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If a topological space is Hausdorff then arbitrary intersection of compact sets is compact.

How to find examples of compact subsets $A,B$ of a topological space $X$ such that $A\cap B$ is not compact

Learnmore
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2 Answers2

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Let $X$ be the real line with doubled origin, i.e., $X=\mathbb R\cup\{0'\}$ where the open sets are the open sets of $\mathbb R$ as well as all sets of the form $(U\setminus\{0\})\cup\{0'\}$, where $U$ is an open neighbourhood of $0$. Then $[0,1]$ and $([0,1]\setminus\{0\})\cup\{0'\}$ are compact, but their intersection $(0,1]$ is not.

Hagen von Eitzen
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Consider two different one-point compactifications of the same non-compact space. Each compactification will be compact, but their intersection (the original space) will not be.

For a specific example, take $\mathbb{R} \cup \{\gamma, \delta\}$ whose open sets are as follows:

  • If a subset $U$ does not contain $\gamma$ or $\delta$, then $U$ is open $\iff$ it is open in $\mathbb{R}$.
  • If a subset $U$ does contain either $\gamma$ or $\delta$ (or both), then $U$ is open $\iff$ it contains all of $\mathbb{R}$.

You can check that this topology is legitimate and that $\mathbb{R} \cup \{\delta\}$ and $\mathbb{R} \cup \{\gamma\}$ are both compact. However, $\big( \mathbb{R} \cup \{ \delta \} \big) \cap \big( \mathbb{R} \cup \{\gamma\} \big) = \mathbb{R}$ is not compact.

P.S. this general idea has been very kind to me in the past, so it might be worth bearing in mind. For instance, read here about how this one-point compactification also serves as a one-point connectification.

Kaj Hansen
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  • what is compactification;any easy one from you – Learnmore May 11 '15 at 06:51
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    @learnmore, a compactification of a space $X$ is a topological embedding $X \hookrightarrow X'$, where $X'$ is compact. As a simple example, $(0, 1)$ can be canonically embedded inside $[0, 1]$, and the latter is compact. Of note, you can always add a single point to a space in such a way that the new space is compact. See, e.g., the so-called Alexandroff one-point compactification. – Kaj Hansen May 11 '15 at 06:55
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    The simplest of one-point compactifications I've used above, which goes as follows: Take a space $X$ that is not compact and add a single point to it, say $\delta$. Then in the new space $X \cup {\delta}$, if a subset $U$ does not contain $\delta$, then $U$ is open $\iff$ it is open in $X$. Otherwise if $U$ contains $\delta$, then it is open $\iff$ it also contains all of $X$. You can verify for yourself that the topology we've given $X \cup {\delta }$ is indeed a topology, and further that $X \cup {\delta}$ is compact. – Kaj Hansen May 11 '15 at 06:58
  • i need some time – Learnmore May 11 '15 at 07:01
  • No problem @learnmore. Math is all about struggling through things and getting yourself unstuck. – Kaj Hansen May 11 '15 at 07:03
  • I have verified its a topology but why is it compact? – Learnmore May 11 '15 at 07:10
  • @learnmore, suppose you have an open cover of $\mathbb{R} \cup { \delta }$. For the cover to be a cover, we must have an open set that contains $\delta$. What do you know about an open set that contains $\delta$? – Kaj Hansen May 11 '15 at 07:11
  • Ohh I got it then it must contain whole of $X$ right – Learnmore May 11 '15 at 07:12
  • mhmm, it's kind of cool when you think about it. Hope this helped! – Kaj Hansen May 11 '15 at 07:12