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Let $\mathfrak{C}$ be an open cover of a topological compact space $X$, and let $\mathfrak{B}\subseteq \mathfrak{C}$ be its finite subcover. Then every subset of $X$ also has the same cover and subcover! Shouldn't every such subset also be compact then?

Motivation: I read somewhere that every closed subset of a compact space is compact. I wonder why it shouldn't be the case for every subset of $X$, rather than just the closed ones.

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marked as duplicate by Zev Chonoles, Ayman Hourieh, Henry T. Horton, amWhy, Amzoti Jul 11 '13 at 0:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Is (0,1) compact in [0,1] under the usual Euclidean topology? Also see here for essentially the same question:… – DJBruce May 31 '13 at 16:47
(@DJBruce: but $\mathbb{R}$ isn't compact! You can repair it by looking at $[0,2]$ instead of $\mathbb{R}$.) – Jason DeVito May 31 '13 at 16:49
@JasonDeVito Thanks, a case when I am typing faster than thinking. – DJBruce May 31 '13 at 16:50
an open cover of the subset need not be an open cover of the original! – bleh May 31 '13 at 16:50
@DJBruce: Any time - I find my fingers outrace my brain all the time ;-) – Jason DeVito May 31 '13 at 16:50
up vote 12 down vote accepted

The point is that EVERY cover must admit a finite subcover, and only because this is true for the ambient space, does not mean it is true for every subset.

So given a compact space $X$ and $U \subset X$ arbitrary, you start with an open cover of $U$, and try to find a finite subcover of it.
But not every cover of $U$ comes from a cover of $X$, not even if $U \subset X$ is open. Consider $X=[0,1] \subset \Bbb{R}$, and $U=(0,1) \subset [0,1]$ and the cover $$ U=\bigcup_{i=1}^{\infty}\left(\frac{1}{n},1-\frac{1}{n} \right) $$

There is no open cover of $[0,1]$ which restricts to the above cover on $(0,1)$.

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Then every subset of $X$ also has the same cover and subcover!

Every subset of $X$ has the same cover and subcover. But there are covers of the subset that would not cover $X$. This cover might not have a subcover. The examples are stated above.

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I got that. Let $\mathfrak{R}$ be an open cover of the subset. Won't $\mathfrak{R}\cup \{X\}$ be an open cover of $X$ then? $X$ is considered to be an open set, I believe. – Ayush Khaitan May 31 '13 at 16:56
If $X$ is your whole space, it's both open and closed. However, it seems, and as emphasised below, you did not get that Every cover should have subcover. Finding some or even many cover with subcover does not work. – edit May 31 '13 at 17:00
@AyushKhaitan True but a finite subcover of $\mathfrak{R}\cup \{X\}$ is not necessarily a subcover of $\mathfrak{R}$... So how do you prove that ANY, in particular $R$, cover has a finite subcover? – N. S. May 31 '13 at 17:00
@N.S.- you're right. I'm unable to generate a finite subcover of $\mathfrak{R}\cup\{X\}$ which would be $\subseteq \mathfrak{R}$. – Ayush Khaitan May 31 '13 at 17:07

Not every open cover of subset of $X$ is cover of $X$. For example, let $X=[0, 1]$. Then $(0, 1)$ is a subset of $X$. If we take $A_n =(1/n, 1),\, n=2, 3, \ldots$, then $A_n,\, n=2, 3, 4, \ldots$ is a cover of $(0, 1)$ but not of $X$.

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The big issues here are

  1. That open sets of a subspace are not always open in the larger space.
  2. That the open sets you add to get a cover of the whole space may have to be very large and contain infinitely many of the subspaces open sets. These large open sets are often preferentially chosen in the finite subcover, like in the one point compactification of a discrete space.
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