grew up with them and in A T2 space they should be equivalent. $\endgroup$ – H. Gutsche Jun 18 '18 at 23:29
  • $\begingroup$ The issue I am facing is that I don't know what the closure of G in M actually is. $\endgroup$ – fosho Jun 19 '18 at 8:25
  • $\begingroup$ Why would you need to know that? You know it exists and it is contained in V and is compact in V hence is a compact neighbourhood of x in M. That is the condition for locally compact. $\endgroup$ – H. Gutsche Jun 19 '18 at 17:45

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