On two definitions of locally compact

[psie]

TL;DR Summary

I've come across two definitions of local compactness. One author seems to equate these definitions and I wonder if they are actually equivalent, and if not, if it has any consequences?

I have a hard time accepting definitions that are inequivalent. So the main point of my post is to ask for confirmation that it does not matter having inequivalent definitions, but I'm not sure about this. Maybe these two definitions being inequivalent actually have some consequences.

First, there are two common definitions of a neighborhood:

Defintion 1 For $x \in X$, $V \subset X$ is a neighborhood of $x$ if $x \in U \subset V$ for some open set $U \subset X$. [e.g. Gamelin and Greene]

Definition 2 For $x \in X$, $V \subset X$ is a neighborhood of $x$ if $x \in V$ and $V$ is open. [e.g. Munkres]

Now, here's the following definition of locally compact which I've come across in Topology: An Invitation by Parthasarathy (who uses definition 1):

Definition 3 A topological space $X$ is said to be locally compact if every point $x\in X$ has a compact neighborhood; i.e. there is an open set $V$ such that $x\in V$ and $\overline{V}$ is compact.

I don't follow how every point having a compact neighborhood is equivalent to there is an open set that contains the point whose closure is compact. When I read definition 3 I interpret the "i.e." as equating these two statements, but are they equivalent? If not, does it matter?

 

[pasmith]

The first part of Definition 3 does not specify that the neighbourhood is open, so the second part is incorrect.

 

[psie]

The first part of Definition 3 does not specify that the neighbourhood is open, so the second part is incorrect.

Hmm, I don't think I understand you.

1. The first part of definition 3 specifies that the neighborhood be compact, i.e. that there is a compact set $V\subset X$ such that it contains an open set $U\subset X$ that contains $x$.
2. The second part of definition 3 specifies simply that there is an open set $U$ that contains $x$ and $V=\overline{U}$ is compact.

When I look at these two statements, statement 2 certainly satisfies the definition 1 of a (compact) neighborhood (so statement 2 implies statement 1). But I can not work out how statement 1 implies statement 2.

 

[martinbn]

I don't follow how every point having a compact neighborhood is equivalent to there is an open set that contains the point whose closure is compact.

A compact neighborhood of $x$ means a compact set $K$ such that $x$ is contained in the interior of $K$.

 

[mathwonk]

a closed subset of a compact set is compact. hence if V is a compact neighborhood of x, then the interior U of V is open and contains x, and the closure of U is a closed subset of the compact set V, at least when the space is Hausdorff. But since a compact set need not be closed in a non Hausdorff space, I don't see how the first part of the definition implies the second, in general.

compare: https://en.wikipedia.org/wiki/Locally_compact_space

 

[martinbn]

a closed subset of a compact set is compact. hence if V is a compact neighborhood of x, then the interior U of V is open and contains x, and the closure of U is a closed subset of the compact set V, at least when the space is Hausdorff. But since a compact set need not be closed in a non Hausdorff space, I don't see how the first part of the definition implies the second, in general.

compare: https://en.wikipedia.org/wiki/Locally_compact_space

This is a good point! The closure of the interior of a closed set need not be the closed set itself.