Math 297 Midterm 2 Review

Relevant Classe: Math 297

Content

Worksheets Covered: Worksheet 01 - Worksheet 23

Homework Covered: Homework 01 - Homework 10a

Questions and Inspiration

Worksheet 13 - Topology: Sequential Compactness

Question 170.

Let $S=\{1/n\mid n\in \mathbb{N}\}$.

Show that the collection of open sets $\mathcal{C}=\{B_{1/m(m+1)}(1/m)\mid m\in \mathbb{N}\}$ is an open cover for $S$

Fix any element in $S$; it is in some set in $\mathcal{C}$, so $\mathcal{C}$ is an open cover of $S$

Does $\mathcal{C}$ contain a finite subcover of $S$

Fix an $n\in \mathbb{N}$. Only a single element of $\mathcal{C}$ contains $1/n$ (specifically, the one centered around $1/n$). Therefore, no finite subcover exists, because $\mathbb{N}$ is unbounded and has infinitely many elements.

Question 171

Show that every finite subset of $\mathbb{R}$ is compact in $\mathbb{R}$.

Fix a finite subset of $S⊊\mathbb{R}$. Fix an open cover $C$ for $S$. For each $s\in S$, there is at least one $\mathcal{O}\in C$ such that $s\in \mathcal{O}$. Choose one and denote this $O_s$. If we take the subcover $C^{\prime }\subseteq C$ defined by $C^{\prime }:=\{O_s\mid s\in S\}$, it is finite (since $S$ is finite) and a cover (since each $s\in S$ is covered by at least one ball, namely $O_s$). Therefore, a finite subcover must exist for every open cover $C$. Therefore, $S$ is compact.

Question 172

Show that $\mathbb{N}$ is not compact in $\mathbb{R}$

Consider the open cover $\mathcal{C}=\{B_1(n)\mid n\in \mathbb{N}\}$

Question 173

Let $A$ be an unbounded subset of an inner product space $V$. By considering the open cover $C=\{B_{\ell }(\vec{0})\mid \ell \in \mathbb{N}\}$, show that $A$ cannot be compact. Conclude that every compact subset of $V$ must be bounded.

Fix any finite subset of $C^{\prime }\subseteq C$. Take the max $\ell$ of all the balls. . Because it’s unbounded, there exists an element $a\in A$ that is outside of $B_{\ell }(\vec{0})$ because $A$ is unbounded. Therefore, $C^{\prime }$ is not a finite subcover. Therefore, no finite subcover exists.

By the contrapositive, if $A$ is compact, it is bounded.

Question 174

Suppose $D$ is a subset of an inner product space $V$ that is not closed. Suppose $\vec{v}\in \bar{D}\setminus D$. By considering the open cover $C=\{V\setminus {\bar{B}}_{1/n}(\vec{v})\mid n\in \mathbb{N}\}$ of D, show that $D$ can not be compact. Conclude that every compact subset of $V$ must be closed.

Fix $O_n\in C$. Notice that the complement of $O_n$ is closed (the closure of $B_{1/n}(\vec{v})$), so $O_n$ is open (and by the Archimedean Property, for all elements in $D$ there is an $O_n$ that contains it).

Fix any finite subset $C^{\prime }⊊C$. Choose the $O_n\in C^{\prime }$ corresponding to the largest $n\in \mathbb{N}$. Because the natural numbers are unbounded in the real numbers, there exists a real number $r\in \mathbb{R}$ such that $r<1/n$. This means that all elements in $D\cap B_r(\vec{v})$ are not in any open set $O_n$. Therefore, $C^{\prime }$ is not a finite subcover. This means that all finite subsets of $C$ are not finite subcovers, so no finite subcovers exists.

By the contrapositive, if D is compact, then D is closed.

Question 175

Show that if $K$ is a compact subset of an Inner Product Space, then $K$ is closed and bounded

Just use 173 and 174

Question 176

Let $A$ be an unbounded subset of an Inner Product Space $V$. For each $n\in \mathbb{N}$, choose $\overrightarrow{a_n}\in A\setminus B_n(\vec{0})$. Show that $(\overrightarrow{a_n})$ does not have a convergent subsequence and so $A$ cannot be sequentially compact. Conclude that every sequentially compact subset of $V$ must be bounded.

Fix an unbounded subset $A\subseteq V$. Let $E_n=A\setminus B_n(\vec{0})$. $E_n$ is nonempty because it is unbounded: for any $n\in \mathbb{N}$, there exists an element ${\vec{a}}_n\in A$ such that $\Vert {\vec{a}}_n\Vert \ge n$, which means that ${\vec{a}}_n\notin B_n(\vec{0})$. This means that there exists an ${\vec{a}}_n\in E_n$.

Let $({\vec{a}}_n)$ be a sequence of elements, where ${\vec{a}}_n\in E_n$. Fix any subsequence $(a_{n_k})$ of $({\vec{a}}_n)$. Since $n_k>n$, for each $M$ there exists at least one element (actually, an infinite number of elements) such that $\Vert a_{n_k}\Vert \ge M$ because $n_k\ge M$. Since The subsequence is unbounded, it does not converge. Therefore, no convergent subsequences exist, so $A$ cannot be sequentially compact.

By the contrapositive, if a subset $A\subseteq V$ is sequentially compact, it must be bounded.

Show that every bounded and closed interval in $\mathbb{R}$ is sequentially compact.

Fix any bounded and closed interval, $I\subseteq \mathbb{R}$. Fix any sequence $(i_n)$ on $I$. Because it is bounded, by the Bolzano-Weierstrass Theorem, it contains a convergent subsequence $(i_{n_k})\to i$. Since $I$ is closed, all convergent sequences must converge to an element in $I$, so $i\in I$. Since $(i_n)$ was arbitrary, all sequences in $I$ contain a convergent subsequence that converges to something in $I$. Therefore, $I$ is sequentially compact.

Question 177

Recall that in Prompt 164 you showed that a subset of a finite dimensional Inner Product Space is closed if and only if every Cauchy Sequence in the subset converges to an element of the subset. Use this result to show that a sequentially compact subset of an Inner Product Space is closed.

Let $(V,⟨,⟩)$ be an inner product space. Let $A\subseteq V$ be a sequentially compact subset of $V$.

Fix a Cauchy Sequence $(a_n)$ in $A$. Because $A$ is sequentially compact, it contains a convergent subsequence $(a_{n_k})\to a$, where $a\in A$. Because $(a_n)$ is Cauchy, it also converges; therefore, it must also converge to $a$. So it converges to an element in $A$. Therefore, $A$ is sequentially closed, so $A$ is also closed.

Question 178: Heine-Borel Theorem

A Subset $C$ of a finite dimensional inner product space $(V,⟨,⟩)$ is sequentially compact if and only if it is both closed and bounded.

$⟹$ Suppose $C$ is sequentially compact. By 177, $C$ must be closed. By 176, $C$ must be bounded. Therefore, $C$ is closed and bounded.

$⟸$ Suppose $C$ is closed and bounded. By 176, $C$ must be sequentially compact.

Question 179

Show that every nonempty sequentially compact subset of $\mathbb{R}$ has both a maximum and a minimum.

Fix $C⊊R$ such that $C$ is sequentially compact. Therefore, $C$ is closed an bounded.

Because $C$ is bounded (and $\mathbb{R}$ is complete), $\sup C$ and $\inf C$ exist. And because $C$ is closed, if $\sup C$ and $\inf C$ exist, they are both respectively in $C$. Therefore, since $\sup C\in C$, $C$ has a maximum (and conversely, since $\inf C\in C$, $C$ has a minimum).

Worksheet 14 - Connected Sets

Question 182

Suppose $(V,⟨,⟩)$ is a finite dimensional Inner Product Space. Let $A$ and $B$ be nonempty subsets of $V$ such that $V=A\cup B$. We will construct Cauchy Sequences $({\vec{a}}_n)$ in $A$ and ${\vec{b}}_n$ in $B$ that both converge to the same element of $V$.

Since $A$ and $B$ are nonempty, we can choose ${\vec{a}}_1\in A$ and ${\vec{b}}_1\in B$. Let ${\vec{c}}_1=({\vec{a}}_1+{\vec{b}}_1)/2\in V$. Since ${\vec{c}}_1\in V=A\cup B$, we know that that ${\vec{c}}_1\in A$ or ${\vec{c}}_1\in B$. If ${\vec{c}}_1$ belongs to $A$, then define ${\vec{a}}_2={\vec{c}}_1$ and ${\vec{b}}_2={\vec{b}}_1$. Otherwise, set ${\vec{a}}_2={\vec{a}}_1$ and ${\vec{b}}_2={\vec{c}}_1$. How would we define ${\vec{c}}_2$? ${\vec{a}}_3$?…

Repeat the same process, but generalize for an arbitrary ${\vec{a}}_i$ and ${\vec{b}}_i$.

Show that both $({\vec{a}}_n)$ and $({\vec{b}}_n)$ are Cauchy.

You can prove this more rigorously, but essentially the gap between $a_i$ and $a_{i+1}$ is at least one half of the distance between $a_i$ and $b_i$. This means that, inductively, the distance is less than $1/2^i\cdot |b_1-a_1|$. Since the distance between consecutive elements is decreasing geometrically, we have already proved that it must be Cauchy (using the geometric series formula)

Show that there is a vector $\vec{v}\in V$ such that ${\vec{a}}_n\to \vec{v}$ and ${\vec{b}}_n\to \vec{v}$.

Because $(a_n)$ and $(b_n)$ are Cauchy, they are convergent. Denote $\lim a_n=v$ and $\lim b_n=\ell$.

Consider the sequence $(d_n)=(b_n-a_n)$. Fix any $\epsilon >0$. There exists an $N\in \mathbb{N}$ such that $\forall n\in {\mathbb{N}}_{>N}$, $b_n-a_n=|b_1-a_1|\cdot \frac{1}{2^{n-1}}<\epsilon$. Therefore, $d_n\to 0$.

Both $(a_n)$ and $(b_n)$ are monotonic and bounded between by $a_1,v$ and $b_1,\ell$ respectively. Since the sequences get arbitrarily close to each other and they converge, their limits must be the same.

Conclude that every finite dimensional Inner Product Space is connected.

By the above construction, we can see that either $\bar{A}\cap B$ is nonempty (a sequence in $A$ converges to something in $B$) or $\bar{B}\cap A$ is nonempty by the same reasoning.

Question 183

The above problem suggests a general approach for determining if a subset of a finite dimensional Inner Product Space is connected. Suppose $(V,⟨,⟩)$ is a finite dimensional inner product space. Show that a subset $E$ of $V$ is connected if and only if for all nonempty disjoint subsets $A,B\subseteq V$ for which $E=A\cup B$ there exists a Cauchy sequence contained in one of $A$ or $B$ that converges to an element of the other.

$⟹$ Suppose $E$ is connected.

Fix any disjoint subsets $A\cup B$ such that $E=A\cup B$. Without a loss of generality $\bar{A}\cap B\ne \varnothing$. Fix $\ell \in \bar{A}\cap B$ because it is nonempty. Since $\ell \in \bar{A}$, there exists a Cauchy sequence in $A$ that converges to $\ell$. $\ell \in B$ as well, so this sequence converges to an element of $B$.

$⟸$ Suppose for all nonempty disjoint subsets $A,B\subseteq V$ for which $E=A\cup B$ there exists a Cauchy sequence contained in one of $A$ or $B$ that converges to an element of the other.

Fix $A,B$ such that $A\bigsqcup B=E$. Without a loss of generality, say there exists $(a_n)$ in $A$ that converges to an $\ell \in B$. That means that $\ell \in \bar{A}$ because the Closure is equal to the Sequential Closure in finite dimensional inner product spaces. $\ell \in \bar{A}\wedge \ell \in B⟹\ell \in \bar{A}\cap B$.

This means that $E$ is not disconnected, and therefore must be connected.

Question 185, 186, 187

Show that for every subset $D$ of $\mathbb{R}$, $D$ is connected if and only if $D$ is an interval.

$⟹$ Let us prove the contrapositive: Suppose $D$ is not an interval.

There exists $x,y,z\in R$ such that $x<y<z$; $x,z\in D$; and $y\notin D$.

Let $X=D\cap (-\infty ,y)$ and let $Y=D\cap (y,\infty )$. Because $y\notin D$, we can easily see that $X\bigsqcup Y=D$.

$\bar{X}=\overline{D\cap (-\infty ,y)}\subseteq \overline{D}\cap \overline{(-\infty ,y)}=\bar{D}\cap [-\infty ,y]$. $\bar{X}\cap Y\subseteq D\cap \bar{D}\cap [-\infty ,y]\cap (y,infty)=\varnothing$, so $\bar{X}\cap Y=\varnothing$.

Following similar logic, the opposite is also true. Therefore, $D$ is not connected.

$⟸$ Let us prove via contradiction: Suppose that $D$ is not connected.

There exists two nonempty disjoint sets $X,Y\subseteq V$ such that $X\bigsqcup Y=D$ and $\bar{X}\cap Y=\varnothing =X\cap \bar{Y}$. Fix $x\in X$ and $y\in Y$. Without a loss of generality, $x<y$.

Consider the set $G=\{a\in X\mid a\le y\}$. Let $\alpha =\sup G$. We can see that $x\le \alpha \le y$. By the equality of the closure and sequential closure, $\alpha \in \bar{X}$. Now, because $\bar{X}\cap Y=\varnothing$, $\alpha \notin Y⟹\alpha <y$. Since $X\bigsqcup Y=D$ and $\alpha \notin Y$, $\alpha \in X$ or $D$ is not an interval. Let us suppose $\alpha \in X$.

Now consider the set $Z=\{b\in Y\mid b\ge \alpha \}$, and let $\beta =\inf Z$. Similarly, we can see that $\alpha \le \beta \le y$. Since $\alpha \in X$ and $\beta \in \bar{Y}$, and $X\cap \bar{Y}=\varnothing$, $\alpha \ne \beta ⟹\alpha <\beta$,

Since the reals are dense in the reals, there exists a number $c$ in $(\alpha ,\beta )$ such that, but $c\notin X$ (because $c\le y$ but $c\ne \sup G$) and $c\notin Y$ (because $c\ge \alpha$ but $c\ne \inf Z$). This means that $c\notin X\cup Y=D$, so $D$ is not an interval.

Worksheet 15 - Limits

Question 189

Show that 3 is a limit point of both $\mathbb{R}$ and $\mathbb{R}\setminus \{3\}$, but 3 is not a limit point of $\mathbb{N}$

$\mathbb{R}$ $\forall \delta \in {\mathbb{R}}_{>0}\exists x\in (3-\delta ,3)\cap \mathbb{R}$ (which already doesn’t include 3) because the reals are dense in the reals

$\mathbb{N}$

Consider $\delta =1/67$. There are no natural numbers that are less than 1 away from 3, so there exists no non-3 elements

Question 190

Suppose $(V,⟨,⟩)$ is a finite dimensional Inner Product Space and $A\subseteq V$. Show that $\vec{v}$ is a limit point of $A$ if and only if there is a sequence in $A\setminus \{\vec{v}\}$ that converges to $\vec{v}$.

$⟹$ Suppose that $\vec{v}$ is a limit point of $A$.

Consider the sequence $(a_n)$ where $a_n$ is an arbitrary element in $B_{1/n}(\vec{v})\setminus \{\vec{v}\}$. This clearly converges to $\vec{v}$.

$⟸$ Suppose that there is a sequence $(a_n)$ in $A\setminus \{\vec{v}\}$ that converges to $\vec{v}$

Fix $\delta >0$. Because $(a_n)$ is a convergent sequence, there exists an $N\in \mathbb{N}$ such that $\forall n\in {\mathbb{N}}_{>N}$, $\Vert a_n-\vec{v}\Vert <\delta$. That means that $B_{\delta }(\vec{v})$ is nonempty.

Question 191

Suppose $(V,⟨,⟩)$ is a finite dimensional Inner Product Space and $B\subseteq V$. Is every point in $B$ a limit point of $B$? Is every limit point of $B$ an element of $B$?

Not every point of $B$ is a limit point of $B$. Fix $v\in V$. Consider $B=\{v\}$. $v\in B$ but $v$ is not a limit point of $B$.

Not every limit point of $B$ is an element of $B$: consider $\mathbb{R}\setminus \{3\}$. 3 is a limit point of this, but 3 is not in the subset.

Question 192

Suppose $(V,⟨,⟩)$ is a finite dimensional Inner Product Space and $C\subseteq V$. Show that $C$ is closed if and only if every limit point of $C$ belongs to $C$.

$⟹$ Suppose $C$ is closed.

That means $C$ is sequentially closed. That means that every convergent sequence in $C$ converges to an element in $C$.

Fix a limit point $v$ in $C$: by Question 190, there is a sequence in $C\setminus \{v\}$ that converges to $v$. Since $C$ is closed, that means $v$ must be in $C$.

$⟸$ Suppose every limit point of $C$ belongs to $C$.

Fix a convergent sequence $(c_n)\to v$ in $C$. This means that $v$ is a limit point of $C$ (by Question 190). That means that $v$ is in $C$. Therefore, every Cauchy sequence converges to something in $C$, which means $C$ is sequentially closed $⟺$ C is closed.

Question 193

With the notation of the definition of the limit, what does it mean to say ${\lim }_{\vec{x}\to \vec{v}}f(\vec{x})\ne \vec{w}$.

It means that there is a ball around $\vec{w}$ which $f(\vec{x})$ never reaches.

Question 194

Suppose $(V,⟨,{⟩}_V)$ and $(W,⟨,{⟩}_W)$ are finite dimensional inner product spaces. Fix a linear transformation $T:V\to W$ and an Orthonormal Basis $\{{\vec{e}}_i:1\le i\le n=\dim (V)\}$ for $V$. Set $c=\sqrt{n}\cdot \max \{\Vert T({\vec{e}}_1\Vert :1\le i\le n\}$.

Show that $\Vert T(\vec{v}){\Vert }_W\le c\cdot \Vert \vec{v}{\Vert }_V$ for all $\vec{v}$ in $V$.

INCOMPLETE

Fix $v\in V$.

$v={\sum }_i⟨v,{\vec{e}}_i⟩{\vec{e}}_i$ in orthonormal coordinates. That means that

> $$ \begin{align*} \lVert T(\vec v)\rVert_W &= \lVert \langle v, \vec e_1 \rangle T(\vec e_1) + \cdots + \langle v, \vec e_n \rangle T(\vec e_n)\rVert_W\\ &\leq |\langle v, \vec e_1 \rangle| \lVert T(\vec e_1)\rVert + \cdots + |\langle v, \vec e_n \rangle| \lVert T(\vec e_n)\rVert_W\\ &\leq |\langle v, \vec e_1 \rangle| \frac{c}{\sqrt n} + \cdots + |\langle v, \vec e_n \rangle| \frac{c}{\sqrt{n}}\\ &= \frac{c}{\sqrt n}\left(|\langle v, \vec e_1 \rangle| + \cdots + |\langle v, \vec e_n\rangle|\right)\\ \end{align*}

Conclude that ${\lim }_{\vec{x}\to \vec{0}}T(\vec{x})=\vec{0}$

Question 195, 196, 197

Show: if ${\lim }_{\vec{x}\to \vec{v}}f(\vec{x})=\vec{w}$, then for all sequences $({\vec{a}}_n)$ in $A\setminus \{\vec{v}\}$ that converge to $\vec{v}$ we have that the sequence $(f({\vec{a}}_n))$ converges to $\vec{w}$.

Fix a sequence, epsilon delta to prove that it converges (they are equivalent)

**Show: if ${\lim }_{\vec{x}\to \vec{v}}f(\vec{x})\ne \vec{w}$, then there is a sequence $({\vec{a}}_n)$ in $A\setminus \{\vec{v}\}$ that converge to $\vec{v}$ yet $(f({\vec{a}}_n))$ does not converge to $\vec{w}$.

There is an epsilon for which no matter how small your delta gets, there is no $N$ in your sequence for which all subsequent elements also map to an element within epsilon of the target.

Important

Conclude: ${\lim }_{\vec{x}\to \vec{v}}f(\vec{x})=\vec{w}$, if and only if for all sequences $({\vec{a}}_n)$ in $A\setminus \{\vec{v}\}$ that converge to $\vec{v}$ we have that the sequence $(f({\vec{a}}_n))$ converges to $\vec{w}$.