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I already searched and saw some questions about definitions of locally compact spaces (like Two definitions of locally compact space), but I could NOT find the exact comparison of the two definitions I want to discuss below:

Definition 1: Let T=(S,τ) be a topological space. Then T is locally compact iff every point of S has a local basis $\mathscr B$ such that all elements of $\mathscr B$ are compact. (from https://proofwiki.org/wiki/Definition:Locally_Compact)

Definition 2: Let T=(S,τ) be a topological space. Then T is locally compact iff - a) T is Hausdorff; and b) Every point has a compact neighbourhood (from https://www.math.ksu.edu/~nagy/real-an/1-05-top-loc-comp.pdf)

I could see that from Definition 1, it is easy to get property b of Definition 2

From https://proofwiki.org/wiki/Definition:Locally_Compact, it comments that "if T is a Hausdorff space, then Definition 1 and Definition 2 are equivalent" - could anyone help to prove the direction from Definition 2 to Definition 1 when T is a Hausdorff space?

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  • $\begingroup$ I don't understand the definition: "every point has a compact neighborhood" because the singletons are already compact. So, every space becomes locally compact. Anyways... I think that $\tau = \{ \emptyset \} \cup \{ (-n,n) : n \in \mathbb N \}$ is an interesting topology on $\mathbb R$. $\endgroup$ – ThePortakal Jan 2 '17 at 20:04
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    $\begingroup$ @ThePortakal Singletons are usually not neighbourhoods. $\endgroup$ – Daniel Fischer Jan 2 '17 at 20:16
  • $\begingroup$ Thanks @ThePortkal and Daniel Fischer! From the definition of a neighbourhood, A set U is a neighbourhood of x if x $\in$ int(U), that is, if there exists an open set G s.t. x $\in$ G $\subset$ U. Thus I wonder could we say whether a singleton set is a neighborhood or not depends on if we define that signleton set as open in our topology (since it needs to contain an open set in order to be considered a neighborhood)? $\endgroup$ – Yujie Zha Jan 2 '17 at 21:33
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To get from Def'n 2 to Def'n 1 for a Hausdorff space. For $p\in S$ let $U$ be a compact nbhd of $p.$ Let $V\in \tau$ with $p\in V\subset U.$ Then $Cl_X(V)$ is compact, because it is a closed subset of the compact Hausdorff space $U.$

(i). Observe that for all $W\subset V$ we have ($W$ is open in the space $U$) iff $W\in \tau.$

Let $B$ be an open local base (basis) at $p$ in the space $X.$ Then $C=\{V\cap b:b\in B\}$ is an open local base at $p$ in the space $X$ and also in the space $U$, by (i).

Now a compact Hausdorff space is a regular space. So $U$ is regular.

So for each $c\in C$ we can choose $c'$ where $c'$ is open in $U$ and $p\in c'\subset Cl_U(c')\subset c.$ Let $C'=\{c': c\in C\}.$ By (i), $C'$ is a local open base at $p.$ And $Cl_U(c')$ (which is equal to $Cl_X(c'))$ is a closed subset of the compact Hausdorff space space $U$ so $Cl_U(c')$ is compact.

Now for $p\in Y\in T,$ take $c\in C$ with $p\in c \subset Y.$ Then $\;\overline {c'}\;$ is a compact nbhd of $p,$ and is a subset of $Y. $ And we are done.

NOTE: To show that a compact Hausdorff $S$ space is regular, let $p\in S$ with $p\not \in A=\bar A\subset S.$ To find disjoint open $U,V$ with $p\in U$ and $A\subset V$:

(i). If $A$ is empty let $U=S$ and $V=\emptyset.$

(ii). If $A$ is not empty then for each $a\in A$ let $U_a,V_a$ be disjoint open sets with $p\in U_a$ and $a\in B_a.$ Now $A$ is compact (as it is a closed subset of the compact Hausdorff space $S$) so the open cover $\{B_a:a\in A\}$ of $A$ has a finite non-empty sub-cover $\{B_a:a\in F\}$ where $F$ is a finite subset of $A.$

Let $U=\cap_{a\in F}U_a$ and $V=\cup_{a\in F}B_a.$

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  • $\begingroup$ Thanks, this is very strict and thorough! $\endgroup$ – Yujie Zha Jan 5 '17 at 3:06
  • $\begingroup$ Different books have different def'ns of compact... General Topology by Engelking uses compact to mean compact Hausdorff, and a space (not necessarily Hausdorff) for which every open cover has a finite sub-cover is called pseudo-compact. But he's very clear in his def'ns. $\endgroup$ – DanielWainfleet Jan 5 '17 at 13:38
  • $\begingroup$ Thanks, this is good to know. But @user254665, in the Definition 1 that I referenced in my question above, compact does not mean compact Hausdorff, right? $\endgroup$ – Yujie Zha Jan 5 '17 at 17:46
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    $\begingroup$ Probably not . The modern style is that "compact" does not assume any separation properties. But I would check the author's def'n. $\endgroup$ – DanielWainfleet Jan 5 '17 at 22:26
  • $\begingroup$ Ok, thanks for the comments $\endgroup$ – Yujie Zha Jan 6 '17 at 2:28
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For some point $x$ take its compact neighborhood $K$. We want to show that for any open neighborhood $U$ of $x$, there is some compact neighborhood $K'\subseteq U$ and we will use $K$ to construct it.

So, take any open neighborhood $U$ of $x$. If $K\subseteq U$, we are done. Otherwise, $K\cap U^c$ is closed subset of $K$, thus compact. Since space is Hausdorff, we can find open $V$ and $V'$ such that $K\cap U^c\subseteq V$ and $x\in V'$, $V\cap V'=\emptyset$. Define $K' = K\cap V^c \subseteq K$. You can easily check that $K'$ is compact neighborhood of $x$ contained in $U$.

Edit:

Let me clarify, in Hausdorff space you can separate compact sets from points. Let $K$ be compact and $x\not\in K$. For each $y\in K$ choose open $y\in U_y$, $x\in V_y$ such that $U_y\cap V_y=\emptyset$. $\{U_y\mid y\in K\}$ is open cover for $K$, so we can find finite subcover $\{U_{y_1},\ldots,U_{y_n}\}$ of $K$. Define $U=\cup_{i=1}^n U_{y_i}$, $V=\cap_{i=1}^n V_{y_i}$. Then $K\subseteq U$, $x\in V$, $U\cap V =\emptyset$.

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  • $\begingroup$ Thanks Ennar. "we can find open $V$ and $V'$ such that $K\cap U^c\subseteq V$ and $x\in V'$, $V\cap V'=\emptyset$." - Hausdorff space does not have this property, but Regular space does. Here is a counter example that a Hausdorff space is not regular. $\endgroup$ – Yujie Zha Jan 4 '17 at 5:00
  • $\begingroup$ But maybe (Hausdorff space + every point has a compact neighborhood) will lead to above property? Could you elaborate and explain for that step of your proof? Thanks $\endgroup$ – Yujie Zha Jan 4 '17 at 5:10
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    $\begingroup$ @YujieZha, Hausdorff space has the property that compact sets can be separated from points, please see my edit. Although, locally compact Hausdorff spaces are indeed completely regular (but if I'm not mistaken, you need to have local base of compact sets to prove regularity). $\endgroup$ – Ennar Jan 4 '17 at 9:03
  • $\begingroup$ Thanks Ennar for the explanation, yes, you are right, it could be separated if it is compact $\endgroup$ – Yujie Zha Jan 5 '17 at 2:34

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