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I'm given the following problem:

Suppose that for every $n\in \mathbb{N}$ $V_n$ is a non-empty, closed subset of a compact space $X$, with $V_n \supseteq V_{n+1}$.

Now I have to show that $V_{\infty}= \bigcap_{n=1}^{\infty} V_n \neq \emptyset$.

How can I do that? I know the nested interval property from real analysis...

The 'answer' should be that the family $\{V_n: n\in \mathbb{N} \}$ has the finite intersection property - the intersection of any finite subfamily $\{V_{n_1}, V_{n_2}, ..., V_{n_r} \}$ is $V_N$, where $N$ is $\max \{n_1,n_2, ..., n_r \}$ and $V_N \neq \emptyset$. Hence since $X$ is compact another exercise $(*)$ says that $\bigcap_{n=1}^{\infty} V_n$ is non-empty.

Exercise $(*)$ was about to prove that for a space $X$ to be compact, it is necessary and sufficient condition that if $\{V_i: i\in I \}$ is any indexed family of closed subsets of $X$ such that $\bigcap_{j\in J}V_j$ is non-empty for any finite subset $J \subseteq I$, then $\bigcap_{i\in I}V_i$ is non-empty.

So I don't understand the proof now.... Can somebody clarify this stuff? :-)

Thanks for your trouble !

Definition of open cover:

Let $A$ be a subset in $X$.

A family $\mathcal{U}=\{U_i: i\in I\}$ of subsets of $X$ is called a cover for $A$ is $A\subseteq \bigcup U_i$.

If each $\{U_k\}$ is open in $X$, then $\mathcal{U}$ is an open cover for $A$

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    $\begingroup$ If the intersection were empty, then there would be a finite subset $M$ of $\mathbb N$ such that the intersection $\bigcap_{m\in M}V_m=\emptyset$. $\endgroup$ – Stefan Hamcke Mar 21 '13 at 23:26
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    $\begingroup$ And I don't understand what you don't understand: the current claim's proof or the past (*) proof? It is a rather important property of compact spaces... $\endgroup$ – DonAntonio Mar 21 '13 at 23:27
  • $\begingroup$ Well it would be nice if anyone could tell me how to prove the second $(*)$ proof. $\endgroup$ – Applied mathematician Mar 21 '13 at 23:30
  • $\begingroup$ It's just a dualization of the definition of compactness. Well, actually it is the contrapositive of this dualization. $\endgroup$ – Stefan Hamcke Mar 21 '13 at 23:32
  • $\begingroup$ Sorry, I don't know what you mean by dualization of a definition.. :-) $\endgroup$ – Applied mathematician Mar 21 '13 at 23:34
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Claim: A topological space $\,X\,$ is compact iff it has the Finite Intersection Property (=FIP):

Proof: (1) Suppose $\,X\,$ is compact and let $\,\{V_i\}\,$ be a family of closed subsets s.t. that $\,\displaystyle{\bigcap_{i}V_i=\emptyset}\,$. Putting now $\,A_i:=X-V_i\,$ , we get that $\,\{A_i\}\,$ is a family of open subsets , and

$$\bigcap_{i}V_i=\emptyset\Longrightarrow \;X=X-\emptyset=X-\left(\bigcap_iV_i\right)=\bigcup_i\left(X-V_i\right)=\bigcup_iA_i\Longrightarrow\;\{A_i\}$$

is an open cover of $\,X\,$ and thus there exists a finite subcover of it:

$$X=\bigcup_{i\in I\,,\,|I|<\aleph_0}A_i=\bigcup_{i\in I\,,\,|I|<\aleph_0}(X-V_i)=X-\left(\bigcap_{i\in I\,,\,|I|<\aleph_0}V_i\right)\Longrightarrow \bigcap_{i\in I\,,\,|I|<\aleph_0}V_i=\emptyset\Longrightarrow$$

The family $\,\{V_i\}\,$ has the FIP

(2) Suppose now that every family of closed subsets of $\,X\,$ hast the FIP, and let $\,\{A_i\}\,$ be an open cover of it. Put $\,U_i:=X-A_i\,$ , so $\,U_i\, $ is closed for every $\,i\,$:

$$\bigcap_iU_i=\bigcap_i(X-A_i)=X-\bigcup_i A_i=X-X=\emptyset$$

By assumption, there exists a finite set $\,J\,$ s.t. $\,\displaystyle{\bigcap_{i\in J}U_i=\emptyset}\,$ , but then

$$X=X-\emptyset=X-\bigcap_{i\in J}U_i=\bigcup_{i\in J}X-U_i)=\bigcup_{i\in J}A_i\Longrightarrow$$

$\,\{A_i\}_{i\in J}$ is a finite subcover for $\,X\,$ and thus it is compact....QED.

Please be sure you can follow the above and justify all steps. Check where we used Morgan Rules, for example, and note that we used the contrapositive of the FIP's definition...

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  • $\begingroup$ @Andrew Hi and thanks for your kind words. I am flagging this comment of mine if you don't mind. I totally forgot when my account was going to be reactivated. Once again, I was put out for one month or so because of "sockpuppet" accounts and whatever. I declare on my honor of mathematician I had none, but that's nothing those moderators care about. Funny, during this last month I got HUNDREDS of upvotes, many of which were deleted later because "User was Removed"....I told them about this, I told about many serial downvotes...They didn't care. ...and I still am geting upvotes! How come? $\endgroup$ – DonAntonio Jul 26 '17 at 17:19
  • $\begingroup$ @Andrew Perhaps they didn't find all my sockpuppets accounts? Perhaps they were so dumb that they thought some actual, real accounts were sockpuppets. I don't know, by the life of me. But I really don't care anymore. This last time I've been very, very busy: grading final exams, getting next semester's list of courses to teach, etc. ...etc. I even sold my car and I haven't yet bought a new one as I've been enjoying my freedom from lecturing, studying for classes, etc. I think the behavior of the mods here is below any reasonable, logical acceptable level and I think I shalln't be back. $\endgroup$ – DonAntonio Jul 26 '17 at 17:21
  • $\begingroup$ @Andrew ...at least for a while. So good luck to you and all the honest, fair-play abiding members of this site. It can be very good, in particular for struggling students, and it can be very frustrating... Really, I did warn them about the crazy up/down votes I was receiving...and they though tit'd be a good idea to ban me from the site for a month. To hell with it. Farewell...and may you continue to enjoy the best challenge to our minds there is: mathematics. :) $\endgroup$ – DonAntonio Jul 26 '17 at 17:25
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If you are working in a metric space then this can be proved using the sequential definition of compactness.

For each $V_n$ choose $x_n \in V_n$. Then all the $x_n's$ live in $V_1$, since this is compact there is a subsequence $x_{n_k}$ converging to some $x \in V_1$. Now we claim that in fact $x \in V_i$ for all $i$. Suppose not; so that there exists $ i> 0$ such that $x \notin V_i$.

Then $V_i$ being closed implies that we can find an open set $U$ about $x$ such that $x \in U$ and $U \cap V_i = \emptyset$. Then $U$ can only meet finitely many terms of the sequence $x_{n_k}$, contradicting $x_{n_k} \to x$.

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Since $V_{n+1}\subset V_n$ and $V_n\neq\varnothing$ for each $n\in\mathbb{N}$, we have that the family of closed subsets of $X$, $\{V_n\}_{n\in\mathbb{N}}$, has the finite intersection property. Since $X$ is compact and $\{V_n\}_{n\in\mathbb{N}}$ has the finite intersection property, $\bigcap\limits_{n\in\mathbb{N}} V_n\neq\varnothing$.

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