Compactness
Last time, we discussed the possibility of generalizing the ideas of open and closed intervals; we hit on the idea of open and closed sets.
There's a fundamental difference, though, between $$ [a,b] = \{x\in\mathbb R: a \leq x \leq b\} $$ and $\mathbb R$ or $$ [a,\infty) = \{x\in\mathbb R: a \leq x\}, $$ even though all of these sets are closed.
Characterizations
Often, the more ways to characterize a concept, the more important it is. I guess that should give us
- More ways to think about the concept and
- more applications of the concept.
Compactness illustrates this nicely.
Definition
Def 3.3.1 A set $K\subset \mathbb R$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.
Example: Closed intervals are compact.
I guess this is a consequence of the Bolzano Weierstrass theorem together with the fact that closed intervals contain their limit points.
Alternative characterization
Thm 3.3.4 A set $K\subset \mathbb R$ is compact if and only if it is closed and bounded.
By the way - Def 3.3.3: A set $A\subset \mathbb R$ is bounded if there exists $M > 0$ such that $|a|\leq M$ for all $a\in A$.
Yet another characterization
A set $K\subset \mathbb R$ is compact if and only if every open cover of $K$ has a finite sub-cover.
We'll get to more on this in a few minutes!
Compact sets and limit points
In this column of slides, we're going to prove half of Theorem 3.3.4. In particular, we'll prove the following:
Claim: If $K\subset \mathbb R$ is compact, then $K$ is closed.
This technique could be useful on the HW.
Assumptions
We're simply assuming that $K$ is compact. That is, every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.
Target
We want to show that $K$ is closed. That is, $K$ contains all of it's limit points.
Of course (from Def 3.2.4 last time), $x\in\mathbb R$ is a limit point of $K$ if every $\varepsilon$-neighborhood $V_{\varepsilon}(x)$ of $x$ intersects the set $K$ at some point other than $x$.
We should be able to use that to construct a sequence in $K$ that converges to $x$.
Proof
Suppose that $K\subset\mathbb R$ is compact and that $x\in\mathbb R$ is a limit point of $K$. We must show that $x\in K$. It then follows that $K$ contains all it's limit points (since $x$ is arbirary) so that $K$ would be closed.
We first generate a sequence in $K$ that converges to $x$. To do so, simply choose (for every $n\in\mathbb N$)
$$x_n \in V_{1/n}(x) \cap K.$$
Then, by construction, each $x_n\in K$ and $$|x_n-x| < \frac{1}{n}$$ so that $x_n\to x$ as $n\to\infty$.
Proof (cont)
We now have a sequence $\{x_n\}$ such that $x_n\to x$ and $x_n\in K$ for every $n\in\mathbb N$.
Since $K$ is compact, $\{x_n\}$ has a further subsequence that converges to an element of $K$. Since $\{x_n\}$ is already convergent, though, the only choice for that limit is $x$ itself.
Thus, $x\in K$. $\Box$
Open covers
Finally, let's try to get a grip on the characterization in terms of open covers.
Definitions
Def 3.3.6 Let $A\subset R$. An open cover for $A$ is a (possibly infinite) collection of open sets $$\{O_{\lambda}: \lambda \in \Lambda\},$$ whose union contains the set $A$; that is, $$A \subset \cup_{\lambda\in\Lambda} O_{\lambda}.$$ Given an open cover for $A$, a finite subcover is a finite sub- collection of open sets from the original open cover whose union still manages to completely contain $A$.
Example
Consider the set $$K = \{0,1,1/2,1/3,1/4,1/5,\ldots\}.$$
Example (cont)
Any open cover has to contain an interval covering the origin. After that, there's only finitely many points left!
Non-example
Consider the open unit interval $$(0,1) = \{x\in\mathbb R: 0 < x < 1\}.$$ Can you find an open cover without a finite sub-cover??
How about $$\left\{\left(\frac{1}{n}, 1\right)\right\}_{n=2}^{\infty}?$$