33 Compact Sets Practice

#todo/school : Read §3.3 📅 2024-11-03 ✅ 2024-11-03
#todo/school : Do problems 3.3.1, 2, 4, 5, 6, 7, 11; 📅 2024-11-05 ✅ 2024-11-06

3.3.1

Question

Show that if $K$ is compact and nonempty, then $sup K$ and $inf K$ both exist and are elements of $K$ .

Proof

These both exist iff $K$ is bounded from above and below. Since $K$ is compact then by Characterization of Compactness in R then $K$ is closed and bounded, so then it's bounded above and below as well as the suprema/infima are in $K$ , so thus both the suprema and infima exist.

Now let $s = sup K$ . We want to show that $s \in K$ . By the Suprema Lemma then for any $ε > 0$ we choose then $\exists k \in K$ such that $s - ε < k$ . If we choose $ε = \frac{1}{n}$ for all $n \geq 1$ then this works, so then:

s - \frac{1}{n} < k

Now construct the sequence $s - \frac{1}{n}$ . Notice that because the sequence converges to $s$ then since $s - \frac{1}{n} \neq s$ for all $n$ then $s$ must be a limit point of $K$ . Now because $K$ is closed, then $s \in K$ .

The argument for $inf K$ is done in a similar way, using $ε = \frac{1}{n}$ just like we did here.

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3.3.2

Question

Decide which of the following sets are compact. For those that are not compact, show how compactness's definition breaks down (give an example of a sequence contained in the given set that doesn't have a subsequence converging to a limit point in the set).
a. $N$
b. $Q \cap [0, 1]$
c. The Cantor Set
d. ${1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots + \frac{1}{n^{2}} : n \in N}$
e. ${1, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \dots}$

Proof

a. False. The sequence of counting numbers $a_{n} = {0, 1, 2, 3, \dots}$ is a sequence such that any subsequence doesn't have a limit point converging in the set. The thing is that any subsequence here is always increasing, so it must diverge. As such it can never converge to limit point in the set, as the set doesn't have any limit points.

b. False. We can construct a sequence $x_{n} \to \frac{1}{\sqrt{2}} \subseteq Q \cap [0, 1]$ implying that all subsequences converge to the same limit, but this value $\frac{1}{\sqrt{2}} \notin Q \cap [0, 1]$ . Thus $Q \cap [0, 1]$ is not closed here so then the set cannot be compact.

c. True. Clearly $C$ is bounded (within $[0, 1]$ ). Notice that the limit points of $C$ are the entirety of $C$ (because for any $\frac{m}{3^{n}}$ limit point you can choose the sequence of opposite endpoints that converge to the limit point in question), so then $C$ is closed. (Alternatively, it is the infinite intersection of closed sets $C = ⋂_{n = 1}^{\infty} C_{n}$ so by Unions & Intersection of Closed Sets then $C$ is closed).

As a result then $C$ is compact.

d. False. Recall that since $\sum \frac{1}{n^{2}}$ is a convergent $p = 2$ series, then the sequence of partial sums $s_{n}$ converges (which is the sequence given). Notice then that all subsequences $(s_{n_{k}})$ must all converge to the same value (let's denote it $ℓ$ for clarity).

Notice that since $\frac{1}{n^{2}} \neq 0$ for any $n$ implies that $ℓ$ isn't in the set (if it was, then eventually $s_{n} = ℓ$ so then we'd have to be adding $0$ to not go past it, which is a contradiction). Thus then the set is not closed so then the set cannot be compact.

e. True. Notice that the set is bounded between 0 and 1. For closed notice that it is a sequence that converges to $1$ so then all subsequences must all converge to $1$ . This value is explicitly in the set, so then it's closed. As a result, the set is compact.

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3.3.4

Question

Suppose $K$ is compact and $F$ is closed. Decide if the following sets are definitely compact, definitely closed, both, or neither.
a. $K \cap F$
b. $\overset{―}{F^{c} \cup K^{c}}$
c. $K - F$
d. $\overset{―}{K \cap F^{c}}$

Notice that since $K$ is compact then $K$ is closed by definition (and since $K$ is bounded, then $K \neq R$ ). See Unions & Intersection of Closed Sets along with Union & Intersection of Collections of Open Sets for most properties used here.

Thus also $K^{c}, F^{c}$ ( $K^{c} \neq \emptyset$ ) are both definitely open. Notice $K^{c}$ is definitely unbounded.

Proof

a. Since $K, F$ are both compact then their intersection is definitely closed. But since it's a subset of a bounded set, then the new set must be bounded. Thus then the set is definitely compact.

b. By definition of closure no matter what the set has to be definitely closed. Notice especially that since $K^{c}, F^{c}$ are open then $F^{c} \cup K^{c} \neq \emptyset$ is definitely open, agreeing with our finding. Is the set bounded? Consider the worst case where $K = \emptyset$ . Then the set overall becomes $R$ which isn't bounded. Thus we only know this is definitely closed.

c. Notice $K - F = K \cap F^{c}$ . Since $K$ is closed then an arbitrary intersection with another set $F^{c}$ means that this set is definitely closed. Since $K - F \subseteq K$ then it is bounded, so then this set is definitely compact.

d. Again by the definition of closure then the set is definitely closed. Notice that since $K \cap F^{c}$ is bounded, then $\overset{―}{K \cap F^{c}}$ is bounded (if $A$ is bounded then there's some closed ball $B$ such that $A \subseteq B$ , and by definition of $\overset{―}{A}$ as the Closure is the Smallest Closed Set then $\overset{―}{A} \subseteq B$ , thus $\overset{―}{A}$ is bounded). Thus the set is definitely compact.

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3.3.5

Question

Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
a. The arbitrary intersection of compact sets is compact.
b. The arbitrary union of compact sets is compact.
c. Let $A$ be arbitrary, and let $K$ be compact. Then, the intersection $A \cap K$ is compact.
d. If $F_{1} \supseteq F_{2} \supseteq F_{3} \supseteq \dots$ is a nested sequence of nonempty closed sets, then the intersection $⋂_{n = 1}^{\infty} F_{n} \neq \emptyset$ .

Proof

a. True. Using Unions & Intersection of Closed Sets, an arbitrary intersection of closed sets is still closed, so for compact sets their arbitrary intersection must be closed. Similarly, take any $U_{λ_{0}}$ that we are intersecting with. It is compact and bounded, so then:

⋂_{λ \in Λ} U_{λ} \subseteq U_{λ_{0}}

Shows that their intersection must also be bounded. Thus the arbitrary intersection of compact sets is compact.

b. False. Using Unions & Intersection of Closed Sets it shows that a finite union of compact sets is compact, suggesting a counterexample using an infinite collection. Notice that if we use the countable collection:

{[- n, n] | n \in N}

Then their union would be:

⋃_{n = 1}^{\infty} [- n, n] = R

which is an unbounded set (and thus not compact).

c. True. Notice that $A \cap K \subseteq K$ so then since $K$ is bounded then their intersection must be bounded. By Unions & Intersection of Closed Sets since $K$ is closed then an arbitrary union, such as with $A$ here, must be closed.

d. False. The idea here is to compare this theorem with the Nested Compact Set Property and see that this is less restricted, only caring about closed sets. The idea here is to find an interval that is closed but unbounded. For example $[1, \infty)$ is such an interval. Thus if we consider the collection:

F_{n} = [n, \infty)

Then clearly $F_{n} \supseteq F_{n + 1}$ but their intersection:

⋂_{n = 1}^{\infty} F_{n} = \emptyset

because if there was such $x \in ⋂_{n = 1}^{\infty}$ then $x > 1$ , $x > 2$ , ... so on. We know by the Archimedian Property that $\exists k \in N$ where $x < k$ so then $\exists F_{k}$ where $x \notin F_{k}$ , thus creating a contradiction.

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3.3.6

Question

Verify that the following three statements are true if every blank is filled with the word "finite". Which are true if every blank is filled in with the word "compact"? Which are true if every blank is filled with the word "closed".
a. Every ______ set has a maximum.
b. If $A, B$ are ______ then $A + B = {a + b : a \in A, b \in B}$ is also ______
c. If ${A_{n} : n \in N}$ is a collection of ______ sets with the property that every finite subcollection has a nonempty intersection, then $⋂_{n = 1}^{\infty} A_{n}$ is nonempty as well.

Proof

a. The only case this wouldn't be true is when $K = \emptyset$ (in which this argument doesn't work).

Finite works since we can always take a max if a finite set.

This works for compact since $K$ , the compact set, is bounded, so then $sup K$ exists. Further since $K$ is closed, then $sup K \in K$ as a result (see 33 Compact Sets Practice#3.3.1 for a proof of this). Thus $sup K$ is the maximal element to choose.

This doesn't work for closed sets $U$ since we may not have a suprema. Consider the closed set $[1, \infty)$ . This has no maximum.

b. Clearly if $A, B$ are finite (say $| A | = n_{A}, | B | = n_{B}$ ) then $| A + B | \leq n_{A} n_{B}$ is still finite (a short inductive argument shows this, but this is just the idea of pairing each $a$ with a $b$ , with potential duplicates for their sum hence the $\leq$ ). Thus this is a finite number showing $A + B$ is finite.

This works for compact since if $A, B$ are compact then their suprema exist $sup A, sup B$ . Further they are in each set (since compact sets are closed). As a result, then $sup A + sup B = sup (A + B) \in A + B$ intuitively (without proof here, but it wouldn't be too hard to show). This shows that $A + B$ is bounded (by the definition of suprema) as well as $A + B$ is closed (since it has all the limit points, namely $sup (A + B)$ .

This doesn't work for closed sets. Namely, if $A, B$ are closed sets, then all limit points are in $A, B$ respectively. Now consider a limit point $x$ for $A + B$ . Then $\exists (x_{n}) \subseteq A + B - {x}$ where $x_{n} \to x$ . Now $x \in A + B$ so then $\exists a, b$ where $a \in A$ $b \in B$ such that $x = a + b$ .

Now then for any $n \in N$ we have $x_{n} = a_{n} + b_{n}$ where each $a_{n} \in A, b_{n} \in B$ . These are sequences $a_{n} \subseteq A - {x}$ and similar for $b$ . But these $a_{n}, b_{n}$ may not converge! As such use the following counterexample $A = {n : n \in N}$ and $B = {- n + \frac{1}{n} : n \in N}$ . Here the sequence $\frac{1}{n} \to 0$ so the sequence $n + (- n + \frac{1}{n}) \to 0$ but that is not in $A + B$ .

Namely, notice that the boundedness argument is needed to show that $(a_{n}), (b_{n})$ converges, and then after that the closed argument would work.

c. Finite works since the collection itself is a finite sub-collection, so by the given then its intersection of all the sets must be nonempty.

This works for compact sets. Namely take any countable collection $A = {A_{n} : n \in N}$ as a collection of closed sets (each $A_{i}$ is closed). Because each finite intersection is non-empty, then $A_{1} \cap A_{2} \neq \emptyset, A_{1} \cap A_{2} \cap A_{3} \neq \emptyset, \dots$ . That implies there is some $x_{1} \in A_{1} \cap A_{2}, \dots$ namely that there is some $x_{n} \in ⋂_{i = 1}^{n} A_{i}$ for all $n$ .

But notice that $⋂_{i = 1}^{n + 1} A_{i} \subseteq ⋂_{i = 1}^{n} A_{i}$ . As a result then let $I_{n} = ⋂_{i = 1}^{n} A_{i}$ , so then $I_{n + 1} \subseteq I_{n}$ . Now notice that because the arbitrary intersection of closed sets are closed, then each set $⋂_{i = 1}^{n} A_{i}$ must also be closed. Clearly they are subsets of the previous set, so since $I_{1}$ is bounded then any $I_{n}$ is also bounded. Thus all $I_{n}$ is compact, so by the Nested Compact Set Property then $⋂_{n = 1}^{\infty} A_{i} \neq \emptyset$ .

This doesn't work for purely closed sets. Akin to 33 Compact Sets Practice#3.3.5 (d) consider the collection $A = {[n, \infty) : n \in N}$ . Clearly each finite intersection is nonempty, since if we consider $⋂_{i \in I} A_{i}$ (here $I$ is finite) then you just $ε = max I \in N$ to give that $⋂_{i \in I} A_{i} = [ε, \infty)$ which is nonempty.

However, notice that similar to that example we assume that $\emptyset \neq ⋂_{n = 1}^{\infty} A_{i}$ then we get a problem. If $x$ is in this set then it must be greater than all natural numbers. But this is a contradiction by the Archimedian Property. Thus the infinite intersection must be empty.

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3.3.7

Question

As some more evidence of the surprising nature of the Cantor Set, follow these steps to show that the sum $C + C = {x + y : x, y \in C}$ is equal to the closed interval $[0, 2]$ . (Keep in mind that $C$ has zero length and contains no intervals).
Because $C \subseteq [0, 1]$ and $C + C \subseteq [0, 2]$ we only need to prove the reverse include $[0, 2] \subseteq C + C$ . Thus, given $s \in [0, 2]$ we must find two elements $x, y \in C$ where $s = x + y$ .
a. Show that there exist $x_{1}, y_{1} \in C_{1}$ for which $x_{1} + y_{1} = s$ . Show that in general for any $n \in N$ we can always find $x_{n}, y_{n} \in C_{n}$ for which $x_{n} + y_{n} = s$ .
b. Keeping in mind that the sequences $(x_{n})$ and $(y_{n})$ do not necessarily converge, show how they can nevertheless be used to produce the desired $x, y \in C$ satisfying $x + y = s$ .

Proof

a. We would do an induction to show the $n \in N$ case. Note that $[a, b] + [c, d] = [a + c, b + d]$ and $c [a, b] = [c a, c b]$ . Then since $C_{1} = [0, \frac{1}{3}] \cup [\frac{2}{3}, 1]$ :

\begin{aligned} C_{1} + C_{1} & = ([0, \frac{1}{3}] \cup [\frac{2}{3}, 1]) + ([0, \frac{1}{3}] \cup [\frac{2}{3}, 1]) \\ = ([0, \frac{1}{3}] + [0, \frac{1}{3}]) \cup ([0, \frac{1}{3}] + [\frac{2}{3}, 1]) \cup ([\frac{2}{3}, 1] + [\frac{2}{3}, 1]) \\ = [0, \frac{2}{3}] \cup [\frac{2}{3}, \frac{4}{3}] \cup [\frac{4}{3}, 2] \\ = [0, 2] \end{aligned}

Notice that we should prove the following lemma to get the above finding:

Distributive Property of Set Unions/Additions

$(A \cup B) + (C \cup D) = (A + C) \cup (A + D) \cup (B + C) \cup (B + D)$ .

Proof

( $\subseteq$ ): Let $x$ in the left set. Then $\exists l \in A \cup B$ and $r \in C \cup D$ where $x = l + r$ . Then we have the following cases:

$l \in A, r \in C$ then $x = l + r \Rightarrow x \in A + C$ .
$l \in B, r \in C$ then $x \in B + C$ .
$l \in A, r \in D$ then $x \in A + D$
$l \in B, r \in D$ then $x \in B + D$

( $\supseteq$ ): Let $x$ in the right set. Then we have 4 cases:

$x \in A + C$ . Then $x = l + r$ where $l \in A$ and $r \in C$ . Then clearly $x \in (A \cup B) + (C \cup D)$ .
Repeat for all cases.
...
...

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Corollary

$(A \cup B) + (A \cup B) = (A + A) \cup (B + B) \cup (A + B)$

Now let $C_{n} + C_{n} = [0, 2]$ . It goes to show that $C_{n + 1} + C_{n + 1} = [0, 2]$ . Notice that the recursive definition of $C_{n}$ is:

C_{n + 1} = \frac{1}{3} C_{n} \cup (\frac{2}{3} + \frac{1}{3} C_{n})

It goes to note that $c A = {c a : a \in A}$ (like we talked about above) and $c + A = {c + a : a \in A}$ , and $c (A + B) = c A + c B$ (this allows our identities to be derived from this definition).

Now notice:

\begin{aligned} C_{n + 1} + C_{n + 1} & = (\frac{1}{3} C_{n} \cup [\frac{2}{3} + \frac{1}{3} C_{n}]) + (\frac{1}{3} C_{n} \cup [\frac{2}{3} + \frac{1}{3} C_{n}]) \\ = (\frac{1}{3} C_{n} + \frac{1}{3} C_{n}) \cup (\frac{1}{3} C_{n} + [\frac{2}{3} + \frac{1}{3} C_{n}]) \cup ([\frac{2}{3} + \frac{1}{3} C_{n}] + [\frac{2}{3} + \frac{1}{3} C_{n}]) \\ = \frac{1}{3} (C_{n} + C_{n}) \cup \frac{1}{3} (C_{n} + C_{n} + 2) \cup \frac{1}{3} (C_{n} + C_{n} + 4) \\ = \frac{1}{3} [0, 2] \cup \frac{1}{3} ([0, 2] + 2) \cup \frac{1}{3} ([0, 2] + 4) \\ = \frac{1}{3} [0, 2] \cup \frac{1}{3} [2, 4] \cup \frac{1}{3} [4, 6] \\ = [0, \frac{2}{3}] \cup [\frac{2}{3}, \frac{4}{3}] \cup [\frac{4}{3}, 2] \\ = [0, 2] \end{aligned}

As a result we've shown equality, so then clearly we get that $\forall n \in N$ then $s = x_{n} + y_{n}$ where $x_{n}, y_{n} \in C_{n}$ .

b. Since $C$ is compact then it is closed, so then since our sequence $(x_{n})$ has a convergent subsequence (see Balzano-Weierstrass Theorem) then $\exists (x_{n_{k}}) \subseteq C$ where $x_{n_{k}} \to x$ with $x \in C$ . Now since $x_{n_{k}} + y_{n_{k}} = s$ for all $k$ by (a), then $lim_{n_{k} \to \infty} y_{n_{k}} = lim_{n_{k} \to \infty} (s - x_{n_{k}}) = s - lim_{n_{k} \to \infty} x_{n_{k}} = s - x$ . Since $(y_{n_{k}}) \subseteq C$ then the limit $y = s - x \in C$ is as well. Thus we've found $x, y \in C$ where $x + y = s$ .

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3.3.11

Question

Consider each of the sets listed in 33 Compact Sets Practice#3.3.2. For each one that is not compact, find an open cover for which there is no finite subcover.

Proof

(a) $N$ , (b) $Q \cap [0, 1]$ , and (d) ${1 + \frac{1}{2^{2}} + \dots + \frac{1}{n^{2}} : n \in N}$ were all not compact.

a. Use the collection $A = {(0, n) : n \in N}$ . Here any finite subcover ${A_{n_{1}}, A_{n_{2}}, \dots, A_{n_{k}}}$ (here each $A_{n_{i}} = {n_{i}}$ ) would miss the element $max {n_{1}, n_{2}, \dots, n_{k}} + 1 \in N$ . .

b. Consider using $U_{x} = (- \infty, x)$ and $W = (\frac{\sqrt{2}}{2}, \infty)$ . Then $W \cup ⋃_{x < \frac{\sqrt{2}}{2}} U_{x} \supseteq Q \cap [0, 1]$ but $x = \frac{\sqrt{2}}{2}$ is not in any finite subcover.

c. Use the collection $A = {{(0, s_{m}) : 0 < m \leq n} : n \in N}$ ( $s_{m}$ is the $m$ -th partial sum of $\sum \frac{1}{n^{2}}$ ). Notice here that $A_{1} \subset A_{2} \subset \dots$ . As a result any finite subcover ${A_{n_{1}}, A_{n_{2}}, \dots, A_{n_{k}}} = A_{max {n_{1}, \dots, n_{k}}} \subset A_{max {n_{1}, \dots, n_{k}} + 1}$ would be missing something from that right set.

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