2
$\begingroup$

This question is a continuation of this. My book says that a metric space is compact if and only if:

$$M=\cup A_{\lambda}\implies M = A_{\lambda1}\cup\cdots\cup A_{\lambda_n}$$

where each $A_{\lambda}$ is open. Then, it says that the definition can be extended to closed sets, since if $A_\lambda$ is a family of open sets in $M$, then the complements $F_\lambda = F-A_\lambda$ forms a family of closed sets in $M$. We have, then: $$M=\cup A_\lambda\iff \cap F_\lambda = \emptyset.$$ So, by the last question, I understood the reason behind this. No problem. The book then proceeds:

A metric space $M$ is compact, $\iff$ all families $(F_\lambda)_{\lambda\in L}$ of closed sets with empty intersection has a finite subfamily with empty intersection: $F_{\lambda_1}\cap \cdots \cap F_{\lambda_n}=\emptyset$.

Then, the book proceeds and says:

A family $(F_\lambda)_{\lambda\in L}$ has the property of finite intersection when for any finite subset $\{\lambda_1,\cdots,\lambda_n \}\subset L$ we have $F_{\lambda_1}\cap\cdots\cap F_{\lambda_n}\neq \emptyset$.

The following condition is necessary and sufficient to tell that a metric space $M$ is compact:

If $(F_\lambda)_{\lambda\in L}$ if a family of closed sets with the property of finite intersection, then $\cap_{\lambda\in L}F_\lambda \neq \emptyset$

Is the last one the negation of the first one? I mean, for $M$ to be compact, either we have:

family of closed sets with empty intersection has finite subfamily with empty intersection

or

family of closed sets with the property of finite intersection (that is, every intersection of finite subfamilies is not empty) does not have a empty intersection.

Is this correct?

I ask this, because the book proceeds for a proof of Dini's theorem like this:

An example of a family of closed subsets with the property of finite intersection is a decrescent sequence $F_1\supset F_2\supset \cdots\supset F_n\supset \cdots$ of nonempty closed subsets of $M$. If $M$ is compact, it follows that $\cap_{n=1}^{\infty}F_n\neq \emptyset$. The following demonstration shows these ideas:

Dini's theorem: If a sequence of continuous real functions $f_n:M\to \mathbb{R}$ defined in a compact metric space $M$, converges simply to a function $f:M\to \mathbb{R}$, and if $f_1(x)\le f_2(x)\le \cdots\le f_n(x)\le \cdots$ for all $x\in M$, then the convergence $f_n\to f$ is uniform in $M$.

Demonstration: Given $\epsilon>0$, choose, for each $n\in \mathbb{N}$:

$$F_n=\{x\in M: |f_n(x)-f(x)|\ge \epsilon\}$$

Then, $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ and each $F_n$ is closed in $M$. We must prove that there is $n_0\in \mathbb{N}$ such that $F_{n_0}=\emptyset$ (then $n>n_0\implies |f_n(x)-f(x)|<\epsilon$ for all $x\in M$). Well, since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n = \emptyset$ (WHY?). Being $M$ compact, we must have $F_n=\emptyset$ for some $n$

And besides the chain of sets $F_1\supset F_2\supset \cdots \supset F_n\supset \cdots$ having the property of finite intersection (every subfamily has a not empty intersection), the proof concludes that $\cap_{n=1}^{\infty}F_n = \emptyset$

$\endgroup$
  • $\begingroup$ Your two bolded statements are contrapositive to each other. $\endgroup$ – Omnomnomnom Jun 24 '16 at 19:53
1
$\begingroup$

It seems like you have two questions, one of which I addressed in a comment. The other is how to justify the statement

Since $\lim_{n\to \infty} f_n(x)=f(x)$ for all $x\in M$, it follows that $\cap_{n=1}^{\infty} F_n =\emptyset$

Note that $$ \bigcap_{n=1}^\infty F_n = \left\{x \in M : |f_n(x) - f(x)| \geq \epsilon \text{ for all } n\right\} $$ However, if $x$ is such that $f_n(x) \to f(x)$, then (by definition of convergence) $x$ is not an element of this set. So, if $f_n(x) \to f(x)$ for every $x \in M$, then $\bigcap_{n=1}^\infty F_n = \emptyset$.

$\endgroup$
  • $\begingroup$ I'd recommend to include the information from the comment in the answer as well. $\endgroup$ – quid Jun 24 '16 at 20:03
  • $\begingroup$ ok, now, in the end, it says that being $M$ compact, we must have $F_n=\emptyset$ for some $n$. Why? And which one of the bolded definitions it's using? I ask because in the beggining, it says that the chain of descending closed sets has the property of finite intersection, that is, any subfamily of them has a not empty intersection. How is it possible to conclude that the intersection IS empty? $\endgroup$ – Guerlando OCs Jun 24 '16 at 20:14
  • $\begingroup$ The statement in the end follows directly from the first of your bolded definitions. As I stated, however, the two bolded definitions are equivalent, since they are contrapositives of one another. That is, if we find that the total intersection is empty, then the collection can't satisfy the finite intersection property. We used the infinite intersection to conclude that there exists a satisfactory finite intersection. $\endgroup$ – Omnomnomnom Jun 24 '16 at 20:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.