• #todo/school : Do problems 3.2.1-8, 10, 13, 14 📅 2024-10-30 ✅ 2024-11-03

3.2.1

Proof

a. It's the part where we have $U_1,\dots ,U_n$. This is a finite list of $n$-elements, and later when we do $\epsilon =min({\epsilon }_1,\dots ,{\epsilon }_n)$ we can only do this with a finite list.

b. Consider the collection:

Here $\bigcap \mathcal{O}=\{0\}$ is clearly closed and $\ne \mathrm{\varnothing },\ne \mathbb{R}$. Let's show the equality holds and that the resulting set is closed.

For the former, notice that for $O_i=[0,\frac{1}{n}]$ that it always has $0\in O_i$ for all $i$. We'll show that there are no other elements $x\in O_i$ for all $i$. Assume for contradiction that one such value exists $x\ne 0$. Notice for all $i$ that $0\le x\le \frac{1}{i}$. The right is a sequence and the left is too, so using the Squeeze Theorem then that implies $(x)\to 0$. But that implies that $x=0$ which is a contradiction.

Now to show it's closed, we just have to show that $\{0\}$ contains it's limit points. The only limit points here are $0$ which are already in the set.

☐

3.2.2

Proof

a. The limit points of $A$ are just $1,-1$ (you would intuitively take the limit of the sequence, but actually prove this via the definition). The limit points of $B$ are any $[0,1]$ points.

b. By intuition it seems that since $A$ is a discrete set of points then it isn't open. It definitely isn't closed since $1\in A$ but $-1\notin A$ showing it's is not closed. It isn't open since if we say have $x=2$ for example, then any $ϵ>0$ has it that:

Because the largest element in $A$ is itself $2$, so then $2+\frac{ϵ}{2}\notin A$.

For $B$ notice that it's similar to $(0,1)\subseteq \mathbb{Q}$. Notice that because of this then the limit points $0,1\notin B$ so then $B$ is not closed. Is it open? Let $x\in B$ be arbitrary. We want to choose some $ϵ>0$ such that any $v\in V_{ϵ}(x)$ has $v\in B$. This happens where:

If we want $v\in B$ then we need to show $0<x-ϵ\Rightarrow ϵ<x$ and $x+ϵ<1\Rightarrow ϵ<1-x$. Choose $ϵ=min\{1-x,x\}$ such that $ϵ>0$ to make this happen (this works since $x\ne 0,1$ so we can always choose then minimum). But we can't ensure $v\in \mathbb{Q}$ even though we can use The Density of Q in R. The $v$ is arbitrary, so then we could always find some $v\in \mathbb{I}$ where $v\notin B$ by construction. Thus $B$ is not open.

c. $A$ has infinitely many isolated points in the set $\{(-1{)}^n+\frac{2}{n}|n\ne 1\}$. $B$ doesn't have isolated points since the set of limit points is a super-set of $B$.

d. $\bar{A}=A\cup \{-1\}$. $\bar{B}=(0,1)\cup \{0,1\}=[0,1]\subseteq \mathbb{Q}$.

☐

3.2.3

Proof

a. $\mathbb{Q}$ is not open since if we choose $x=2$ for example then for any $\epsilon >0$ then since there's always some $\mathbb{I}$ between two rationals (see Suprema + Density Practice#1.4.1). Thus that irrational is not in $\mathbb{Q}$ by definition, so then $V_{\epsilon }(2)⊈\mathbb{Q}$.

$\mathbb{Q}$ is not closed since the limit points of $\mathbb{Q}$ is $\mathbb{R}$ (this is the Axiom of completeness) are not all within $\mathbb{Q}$ (for example, $\sqrt{2}$).

b. $\mathbb{N}$ is not open since literally choosing say $n=1$ then any $\epsilon >0$ gives that $V_{\epsilon }(1)⊈\mathbb{N}$ as say $1+\frac{\epsilon }{2}$ is in the neighborhood but not the natural numbers.

$\mathbb{N}$ is closed since there are no limit points.

c. The set is open since we can set our $\epsilon$-neighborhood to not include the missing $0$. Namely, if we let $x\ne 0$ be arbitrary, then choose:

• If $x>0$ then choose any $\epsilon >x>0$ where then any $p\in V_{\epsilon }(x)$ implies $|p-x|<\epsilon$ so considering that:

But $x-\epsilon >0$ so then $p\in \mathbb{R}$ as desired.

• If $x<0$ then choose $\epsilon >-x>0$ where a similar argument shows openness.

The set is not closed since the limit point $0$ is not in the set by construction.

d. For some context we know that $\sum \frac{1}{n^2}$ is a convergent $p=2$ series, so then it is some limit point not in the set $L$.

Thus the set is not open since if we choose say $x=1$ then any $V_{\epsilon }(1)$ will have real numbers between $1$ and $1+\frac{1}{4}=\frac{5}{4}$ that are not in the set.

The set is not closed since the limit point $L$ is not in the set (if it was, then any future terms would have to be 0, but $\frac{1}{n^2}\ne 0$ for any $n$).

e. For context we know that $\sum \frac{1}{n}$ is a divergent $p=1$ series, so then there are no limit points that aren't in the set.

The set is not open similar in the way that choosing $x=1$ lets $V_{\epsilon }(x)$ have real numbers between $1$ and $1+\frac{1}{2}=\frac{3}{2}$ that are definitely not in the set.

The set is closed since there are no limit points for the set (as the sequence diverges, so every sequence diverges).

☐

3.2.4

Proof

a. It's enough to show either $s\in A$ or $s$ is a limit point of $A$. If $s\in A$ then we are done so suppose $s\notin A$. To show $s$ is a limit point of $A$ it goes to show every $\epsilon$-neighborhood around $s$ intersects $A$ at a point other than $s$ (which we don't have to worry about since $s\notin A$).

Let $\epsilon >0$. We'll show that $A\cap V_{\epsilon }(s)$ is non-empty by choosing an element $a=s-\frac{\epsilon }{2}$ and showing it's in this set.

First notice that $a\in V_{\epsilon }(s)$ since $|a-s|=|s-\frac{\epsilon }{2}-s|=\frac{|\epsilon |}{2}<\epsilon$. Also $a<s$ so then since $s$ is the suprema of $A$ then $a\in A$ (otherwise $a>s$ which is a contradiction). Thus $a\in A\cap V_{\epsilon }(s)$.

b. Nope. We'll show it's impossible. Say some open set $U$ exists such that $s=supU\in U$. Then since $s\in U$ then $\mathrm{\exists }\epsilon >0$ such that $V_{\epsilon }(s)\subseteq U$. But that would imply that the points $(s,s+\epsilon )\subseteq U$ but clearly if $s+\epsilon \in U$ then that contradicts $s$ being the supremum.

☐

3.2.5

Proof

($\Rightarrow$): Suppose $A$ is closed, so then $A$ contains it's limit points. Let $(x_n)\subseteq A$ be a Cauchy Sequence. Then $(x_n)$ converges by being Cauchy. Say it converges to $x$, then since $A$ is closed then $x\in A$ as required since $A$ contains its limit points, which are equivalent to the limits of the convergent sequences of $A$.

($\Leftarrow$): Suppose $\mathrm{\forall }(x_n)\subseteq A$ where $(x_n)\to x$ (ie: the sequence $(x_n)$ is Cauchy because it converges) where $x\in A$. We need to show that $A$ contains it's limit points.

Let $\ell$ be a limit point of $A$. We want to show that $\ell \in A$. Since $\ell$ is a limit point of $A$ then equivalently there is some sequence of points in $A-\{\ell \}$ where it converges to $\ell$. Denote this sequence $(a_n)\in A-\{\ell \}$ where $(a_n)\to \ell$. then since all convergent sequences in $A$ converge to a point in $A$ then $\ell \in A$.

☐

3.2.6

Proof

a. False. The set $\{x|x\ne \sqrt{2}\}\subseteq \mathbb{R}$ contains every rational number, but isn't all of $\mathbb{R}$.

b. False. Let $A_n=[n,\mathrm{\infty })$. Then each $A_n\ne \mathrm{\varnothing }$ and $A_{n+1}\subseteq A_n$ but $\bigcap _{n=1}^{\mathrm{\infty }}{A}_n=\mathrm{\varnothing }$

c. True. Any nonempty open set has some $a\in A$ and since $A$ is open then $\mathrm{\exists }\epsilon >0$ such that $V_{\epsilon }(a)\subseteq A$. Now clearly $a-\epsilon ,a+\epsilon$ are real numbers so since $\mathbb{Q}$ is dense in $\mathbb{R}$ then there is some rational number $r$ such that $a-\epsilon <r<a+\epsilon$. Now that implies $r\in V_{\epsilon }(a)\subseteq A$ so then $r\in A$, so $A$ has some rational number.

d. False. Consider $A=\{\sqrt{2}+\frac{1}{n}:n\in \mathbb{N}\}\cup \{\sqrt{2}\}$. This set has the limit point of $\sqrt{2}$ which is in the set, so it's closed. Notice also $A$ is bounded between $\sqrt{2}+1$ and $\sqrt{2}$. It's also infinite in size (countable infinite to be exact).

e. True. Take any limit point $x$ of $C$. Then we can create a series of binary digits where $0$ represents it being the the left set of $C_i$ (for each cut $i$), and $1$ for the right set. Then $x$ is represented by some series of binary digits $b_1{b}_2\dots$ where each $b_i$ is 0 or 1. Then convert each $b_i$ into either the left endpoint of the related set if it's a $0$, or the right endpoint if it's a $1$. If any $b_i$ should make the sequence ever equal $x$, then choose the opposite endpoints (since all the following $b_i$'s should be the same number, we should have all 0's or 1's to follow, so choosing the opposite endpoints should create a sequence that doesn't equal $x$ but converges to $x$). As a result this sequence of numbers will never equal $x$ but will converge to $x$, thus $x\in C$. Therefore, $C$ is closed.

☐

3.2.7

Proof

a. Let $\ell$ be a limit point of $L$, so then $\mathrm{\exists }(x_n)\subseteq L$ such that each $x_n\in L-\{x\}$ and then $x_n\to \ell$. We need to show that $\ell \in L$ by showing it as a limit of a sequence of some $a_n$'s. Now since $(x_n)\subseteq L$ then each $x_n$ is some limit point for $A$, namely $\mathrm{\forall }n\in \mathbb{N}$ then $\mathrm{\exists }{a}_n\in A$ such that $a_m\to x_n$. Denote $a_{mn}$ being $a_m$ for the corresponding $x_n$. We want to show that $a_{mn}\to \ell$ as $m\to \mathrm{\infty }$.

Let $\epsilon >0$. As a result $\frac{\epsilon }{2}>0$ so then since $a_m\to x_n$ and $x_n\to \ell$ then $\mathrm{\exists }N\in \mathbb{N}$ such that for $m,n>N$ then:

Thus $a_{mn}\to \ell$ as desired, so then $\ell$ is a limit point for $A$ so $\ell \in L$.

b. Let's prove the lemma:

Proof

Let $x$ be a limit point of $A\cup L$. Then $\mathrm{\exists }(x_n)\subseteq A\cup L$ where each $x_n\in (A\cup L)-\{x\}$ while $x_n\to x$. Now we need to construct a sequence such that each element is in $A-\{x\}$ while that sequence converges to $x$. The problem we'll face is that some of the $x_n$'s will be in $L$ and not in $A$ (since $A$ may or may not be closed).

This is easy. Create a subsequence $(x_{n_k})$ such that for each $n$, you do the following:

• If $x_n\in A$ then add it to the end of the subsequence.
• If $x_n\notin A$ (ie: $x_n\in L$) then append the previous $x_n\in A$ to the sequence.
  The only issue is potentially if $x_n$ eventually only has $x_n\in L$ terms. But if this happens then because all the $x_n\in L$ terms have some corresponding sequence $\mathrm{\exists }(a_n)\in A$ then just choose the $n$ and onward terms from that sequence instead.

Since this is a subsequence of a convergent sequence that converges to $x$ then this subsequence must also converge to $x$, showing $x$ is a limit point to $A$.

☐

The actual proof showing the smallest is seen at Closure is the Smallest Closed Set, which uses our lemma.

☐

3.2.8

For all of these it's useful to use the Unions & Intersection of Closed Sets and the Union & Intersection of Collections of Open Sets properties, along with the complement/closure properties. Furthermore, since there are only two sets that are both open and closed then we should only have definitely closed or definitely open.

Proof

a. We know that $\overline{A\cup B}$, by the definition of closure, is a closed set.

b. Notice that $A-B=A\cap B^c$. Notice that since $B$ is closed then $B^c$ is open. Since $A$ is open and $B$ is open then their intersection must be open.

c. This is equivalent to (b).

d. Since $A$ is open then $A^c$ must be closed. Then $A^c\cap B$ must be closed as well, so then the whole set is definitely closed.

e. Note that by closure then $\bar{A}$ is definitely closed, implying ${\bar{A}}^c$ is definitely open. Similarly $\overline{A^c}$ is definitely open. Their intersection implies that the whole thing is definitely open.

☐

3.2.10

Proof

i. This one is impossible. Say some countable set $A=\{a_n:n\in \mathbb{N}\}\subseteq [0,1]$ such that there are no isolated points. Then we can sort our $a_n$'s into a sequence $b_n$ such that the sequence of $b_n$'s are increasing. Since $b_n\le 1$ for all $n$ then the Monotone Convergence Theorem says that $b_n\to x$ converges. But if $x\in A$ then that would imply that eventually the $b_i=x$ which implies that $A$ is finite and thus not countable. Thus $x\notin A$ implying that $x$ is a limit point of $A$ which is also a contradiction.

(alternatively one could use Balzano-Weierstrass Theorem to show there exists a convergent subsequence, implying a limit point).

ii. This one is possible. Use $\mathbb{Q}\cap [0,1]$ has no isolated points while being countable.

iii. This one is impossible. Let $A\subseteq \mathbb{R}$ where $x$ is some isolated point of $A$. From the definition then $\mathrm{\exists }\epsilon >0$ where $V_{\epsilon }(x)\cap A=\{x\}$. Consider the set of points unioning the $V_{\epsilon }(x)$ for each isolated point $x$:

But via Cardinality Practice#1.5.6 suggests that since each $\{\lambda \}$ is disjoint with one another (and are open subintervals) then their collection cannot exist. Hence $A$ cannot exist.

☐

3.2.13

Proof

Notice that ${\mathrm{\varnothing }}^c=\mathbb{R}$ so then if we show that $\mathrm{\varnothing }$ is closed and open, then $\mathbb{R}$ is both closed an open by Closed Implies Complement is Open (Vice Versa).

Notice that $\mathrm{\varnothing }$ being open is vacuously true since there is no element $x\in \mathrm{\varnothing }$ such that we have to construct some $\epsilon$-neighborhood within $\mathrm{\varnothing }$.

Now to show that it is open; that $\mathrm{\varnothing }$ has all of its limit points, notice that there are no limit points to $\mathrm{\varnothing }$, so then $\mathrm{\varnothing }$ is also vacuously open.

Now we should show that any set $A$ that isn't one of these sets must exclusively be either open or closed. Assume for contradiction that the theorem doesn't hold, that instead $A$ here is non-empty (and not $\mathbb{R}$) but is both open and closed. Choose $a\in A$ be an arbitrary element (which we can choose since $A\ne \mathrm{\varnothing }$). Then because $A$ is open then $\mathrm{\exists }\epsilon >0$ where $V_{\epsilon }(a)\subseteq A$.

Consider the cases that $A$ could be:

• If $A$ is finite or countable then since $V_{\epsilon }(a)$ is uncountable, then that would imply $A$ is uncountable which is a contradiction (since $A$ is finite or countable).
• So then $A$ must be uncountable. But if that's the case then $A$ must have some limit point (for example: the sequence $\{a+\frac{(2^n-1)s}{2^n}:n\in \mathbb{N}\}\subseteq V_{\epsilon }(a)$ is a convergent sequence with limit point $a+\epsilon$). Specifically (like we did in the example) we can show $a+\epsilon$ must be a limit point of $A$. But look! We can repeat this process since $A$ is closed so then $a+\epsilon \in A$, so then $\mathrm{\exists }{\epsilon }_1>0$ such that $V_{{\epsilon }_1}(a+\epsilon )\subseteq A$. Denote the points $a_1=a+\epsilon$, $a_2=a+{\epsilon }_1>a+\epsilon =a_2$, $a_3=a+{\epsilon }_2>a+{\epsilon }_1=a_2$, and so one. Notice that the sequence $a_n$ is strictly increasing, where we aren't missing any points between the $a_i$'s. As such then:

A similar process can be done for the negative numbers, giving that $(-\mathrm{\infty },a)\subseteq A$. That means that since $a\in A$ then $A=\mathbb{R}$ which is a contradiction.

☐

3.2.14

Proof

a. Let's do the first proof:

($\Rightarrow$): Suppose $E$ is closed, so then $E$ contains its limit points. As a result then $L\subseteq E$ and $\bar{E}=E\cup L=E$.
($\Leftarrow$): Suppose $\bar{E}=E$. Then $E\cup L=E$ so then $L\subseteq E$ and thus $E$ is closed (any $x\in L$ implies $x\in E$).

For the second proof:

($\Rightarrow$): Suppose $E$ is open. Then we'll show any $x\in E^{\circ }$ implies $x\in E$ and vice versa. Suppose $x\in E^{\circ }$. Then $\mathrm{\exists }\epsilon >0$ such that $V_{\epsilon }(x)\subseteq E$. Clearly $x\in V_{\epsilon }(x)$ so then $x\in E$ as desired. Similarly let $x\in E$. Since $E$ is open then $\mathrm{\exists }\epsilon >0$ such that $V_{\epsilon }(x)\subseteq E$. Clearly then $x\in E^{\circ }$ as a result.

($\Leftarrow$): Suppose $E^{\circ }=E$. To show $E$ is open, then let $x\in E$ be arbitrary. Then $x\in E^{\circ }$ by our given, so $\mathrm{\exists }{V}_{\epsilon }(x)\subseteq E$. By definition of openness then $E$ must be open.

b. For the first proof:

$x\in {\bar{E}}^c$ iff $x\notin E$ and $x$ is not a limit point of $E$. $x\in (E^c{)}^{\circ }$ iff $x\notin E$ and $\mathrm{\exists }{V}_{\epsilon }(x)\subseteq E^c$. All we have to show is that $x$ is not a limit point is equivalent to $\mathrm{\exists }{V}_{\epsilon }(x)\in E^c$, but the definitions give this equivalence (take the opposite of the Limit Point definition, and ignore the $\{x\}$ case since $x\notin E$).

For the second proof try using the finding above, where here $D=E^c$:

☐