Homework 1

Solution.

1. Given $x, y \in \R^n$,

   $$\begin{aligned} \p{d_{\ell^1}\p{x, y}}^2 = \p{\sum_{i=1}^n \abs{x_i - y_i}}^2 &= \sum_{i=1}^n \abs{x_i - y_i}^2 + \underbrace{\sum_{i \neq j} \abs{x_i - y_i} \abs{x_j - y_j}}_{\geq 0} \\ &\geq \sum_{i=1}^n \abs{x_i - y_i}^2 \\ &= \p{d_{\ell^2}\p{x, y}}^2. \end{aligned}$$

   Thus, because metrics are non-negative,

   $$d_{\ell^2}\p{x, y} \leq d_{\ell^1}\p{x, y}.$$

   In particular, if $\lim_{k\to\infty} d_{\ell^1}\p{x^{\p{k}}, x^{\p{0}}} = 0$, then by the squeeze theorem,

   $$d_{\ell^2}\p{x^{\p{k}}, x^{\p{0}}} \leq d_{\ell^1}\p{x^{\p{k}}, x^{\p{0}}} \xrightarrow{k\to\infty} 0.$$

2. Given $x, y \in \R^n$,

   $$\begin{aligned} d_{\ell^1}\p{x, y} = \sum_{i=1}^n \abs{x_i - y_i} \leq \sum_{i=1}^n \max_{j=1,\ldots,n} \abs{x_j - y_j} &= n d_{\ell^\infty}\p{x, y}. \end{aligned}$$

   The claim follows from the squeeze theorem like in part (1).

3. Given $x, y \in \R^n$, for any $1 \leq i \leq n$, we have

   $$\abs{x_i - y_i} \leq \sum_{j=1}^n \abs{x_j - y_j} = d_{\ell^1}\p{x, y}.$$

   Thus, because this inequality holds uniformly for all $1 \leq i \leq n$,

   $$d_{\ell^\infty}\p{x, y} = \max_{i=1,\ldots,n} \abs{x_i - y_i} \leq d_{\ell^1}\p{x, y}.$$

   The claim follows from the squeeze theorem like in part (1).

Solution.

We will show (i) $\iff$ (ii), and then (i) $\implies$ (iii) $\implies$ (i).

(i) $\iff$ (ii):

Recall that $X = \Int\p{E} \cup \partial E \cup \Ext\p{E}$ and that this is a disjoint union. Thus, it suffices to show that $x_0 \notin \Ext\p{E}$. Notice that

In other words,

(i) $\implies$ (iii):

By definition,

Thus, for each $n \in \N$, we have $B\p{x_0, \frac{1}{n}} \cap E \neq \emptyset$. Let $x_n \in B\p{x_0, \frac{1}{n}} \cap E$ (this is where we use (countable) choice, but this detail isn't that important to us in this class.) Thus, we get a sequence $\p{x_n}_n \subseteq E$ such that

so $\p{x_n}_n$ converges to $x_0$ by the squeeze theorem.

(iii) $\implies$ (i):

Let $r > 0$. Since $\p{x_n}_n$ converges to $x_0$, there exists $n \in \N$ such that

Since $x_n \in E$ by assumption, this means $x_n \in B\p{x_0, r} \cap E$, i.e., this is a non-empty set.

Solution.

While Solution 1 is much easier than Solution 2, I highly recommend that you understand Solution 2 as well since it requires more understanding of the metric space definitions--it's very good practice.

Throughout this problem, I'll be using Proposition 1.2.15 (the long one) many times without mentioning it explicitly.

Solution 1 (using part (i))

"$\implies$"

Assume $E$ is relatively closed with respect to $Y$. Then $Y \setminus E$ is open in $Y$, so by part (i), there exists $V \subseteq X$ which is open in $X$ such that $Y \setminus E = V \cap Y$. Notice that

Since $V$ was open in $X$, it follows that $K \coloneqq X \setminus V$ is closed in $X$.

"$\impliedby$"

Assume that $E = K \cap Y$ for some set $K \subseteq X$ which is closed in $X$. Then by a similar computation as above,

Since $K$ is closed in $X$, $V \coloneqq X \setminus K$ is open in $X$. Thus, part (i), $Y \setminus E$ is open in $Y$, and hence $E$ is closed in $Y$.

Solution 2 (from the definitions)

"$\implies$"

Assume that $E$ is closed in $Y$. We let $K \coloneqq \cl{E}$, where the closure is taken with respect to $X$. We immediately have $E \subseteq \cl{E} \cap Y$, so it remains to check the reverse inclusion.

Let $x_0 \in \cl{E} \cap Y$. We will show that $x_0$ is an adherent point of $E$ in $\p{Y, d_{Y \times Y}}$. Let $r > 0$. Notice that because $x_0 \in Y$,

Thus,

Since $r$ was arbitrary, this means $x_0$ is an adherent point of $E$ in $Y$. But $E$ is closed in $Y$, so $x_0 \in E$.

"$\impliedby$"

Assume that $E = K \cap Y$, where $K$ is a closed set in $X$. We will show that $E$ contains all of its adherent points in $Y$.

Let $y_0 \in Y$ be an adherent point of $E$ in $\p{Y, d_{Y \times Y}}$. Since $E = K \cap Y$, we only need to show that $y_0 \in K$. Moreover, $K$ is closed in $\p{X, d}$, so it further suffices to show that $y_0$ is an adherent point of $K$ in $\p{X, d}$. Let $r > 0$ and notice that

In particular,

which is non-empty since $y_0$ is an adherent point of $E$ in $\p{Y, d_{Y \times Y}}$. Thus, $y_0$ is an adherent point of $K$, which completes the proof.