1
$\begingroup$

I was reading van Dantzig's theorem from Terry Tao's notes, where I encountered that he's writing locally compact topological space to mean every point has a compact neighborhood.

What does locally compact mean? Every point having compact neighborhood or every point having neighborhood whose closure is compact?

If anyone explains in detail it would be great.

$\endgroup$
3
  • $\begingroup$ They are equivalent in Hausdorff spaces. Ordinarily one refers to points having compact neighbourhoods as locally compact. See en.m.wikipedia.org/wiki/Locally_compact_space $\endgroup$ Commented Oct 19, 2017 at 13:07
  • $\begingroup$ Does you working definition of "neighborhood" require that it be an open set? (There are two standards here.) $\endgroup$ Commented Oct 19, 2017 at 14:01
  • $\begingroup$ Is your confusion based on whether neighbourhoods are by definition open, or is it based on the distinction of "every point has a (quasi)compact neighbourhood" vs. "every point has a (quasi)compact closed neighbourhood"? $\endgroup$ Commented Oct 19, 2017 at 16:18

1 Answer 1

2
$\begingroup$

$X$ is a topological space throughout.

There are two standard definitions of a neighborhood. Both are used widely. The difference you are highlighting comes down to whether neighborhoods are defined to contain an open subset, or defined to be an open subset.

  • For $x \in X$, $V \subset X$ is a neighborhood(1) of $x$ if $x \in U \subset V$ for some open set $U \subset X$. (note 1)
  • For $x \in X$, $V \subset X$ is a neighborhood(2) of $x$ if $x \in V$ and $V$ is open. (note 2)

In the familiar setting $X = \mathbb{R}$, there are no neighborhood(2)s that are compact. However, for each $x \in X$, there are neighborhood(2)s, $V_x = (x-1/2, x+1/2)$, for instance, whose closure is closed and bounded, hence compact (by the Heine-Borel theorem). In this setting, the closures of the neighborhood(2)s are neighborhood(1)s, so we get these two characterizations:

  • $\mathbb{R}$ is locally compact because every $x \in \mathbb{R}$ has a compact neighborhood(1).
  • $\mathbb{R}$ is locally compact because every $x \in \mathbb{R}$ has a neighborhood(2) whose closure is compact.

(note 1): See, for instance, Bredon, Glen E., Topology and Geometry, ISBN ISBN 0-387-97926-3, p. 4: "If $X$ is a topological space and $x \in X$ then a set $N$ is called a neighborhood of $x$ in $X$ if there is an open set $U \subset N$ with $x \in U$."

(note 2): See, for instance, Munkres, James R., Topology: A First Course, ISBN 0-13-925495-1, p. 96: "Mathematicians often use some special terminology here. They shorten the statement '$U$ is an open set containing $x$' to the phrase '$U$ is a neighborhood of $x$.'" This language also appears in the second edition of this work, ISBN 978-0131816299, still p. 96. Alternatively, Kelley, John L., General Topology, ISBN 0-387-90125-6, p. 14: "A neighborhood of a point is any open set containing this point." Kelley then asserts that analysts and Bourbaki use "neighborhood" in the neighborhood(1) sense.

$\endgroup$
8
  • $\begingroup$ Isn't there a FAQ containing this information? $\endgroup$ Commented Oct 19, 2017 at 14:20
  • $\begingroup$ Can you please provide an example of a standard Topology textbook in which, by definition, all neighborhoods are open sets? $\endgroup$ Commented Oct 19, 2017 at 14:34
  • $\begingroup$ @JoséCarlosSantos : Done. $\endgroup$ Commented Oct 19, 2017 at 15:27
  • $\begingroup$ @JoséCarlosSantos : And done adding a reference for neighborhood(1) which seems much harder to do from standard Topology textbooks. $\endgroup$ Commented Oct 19, 2017 at 15:44
  • 1
    $\begingroup$ I think the OP's problem is with a different issue. In a space where every point has a quasicompact neighbourhood [in the sense (1) of course], it need not be the case that every point has a closed quasicompact neighbourhood. Consider the line with infinitely many origins. Every origin has quasicompact neighbourhoods, but no closed quasicompact neighbourhood. $\endgroup$ Commented Oct 19, 2017 at 15:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .