$\gdef\ucal{\mathcal U}$

Last time we discussed two versions of compactness: the sequential compactness and the open cover compactness.

Theorem: let $X$ be a metric space. Then, $X$ is sequentially compact, if and only if $X$ is open cover compact.

Last time we have shown open cover compactness implies sequential compactness, today we show the converse.

Suppose a metric space $(X,d)$ is sequentially compact, namely, any sequence $(p_n)$ in $X$ has a convergent subsequence. How to show that any open cover $\ucal$ of $X$ has a finite sub-cover? We first introduce

Definition (A Lebesgue number of a cover $\ucal$) we say $\lambda$ is a Lebesgue number of the covering $\ucal$, if for any $p \in X$, there is some $U \in \ucal$, such that $B_\lambda(p) \In U$.

Lemma Every open cover $\ucal$ of a sequentially compact space $X$ has a Lebesgue number $\lambda>0$.
Proof: suppose not, then for every $\lambda$, there exists a point $p$, such that $B_\lambda(p)$ is not contained in any $U \in \ucal$. Let $\lambda$ take values $1/n$ for $n=1,2,\cdots$, and we get a sequence $p_n$ for $\lambda=1/n$. By sequential compactness, this sequence has a convergent subsequence, say converge to $p \in X$. However, $p$ is contained in some open set $U_0 \in \ucal$, thus there is $r>0$ that $B_r(p) \In U_0$. Thus, let $\epsilon = r/3$, and $N$ large enough such that $1/N < r/3$, since $p_n$ sub-converge to $p$, there exists $n > N$ with $d(p_n, p) < \epsilon$. Thus, we have $B_{1/n}(p_n) \In B_{r/3}(p_n) \In B_r(p) \In U_0$ which contradict with $B_{1/n}(p_n)$ is not contained in any $U \in \ucal$. This proves the lemma.

Lemma If $X$ is a sequentially compact space, then for any $r>0$, $X$ can be covered by finitely many open balls of radius $r$. Proof: We claim that $X$ can only contain finitely many disjoint open balls with radius $r/2$. Then, take such a maximal $r/2$-radius ball packing, and replace the radius $r/2$ balls by radius $r$ balls, claim the bigger balls cover $X$. (Discussion: prove the two claims)

Given the above two lemma, one can prove that any open cover of a sequentially compact space $X$ admits a finite subcover. (Discussion: prove it)

1. Compactness is an absolute (or intrinsic) property of a metric space. If $X$ is a metric space, and $K \In X$ is a subset, when we say $K$ is compact, we mean $K$ as a 'stand-alone' metric space (totally forgetting about $X$, but only using the distance function inherited from $X$) is compact.

2. We proved last time: If $K \In X$ is (open cover) compact, then $K$ is closed and bounded. (Discussion: if you replace open cover compact by sequential compactness, can you prove the two conclusions directly (without using the equivalence of the two definitions)?)

3. We know that $[0,1]$ is sequentially compact (by Heine-Borel theorem), can you show that $[0,1]^2$ is sequentially compact? (Hint: given a sequence $(p_n)$ in $[0,1]^2$, first show that you can pass to subsequence to make the sequence of the first coordinate converge, then …)

There are two notions of compactness, they turn out to be equivalent for metric spaces.

Let $X$ be a metric space, $K \In X$ a subset.

• sequential compactness: we say $K$ is compact, if every sequence in $K$ has a convergent subseq.
• compactness: any open cover of $K$ admits a finite subcover.

The two notions turns out are equivalent, see https://courses.wikinana.org/math104-f21/compactness

We will follow Pugh to give a proof. See also Rudin Thm 2.41

We will finish discussion about compactness. In particular, in $\R^n$, we have Heine-Borel theorem, namely, compact subsets are exactly those subsets which are closed and bounded. Note that, this is false for general metric space, e.g. closed and bounded subset in $\Q$ are not compact.

Last time we had some basic notion of metric space (a set with the notion of distances), and topological space (which is a set with some notion of which subsets are open).

Today, we will define continuous functions (or rather continuous maps) between metric spaces and between topological spaces.

Def 1 : Let $(X, d_X)$ and $(Y, d_Y)$ be two metric spaces, a map $f: X \to Y$ is continuous, if for every convergent sequence $p_n \to p$ in $X$, we have convergent sequence $f(p_n) \to f(p)$.

Def 2 : ($\epsilon-\delta$: Let $(X, d_X)$ and $(Y, d_Y)$ be two metric spaces, a map $f: X \to Y$ is continuous, if for every $x \in X$, and every $\epsilon > 0$, we have $\delta>0$, such that $f( B_\delta(x)) \In B_\epsilon(f(x))$.

Def 3 : Let $X, Y$ be two topological spaces, we say a map $f: X \to Y$ is continuous, if for every open sets $V \In Y$, we have $f^{-1}(V)$ is open in $X$.

We can discuss later, why the three definitions are equivalent.

Homeomorphism: if $f: X \to Y$ is continuous and a bijection, and if $f^{-1}: Y \to X$ is also continuous, then we say $f$ is a homeomorphism.

Topologically, we cannot distinguish spaces that are homeomorphic.

We plan to discuss topological space. Read Ross section 13 to see the axioms of 'topology'.

Exam solution given in lecture.

We begin discussing metric space and topological space. The note from last year might be useful.

Can you think of other example of metric spaces?