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Let $T$ be a topological space. Let $K \subset Y \subset T$. Let $Y$ have the subspace topology. Let $K$ be compact in $Y$. Is $K$ compact in $T$ ?

Well, for one hand, i think that if we think about $]0,1] \subset ]0,2[ \subset \mathbb{R} $, we see that since $]0,1]$ is closed and bounded in $]0,2[$, is compact in $Y$. However $]0,1]$ is not compact in $\mathbb{R}$, since is not closed.

But, what fails in this "proof" ? Let $\lbrace U_{\alpha}\rbrace_{\alpha \in A}$ be an open covering of $K$ with open sets of $X$. Let $V_{\alpha} = U_{\alpha} \cap Y$, which is an open set in Y with subspace topology. Since $K$ is compact in $Y$ by hypothesis, we have that $K \subset \cap_{i=1}^{n} V_{i}$. Then, considering $\lbrace U_{i}\rbrace_{i=1}^{n} \subset \lbrace U_{\alpha}\rbrace_{\alpha \in A}$, we have that $K$ is compact in X ???

Thanks for the attention :) [and help]

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$(0,1]$ is not compact. The subspace topology is the same as a subspace of $(0,2)$ or of $\mathbb{R}$. –  egreg Apr 29 '13 at 23:14
    
$]0,1]$ is closed and bounded subset of $]0,2[$. Why doesn't Heine-Borel imply this to be compact ? Or isn't $]0,1]$ closed in subspace topology ? –  thetruth Apr 29 '13 at 23:20
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No, it doesn't. Would you argue that $(0,2)$ is compact, since it is bounded and closed in $(0,2)$? It isn't, because $(0,2)$ is homeomorphic to $\mathbb{R}$ which is not compact. –  egreg Apr 29 '13 at 23:25
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H-B is a Theorem about the spaces $\Bbb R^n$, not about spaces homeomorphic to a $\Bbb R^n$. –  David Mitra Apr 29 '13 at 23:31
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@thetruth The topology on a metrizable space can always be defined by a bounded metric: just define $e(x,y)=\min(d(x,y),1)$, where $d$ is a metric that induces the topology. The correct concept in the general case is total boundedness together with completeness. A metric space is compact if and only if it is totally bounded and complete. In the space $\mathbb{R}^n$ with the usual metric, total boundedness is equivalent to boundedness; moreover the space is complete. But this is a property of the metric, not invariant under homeomorphisms. –  egreg Apr 29 '13 at 23:37

1 Answer 1

up vote 2 down vote accepted

$]0,1]$ is not compact: e.g. the sequence $(1/n)_n$ doesn't have a convergent subsequence within $]0,1]$. The proof is correct.

This example shows, that in a general metric space, it is not true that

compact = bounded and closed.

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Why can't I conclude from Heine-Borel Theorem that $]0,1]$ is compact in $]0,2[$ with subspace topology ? –  thetruth Apr 29 '13 at 23:22
    
How is Heine-Borel stated? The problem is that $]0,1]$ is not closed in $\Bbb R$ (though indeed closed in $]0,2[$). –  Berci Apr 30 '13 at 0:42

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