Proofs from Mendelson 2.4.7

I’ve been working through Mendelson’s Introduction to Topology and I’d like to share a couple of proofs for problem 7 from Chapter 2, Section 4, Open balls and neighborhoods. I’ll also reflect on things I’ve learned from working on this problem.

Let $(X_{1}, d_{1})$, $(X_{2}, d_{2})$, $\dots$, $(X_{n}, d_{n})$ be metric spaces and convert $X = \prod_{i=1}^{n}{X_{i}}$ into a metric space $(X, d)$ in the standard manner. Prove that:

1. An open ball in $(X, d)$ is the product of open balls from $X_{1}, X_{2}, \dots, X_{n}$ respectively.

2. Let $a_i \in X_i$, $i = 1, 2, \dots, n$, and let $\mathcal{B}_{a_i}$ be a basis for the neighborhood system at $a_i$. Let $\mathcal{B}_a$ be the collection of all sets of the form $B_1 \times B_2 \times \dots \times B_n$, $B_i \in \mathcal{B}_{a_i}$. Prove that $\mathcal{B}_a$ is a basis for the neighborhood system at $a = (a_1, a_2, \dots, a_n) \in X$.

3. Let $p_i:X \rightarrow X_i$, $i = 1, 2, \dots, n$ be the projection that maps $p_i(a) = a_i$. Prove that each $p_i$ is continuous.

4. Let $Y$ be a metric space and $f: Y \rightarrow X$ a function. Prove that $f$ is continuous if and only if each of the $n$ functions $p_if$ is continuous.

Proof for 7.1

Define the metric space $(X, d)$ where given $x = (x_1, x_2, \dots, x_n) \in X$ and $y = (y_1, y_2, \dots, y_n) \in X$, $d(x, y) = \max_{1 \leq i \leq n} d_i(x_i, y_i)$. Let $b \in B(a; \delta)$ where $a = (a_1, a_2, \dots, a_n) \in X$ and $\delta \gt 0$. Then for all $i = 1, 2, \dots, n$, $d_i(a_i, b_i) \lt \delta$. Hence $b_i \in B_i(a_i; \delta)$ and by definition of the product of sets, $b \in \prod_{i=1}^{n} B_i(a_i; \delta)$. Since $b$ was arbitrary, it follows that $B(a; \delta) \subseteq \prod_{i=1}^{n} B_i(a_i; \delta)$. Conversely, let $b \in \prod_{i=1}^{n}{B_i(a_i; \delta)}$. Then for all $i = 1, 2, \dots, n$, $d_{i}(a_{i}, b_{i}) \lt \delta$, hence $\max_{1 \leq i \leq n}d_i(a_i, b_i) \lt \delta$. It follows that $b \in B(a; \delta)$. Since $b$ was arbitrary, we have $\prod_{i=1}^{n}{B_i(a_i; \delta)} \subseteq B(a; \delta)$, as required.

Reflection

It’s good to restate what you’ve proven, as the result here is stronger than what the problem’s prompt asks for. I would state it as follows: $B(a; \delta) = \prod_{i = 1}^{n} B_i(a_i; \delta)$, where $a = (a_1, a_2, \dots, a_n) \in X$ and $\delta \gt 0$.

Proof for 7.2

Let $S \in \mathcal{B}_a$. Then $S = (B_1 \times B_2 \times \dots \times B_n)$ such that $B_i \in \mathcal{B}_{a_i}$. Since $\mathcal{B}_{a_i}$ is a basis at $a_i$, $B_i$ is a neighborhood of $a_i$. Hence for some $\delta_i \gt 0$, $B_i(a_i; \delta_i) \subseteq B_i$. Take $\delta = \min\{\delta_1, \delta_2, \dots, \delta_n\}$. Then $B_i(a_i; \delta) \subseteq B_i$, hence $\prod_{i = 1}^{n} B_i(a_i; \delta) \subseteq \prod_{i = 1}^{n} B_i = S$. Then by 7.1, we have $B(a; \delta) = \prod_{i=1}^{n} B_i(a_i; \delta) \subseteq S$. Since $S$ was arbitrary, every element of $\mathcal{B}_a$ is a neighborhood of $a$.

Let $N_a$ be a neighborhood of $a$. Then there exists some $\delta \gt 0$ such that $B(a; \delta) \subseteq N_a$. By 7.1, $\prod_{i=1}^{n} B_i(a_i; \delta) = B(a; \delta) \subseteq N_a$. Since $\mathcal{B}_{a_i}$ is a basis at $a_i$, for $B_i(a_i; \delta)$ there exists $N_{a_i} \in \mathcal{B}_{a_i}$ such that $N_{a_i} \subseteq B_i(a_i; \delta)$. Take $S = \prod_{i=1}^{n} N_{a_i}$ where $N_{a_i} \in \mathcal{B}_{a_i}$ and $N_{a_i} \subseteq B_i(a_i; \delta)$. Then $S \subseteq \prod_{i=1}^{n} B_i(a_i; \delta) = B(a; \delta) \subseteq N_a$. Since $S$ is of the form $(B_1 \times B_2 \times \dots \times B_n)$ such that $B_i \in \mathcal{B}_{a_i}$, it follows that $S \in \mathcal{B}_a$.

QED.

Reflection

The types in this problem were dense. A minor type error initially caused a lot of friction for me and made it difficult to start the proof. I’m not sure there’s a great lesson other than “keep practicing.” This cognitive load of this kind of thing—at least in software engineering—can usually be offloaded to a computer, so it’s no surprise that I’d need to strengthen this muscle.

Proof for 7.3

Let $a \in X$ and $\epsilon \gt 0$ be given. Set $\delta = \epsilon$. Let $b \in X$ and suppose $d(a, b) \lt \delta$. Since $d(a, b) = \max_{1 \leq j \leq n} d_j(a_j, b_j)$ and $d_i(p_i(a), p_i(b)) = d_i(a_i, b_i) \leq d(a,b) \lt \delta = \epsilon$, we have $d_i(a_i, b_i) \lt \epsilon$. Hence $p_i$ is continuous at $a$. Since $a$ was arbitrary, $p_i$ is continuous on $X$.

Reflection

This one wasn’t bad. The only part requiring care is the need to juggle indices ($i$ vs $j$). Otherwise, we introduce ambiguity into the proof.

Proof for 7.4

Let $a \in (Y, d')$ and $\epsilon \gt 0$ be given.

$\rightarrow$ Suppose $f$ is continuous. By 7.3, $p_i$ is continuous for all $1 \leq i \leq n$. Since the composition of two continuous functions is again a continuous function, $p_if$ is continuous.

$\leftarrow$ Suppose $p_if$ is continuous for all $1 \leq i \leq n$. We must show that there exists some $\delta$ such that $d(f(x), f(a)) \lt \epsilon$ whenever $d'(x, a) \lt \delta$. Since $p_if$ is continuous for all $1 \leq i \leq n$, for each $i$ there exists $\delta_i$ such that if $d'(x, a) \lt \delta_i$, then $d_i(p_i(f(x)), p_i(f(a))) \lt \epsilon$. Take $\delta = \min\{\delta_1, \delta_2, \dots, \delta_n\}$ and suppose $d'(x, a) \lt \delta$. Since $d_i(p_i(f(x)), p_i(f(a))) = d_i(f(x)_i, f(a)_i)$ and $d(f(x), f(a)) = \max_{1 \leq i \leq n} d_i(f(x)_i, f(a)_i)$, it follows that $d(f(x), f(a)) \lt \epsilon$. Therefore $f$ is continuous at $a$. Since $a$ was arbitrary, $f$ is continuous.

This completes the proof.

Reflection

Originally, I messed this up. I honed in on the $\leftarrow$ direction and forgot the proof was if and only if (oops). It wasn’t until I was typing the proof up that I realized my error.

Secondly, I originally botched my choice of $\delta$. The name of the game with product spaces is coordinate-wise decomposition followed by global recomposition. Instead of taking $\delta = \min\{\delta_1, \delta_2, \dots, \delta_n\}$, I captured $\delta = \delta'$, where I erroneously assumed $\delta'$ was some value that satisfied continuity for all $p_if$. Technically such a value exists, since it’s the minimum. But being accidentally correct is not a good thing.