ANA1_L16

To-do:

• Study sequential and limit point compactness. Show that their absolute and relative definitions are equivalent, essentially showing that they are intrinsic properties. Prove equivalence of the two notions for metric spaces.
• Examples of compact metric spaces
• Applications of compactness in $\mathbb{R}$
• The extreme value theorem
• Heine-Borel characterization of compact sets in ${\mathbb{R}}^k$

Recall

An introduction to compactness

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Compactness: Absolute and relative definitions

Sequential compactness

Definition 1(Absolute definition for metric space$X$).

Every sequence in $X$ has a convergent subsequence.

Definition 2(Relative definition for$E\subset$ metric space $X$).

Every sequence in $E$ has a convergent subsequence with limit in $E$.

Equivalence

$E$ is sequentially compact as per definition 1 as a metric space in its own right $⟺$ $E$ is a sequentially compact subset of $X$ as per 2.

Thus, sequential compactness is intrinsic, i.e, if $E\subset Y\subset X$,
$E$ is sequentially compact in $X$ $⟺$ $E$ is sequentially compact in $Y$ $⟺$ $E$ is sequentially compact.

Limit point compactness

Definition 3(Absolute definition for metric space$X$).

Every infinite $S\subset X$ has a limit point.

Definition 4(Relative definition for$E\subset$ metric space $X$).

Every infinite $S\subset E$ has a limit point in $E$.

Equivalence

Same argument as in sequential compactness.

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Examples of compact sets

Finite sets

Any set $S$ of finite cardinality is compact. Any sequence in $S$ must have at least one element of $S$ repeating infinitely many times - take the subsequence required by sequential compactness to be a constant sequence of just this element. The condition for limit point compactness is vacuously true, since no infinite subset of $S$ exists.

The Cantor set

The Cantor set is closed and bounded in $\mathbb{R}$. It follows from the Heine Borel Theorem that it is compact.

Open intervals in R (non example)

$(a,b)$, $[a,b)$, $(a,b]$, and $\mathbb{R}$ are not compact. It is easy to find sequences in these sets which violate the definition of sequential compactness. Take the set of terms of these sequences to violate the definition of limit point compactness.

Closed intervals in R

$[a,b]$ is compact.

Sequential compactness: Consider a sequence in $[a,b]$. We know it has a convergent subsequence, courtesy the Bolzano-Weierstrass Theorem. The point to which this subsequence converges is a limit point of $[a,b]$, and must be contained in $[a,b]$ since $[a,b]$ is a closed set.

Limit point compactness: Showing this follows a very similar strategy to Abbot’s proof of the Bolzano-Weierstrass Theorem (bisection search, or what Kulkarni likes to call lion-hunting). Given an infinite $S$, successively bisect the intervals, continuing with the half having infinite elements of $S$ at each step. This should give us nested intervals of the form $[a_0,b_0],[a_1,b_1],[a_2,b_2],\dots$, where the length of $[a_n,b_n]$ is $|a_0-b_0|/2^n$ ($\to 0$), and $|[a_n,b_n]\cap S|=\infty$. From the nested interval property, we know that ${\bigcap }_{i\ge 0}[a_i,b_i]$ is non-empty (in fact, we know it is a singleton, but we just need it to be non-empty for this to work). Let $p$ be an element in this intersection. Let $ϵ>0$. We can find $k$ such that the length of $[a_k,b_k]<ϵ$. Thus, $[a_k,b_k]\subset (p-ϵ,p+ϵ)$. So, any $ϵ$ ball around $p$ contains elements of $S$, i.e, $p\in [a,b]$ is a limit point of $S$.

It follows that “closed boxes” in ${\mathbb{R}}^k$ are limit point compact. This can be shown using an analogue of bisection search in ${\mathbb{R}}^k$, where you divide the set into $2^k$ sections in each step. We will skip showing these are sequentially compact (follows trivially from BWT), since the two notions are actually equivalent, as we will now show.

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Equivalence of sequential and limit point compactness

Theorem 5(Theorem).

Let $X$ be a metric space. Then, $X$ is sequentially compact $⟺$ $X$ is limit point compact.

Proof of $⟹$
Let $S\subset X$, $|S|=\infty$. We want to show that $S$ has a limit point in $X$. Take $T=\{x_1,x_2,\dots \}\subset S$. Since $X$ is sequentially compact, we can extract a sequence $x_{k_1},x_{k_2},\dots$ from $T$ such that $(x_{k_n})\to p\in X$. Observe that $p$ is a limit point of $T$, and hence of $S$. ❏

Proof of $⟸$
Let $(p_n)$ be a sequence in $X$. We want to show that some subsequence of $(p_n)$ converges to a point in $X$.
Let $S=\{p_n|n\in \mathbb{N}\}$. If $S$ is finite, i.e, $(p_n)$ has a finite number of distinct terms, a single element must repeat infinite times in the sequence. Let the constant sequence of this element be the subsequence required, and we are done.
If $S$ is infinite, since $X$ is limit point compact, $S$ must have a limit point in $X$. Call it $p$. Define a sequence of $ϵ$‘s: $({ϵ}_n)=\left(\frac{1}{2^n}\right)$. Construct a subsequence $(p_{k_n})$ of $(p_n)$ like so: Let $p_{k_1}$ be any point in $B_{{ϵ}_1}(p,S)\setminus \{p\}$, and let $p_{k_{n+1}}$ be any point such that $k_{n+1}>k_n$ and $p_{k_{n+1}}\in B_{{ϵ}_{n+1}}(p,S)\setminus \{p\}$ (such a $k_{n+1}$ must exist since every deleted neighborhood of $p$ contains infinitely many elements of $S$). Then, $(p_{n_k})\to p$ since $({ϵ}_n)\to 0$. ❏

Info

We have not yet shown that open cover compactness is equivalent to limit point/sequential compactness, so replace every instance of compact with limit point compact or sequentially compact in the following theorems and proofs if you’re reading these lecture notes in order.

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Image of a compact set under a continuous function

Theorem 6(Theorem).

If $f:X\to Y$ is continuous, and $X$ is compact, $f(X)$ is compact.

Proof
Consider a sequence $f(x_1),f(x_2),\dots$ in $f(X)$. $x_1,x_2,\dots$ is a sequence in $X$. Since $X$ is compact, $x_1,x_2,\dots$ has a subsequence $(x_{k_n})\to p\in X$. Since $f$ is continuous, it follows that $f(x_{k_n})\to f(p)\in f(X)$. ❏

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Characterization of compact sets

Rudin, 2.34

Theorem 7(Theorem).

Every compact set $C$ in a metric space $X$ is closed in $X$ and bounded.

Proof
Assume $C$ is not closed. Then, $|C|=\infty$, and $C$ has a limit point $p$ outside $C$. There exists a sequence $(p_n)$ in $C$ that converges to $p$. Since every subsequence of $(p_n)$ also converges to $p$, $(p_n)$ has no subsequence that converges to a point in $C$, contradicting the hypothesis that $C$ is compact. $\Rightarrow \Leftarrow$
Suppose $C$ is unbounded. Pick $x_0\in C$. Pick $x_{n+1}$ such that $d(x_0,x_{n+1})>1+d(x_0,x_n)$. This construction results in the distance between any two terms in $(x_n)$ being greater than 1. Thus, there does not exist an $N$ such that for all $n\ge N$, all $x_n$ lie in a ball of radius $\frac{1}{2}$. So, no subsequence of $(x_n)$ converges, contradicting the hypothesis that $C$ is compact. $\Rightarrow \Leftarrow$ ❏

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Extreme value theorem

Consider a continuous function $f:X\to \mathbb{R}$, where $X$ is a compact set. We know now that this implies $f(X)$ is compact, which in turn implies $f(X)$ is closed in $\mathbb{R}$ and bounded. Since $f(X)$ is non-empty and bounded, it must have a supremum and infimum, when then must be contained in $f(X)$ since it is closed. Hence the theorem.

Rudin, 4.16

Theorem 8(Extreme value theorem).

A continuous function from a non-empty compact space to a subset of the real numbers attains a maximum and a minimum value.

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Closed subsets of compact sets are compact

Rudin, 2.35

Theorem 9(Theorem).

If $X$ is compact, $C$ is closed in $X$ $⟹$ $C$ is compact.

Proof
Consider a infinite subset $S\subset C$. Since $S\subset X$, $S$ has a limit point $p$ in $X$. However, a limit point of $S$ is also a limit point of $C$, which contains all of its limit points since it is closed. Thus, $p\in C$. ❏

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Heine Borel Theorem

Theorem 10(Theorem).

Let $C$ be a subset of $\mathbb{R}$. Then, $C$ is compact $⟺$ $C$ is closed in $\mathbb{R}$ and bounded.

Proof of $⟹$
Follows from the Characterization of compact sets.

Proof of $⟸$
Since $C$ is bounded, $C\subset [-M,M]$ for some $M$. Now, $C=[-M,M]\cap C$, and thus $C$ is closed in $[-M,M]$ (Subspace topology). We have previously shown that closed intervals in $\mathbb{R}$ are compact, so $[-M,M]$ is compact. It follows form the previous theorem that $C$ is compact. ❏

Warning

In general, $X$ is closed and bounded $/⟹$ $X$ is compact. For example, let $E$ be the set of all $p\in \mathbb{Q}$ such that $2<p^2<3$. $E$ is closed and bounded in $\mathbb{Q}$, but $E$ is not compact.

Important

While stated for $\mathbb{R}$, the Heine Borel theorem is valid for ${\mathbb{R}}^k$. It is easy to show.