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I'm reading principles of mathematical analysis and have a question about a theorem 2.37.

Theorem 2.37

If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$.

The proof is

If no point of $K$ were a limit point of $E$, each $q \in K$ would have a neighborhood $V_q$ which contains at most one point of $E$. It is clear that no finite subcollection of $\{V_q\}$ can cover $E$. The same is true of $K$, since $E \subset K$. This contradicts the compactness of $K$.

I understand the first part that states that no finite subcollection of $\{V_q\}$ can cover $E$, in other words, $E$ is not compact. But I don't understand why it means that no finite subcollection can cover $K$. Is the author saying that if a subset of a set K is not compact, then $K$ is not compact? If that's the case, a compact set may have open subsets which are not compact, so I'm confused.

Thanks in advance.

  • 1
    @WacDonald's: your comment is misinformed and may confuse others (myself included). In fact, there is no assumption that $K$ lives in a metric space. Yes, that means this proof shows that compact implies limit point compact. I am not sure why you believe this statement to be false. If you don't believe this proof, it also follows from the implications compact implies countably compact implies limit point compact. Check out the first two properties listed on https://en.wikipedia.org/wiki/Countably_compact2013-08-28

5 Answers 5

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A set with no limit points is necessarily a closed set. Being a closed subset of a compact space, it is compact. On the other hand, you're looking at a proof that $E$ is not compact, so you've got a contradiction.

Alternatively, look at this set of neighborhoods that cover $E$ and add one more open set to this collection: the complement of $E$. That set is open, since as noted above, $E$ is closed. Now you've got an open cover of $K$. It must therefore have a finite subcover. But every finite subset of this cover fails to cover all of $E$, so again you have a contradiction.

  • 0
    @ChristopherTurnbull : Rudin's book deals with metric spaces but not with other topological spaces -- in particular, not with non-Hausdorff spaces. Maybe that should have been mentioned here.2017-07-10
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Since for all $q$, $V_q \cap E$ has at most one element, for any finite subset $\{q_1,\ldots,q_n\}$,

$\bigcup_{i=1}^n V_{q_i} \cap E$ has at most $n$ elements, i.e., is finite. But $E$ is assumed to be infinite, so we cannot have $\bigcup_{i=1}^n V_{q_i} \supset E$. Since $K \supset E$, therefore we cannot have $\bigcup_{i=1}^n V_{q_i} \supset K$, contradicting the compactness of $K$.

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Here is my comment, it maybe a little far from the topic, however, I hope it will be helpful for you.

The compactness condition is too strong, we only need the countable extent condition in the Theorem.

Countable extent= The cardinality of any closed discrete subspace must be countable.

So, in fact, we have

If $E$ is an infinite subset of a space $K$ which is countable extent, then $E$ has a limit point in $K$.

There are many topological space which is countable extent but not compact. For example, countably compact space, lindelof space etc. It is far from countable extent to compactness.

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    Mea culpa. I confused extent and spread.2012-08-13
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Suppose every $ p\in K $ has an open nbhd $ U_p $ such that $ E\cap U_p $ is finite. Then $ C=\{U_p: p\in K\} $ is an open cover of $ K $ but any finite $ D\subset C $ covers only a finite subset of $ E. $ Note that we do not need to assume that $ K $ is a $ T_1 $ space nor even a $ T_0 $ space.

I said "open nbhd" because some people say that a nbhd $ U $ of $ p $ is an open set with $ p\in U, $ while some say that a nbhd $ U $ of $ p $ is any subset of the space, such that $ p\in V\subset U $ for some open $ V.$

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Suppose $\overline{E}$ doesn't have a limit point. Then $\forall e\in E,\exists N_{r_e}(e)\ s.t\ N_{r_e}(e)\cap E=\{e\}$.

Consider the open cover of the compact set ${\overline E}$: $\bigcup_eN_{r_e}(e)\rightarrow\bigcup_{i=1}^nN_{r_{e_i}}(e_i)$ but $\bigcup_{i=1}^nN_{r_{e_i}}(e_i)\cap E=\bigcup_{i=1}^n(N_{r_{e_i}}(e_i)\cap E)=\bigcup_{i=1}^ne_i\ne E$ which is absurd.