MATH 447: Real Variables - Homework #7
Jerich Lee
December 22, 2024
Problem.
[22.3] Prove that if $E$ is a connected subset of a metric space $(S, d)$, then its closure $E^-$ is also connected.
[22.3] Prove that if $E$ is a connected subset of a metric space $(S, d)$, then its closure $E^-$ is also connected.
Solution.
Proof.
To solve this problem, we will prove the contrapositive, i.e., if $E^{-}$ is disconnected, then $E$ is also disconnected. Because $E^{-}$ is disconnected, there exist $U_1, U_2$ such that:
- $E^{-}\subseteq U_1 \cup U_2$
- $(E^{-}\cap U_1)\cap (E^{-}\cap U_2)=\emptyset$
- $(E^{-}\cap U_1) \neq \emptyset, (E^{-1}\cap U_{2})\neq \emptyset$
- $E \subset E^{-} \subseteq U_1 \cup U_2\implies E \subseteq U_1 \cup U_2$
- $(E\cap U_1)\cap (E\cap U_2)=\emptyset$
- $(E\cap U_1)\neq \emptyset, (E\cap U_{2})\neq \emptyset$
∎
Problem.
Prove that an intersection of convex sets in $\mathbb{R}^n$ is convex.
Prove that an intersection of convex sets in $\mathbb{R}^n$ is convex.
Solution.
Proof.
Suppose we have sets $E,F$ that are both convex. We wish to find $E\cup{} F$ is convex. We know the following:
$$\begin{align}
\forall x, y \in E, 0<t<1 \implies tx+(1-t)y\in E\\[10pt]
\forall u, v \in F, 0<t<1 \implies tu+(1-t)v \in F
\end{align}$$
Choose $\forall{} a,b \in{} E\cup{} F$. Then,
$$\begin{align}
ta+(1-t)b\in E\\[10pt]
ta+(1-t)b \in F
\end{align}$$
By the definition of an intersection of a set, we know that $x\in{} P$ iff $x\in{} E_{\alpha{}}$ for every $\alpha{} \in{} A$. Let $a \in{} E$, and $b \in{} F$, and $E = E_1$, $F = E_2$, $E_1, E_2 \in{} E_{\alpha{}}$. $P=E\cap{} F$. By the above, we know that $ta+(1-t)b\in{} E\cap{} F$. Therefore, the intersection of convex sets $E$ and $F$ is also convex.
∎
Problem.
On the metric space $\mathbb{R}^n$ (with the Euclidean metric $d$), denote by $P_i$ $(1 \leq{} i \leq{} n)$ the projection onto the $i$-th coordinate. Specifically, $P_i : \mathbb{R}^n \to{} \mathbb{R}$ takes $\vec{x} = (x_1, \dots{}, x_n)$ to $x_i$. Prove that $P_i$ is Lipschitz.
On the metric space $\mathbb{R}^n$ (with the Euclidean metric $d$), denote by $P_i$ $(1 \leq{} i \leq{} n)$ the projection onto the $i$-th coordinate. Specifically, $P_i : \mathbb{R}^n \to{} \mathbb{R}$ takes $\vec{x} = (x_1, \dots{}, x_n)$ to $x_i$. Prove that $P_i$ is Lipschitz.
Solution.
Proof.
The idea is to choose the largest value of $P_i(s)$, where $s\in{} \mathbb{R}^{n}$, and $P_i$ is the projection function of $\mathbb{R}^{n}$, i.e., choose the projection with the largest magnitude. By the triangle inequality, we know that the difference between the largest projection of values $s, t \in{} \mathbb{R}^{n}$ is always less than or equal to their respective Euclidean distances. As a result, we can always bound the differences between the difference of the outputs by the difference of the inputs by some constant value $k$, thereby making the function $P_i$ Lipschitz. For elements $ s, t \in{} \mathbb{R}^{n} $, choose $k=1$. Then,
$$\begin{align}
\left\vert s_i - t_i \right\vert < (1)d(s,t) \\[10pt]
=\left( \sum_{i=1}^{n} \left\vert s_i - t_i \right\vert^{2} \right)^{\frac{1}{2}}
\end{align}$$
Therefore, $P_i$ is Lipschitz.
∎
Problem.
Denote by $\ell_1{}$ the set of all absolutely convergent series: the elements of $\ell_1{}$ are sequences $a = (a_i)_{i=1}^{\infty{}}$ with $\sum_{i=1}^{\infty{}} |a_i| < \infty{}$. For $a = (a_i)_{i=1}^{\infty{}}$ and $b = (b_i)_{i=1}^{\infty{}}$, define \[ d(a,b) = \sum_{i=1}^{\infty{}} |a_i - b_i|. \]
Denote by $\ell_1{}$ the set of all absolutely convergent series: the elements of $\ell_1{}$ are sequences $a = (a_i)_{i=1}^{\infty{}}$ with $\sum_{i=1}^{\infty{}} |a_i| < \infty{}$. For $a = (a_i)_{i=1}^{\infty{}}$ and $b = (b_i)_{i=1}^{\infty{}}$, define \[ d(a,b) = \sum_{i=1}^{\infty{}} |a_i - b_i|. \]
- Prove that $d$ is a metric.
- Prove that the function $f : \ell_1 \to \mathbb{R}: (a_i) \mapsto \sum_{i=1}^{\infty} a_i$ is Lipschitz.
- Determine whether the function $g: \mathbb{R} \to \ell_1$, taking $t \in \mathbb{R}$ to the sequence $\left( \frac{t^2}{2^i} \right)_{i=1}^{\infty}$, is uniformly continuous.
Solution.
- To show that $d$ is a metric, we need to show that $d$ satisfies the three criteria of a metric space:
- Non-degeneracy:
$$\begin{align}
d(x,y)=0 \iff x=y \\[10pt]
d(a,b) = \sum_{i=1}^{\infty} \left\vert a_i - b_{i} \right\vert
\end{align}$$
Proof. $\implies{}:$ $$\begin{align} \sum_{i=1}^{\infty} \left\vert a_{i}-b_{i} \right\vert = \sum_{i=1}^{\infty} \left\vert 0 \right\vert =0 \end{align}$$ $\impliedby{}:$ $$\begin{align} \sum_{i=1}^{\infty} \left\vert a_{i}-b_{i} \right\vert =0 \implies \left\vert a_{i}-b_{i} \right\vert =0 \\[10pt] a_{i}=b_{i}=0 \end{align}$$∎
- Symmetry:
$$\begin{align}
\forall x,y \in \ell_1, d(x,y)=d(y,x)
\end{align}$$
Proof. $$\begin{align} \left\vert a_{i}-b_{i} \right\vert =\left\vert b_{i}-a_{i} \right\vert \end{align}$$∎
- Triangle Inequality:
$$\begin{align}
\forall x,y,z \in \ell_1, d(x,y)+d(y,z)\geq d(x,z)
\end{align}$$
Proof. $$\begin{align} \sum_{i=1}^{\infty} \left\vert a_i -b_{i} \right\vert +\sum_{i=1}^{\infty} \left\vert b_{i}-c_{i} \right\vert \geq \sum_{i=1}^{\infty} \left\vert a_i -c_{i} \right\vert \\[10pt] \sum_{i=1}^{\infty} a_i = A, \sum_{i=1}^{\infty} b_i = B, \sum_{i=1}^{\infty} c_i = C \label{hw-5}\\[10pt] \left\vert A-B \right\vert +\left\vert B-C \right\vert \geq \left\vert A -C \right\vert \\[10pt] \left\vert A-B \right\vert +\left\vert B-C \right\vert \geq \left\vert (A-B)+(B-C) \right\vert \label{ti} \end{align}$$ Where \autoref{hw-5} is proven in Homework 5 Problem 3.1, and \autoref{ti} is proven by the triangle inequality.∎
- Non-degeneracy:
$$\begin{align}
d(x,y)=0 \iff x=y \\[10pt]
d(a,b) = \sum_{i=1}^{\infty} \left\vert a_i - b_{i} \right\vert
\end{align}$$
- $$\begin{align}
f: \ell_1 \to \mathbb{R} : (a_{i}) \mapsto \sum_{i=1}^{\infty} a_{i}\\[10pt]
s,t \in \ell_1, d^{*}(f(s), f(t))\leq kd(s,t)
\end{align}$$
Proof. We wish to find: $$\begin{align} \sum_{i=1}^{\infty} s_{i} - \sum_{i=1}^{\infty} t_{i} \leq k \sum_{i=1}^{\infty} \left\vert a_{i}-b_{i} \right\vert \\[10pt] \left\vert S \right\vert -\left\vert T \right\vert \leq k\left\vert S-T \right\vert \end{align}$$ Choose $k=1$. Then: $$\begin{align} \left\vert \left( S-T \right)+T \right\vert \leq \left\vert S-T \right\vert +\left\vert T \right\vert \\[10pt] \left\vert S \right\vert \leq \left\vert S-T \right\vert +\left\vert T \right\vert \end{align}$$ By the triangle inequality established in Problem 4.1, at $k = 1$, $f$ is Lipschitz.∎
- To prove that $g: \mathbb{R} \to \ell_1$ is uniformly continuous, we will use the definition, i.e., $$\begin{align} \forall \varepsilon>0, \exists \delta \ s.t. \ x,y\in S \text{ and } \left\vert x-y \right\vert <\delta \implies \left\vert f(x)-f(y) \right\vert <\varepsilon \end{align}$$ We will determine $\delta$ by the following discussion: $$\begin{align} f(x)-f(y) = \left(\frac{x^{2}}{2^{n}}- \frac{y^{2}}{2^{n}} \right) =\\[10pt] \left( \frac{x^{2}-y^{2}}{2^{n}} \right) = \left( \frac{(x+y)(x-y)}{2^{n}} \right) \\[10pt] \left( \frac{(x+y)(x-y)}{2^{n}} \right) <\varepsilon \\[10pt] (x-y) < \frac{\varepsilon \cdot 2^{n}}{(x+y)} \\[10pt] \end{align}$$ Choose $\delta=\frac{\varepsilon \cdot 2^{n}}{(x+y)}$. Then, $\left\vert x-y \right\vert <\delta $ implies: $$\begin{align} \left\vert f(x)-f(y) \right\vert = \left\vert \left(\frac{x^{2}}{2^{n}}- \frac{y^{2}}{2^{n}} \right) \right\vert =\\[10pt] \left( \frac{x^{2}-y^{2}}{2^{n}} \right) = \left( \frac{(x+y)(x-y)}{2^{n}} \right) < \left( \frac{(x+y)\delta}{2^{n}} \right) \\[10pt] \frac{(x+y)\cdot \varepsilon \cdot 2^{n}}{(x+y)\cdot 2^{n}} = \varepsilon \\[10pt] \end{align}$$ Therefore, $$\begin{align} \left\vert x-y \right\vert < \delta \implies \left\vert f(x)-f(y) \right\vert < \varepsilon \end{align}$$ Then $g$ is uniformly continuous.