universal property of product metric, proof explanation

Question

Let $(Y,d)$ and $(X_1, d_1),\dotso, (X_m,d_m)$ be metric spaces. Let $X=X_1\times\dotso\times X_m$ and $d(u,v)=\sum_{i=1}^m d_j(u_j,v_j)$. Let $f:Y\to X$ a function. Then are equivalent:

I) $f$ is continuous

II) For every $j=1,\dotso, m$ is the function $f_j=pr_j\circ f: Y\to X_j$ continuous.

[Of course $pr_j$ notes here the projection on the $j$-th coordinate.]

I have a question to one detail of the proof, which I can not verify for some odd reason. I added my thought in brackets [...].

Well, the proof I)$\Rightarrow$ II) is trivial.

For II)$\Rightarrow$ I), let $f_1,\dotso, f_m$ be continuous.

Now take $W\subseteq X$ open, and show, that $f^{-1}(W)\subseteq Y$ is open. Since $W$ is open there is for $w\in W$ an $\epsilon >0$ such that $B_\epsilon^{d}(w)\subseteq W$. Now holds $pr_j(W)\supseteq B_\epsilon^{d_j}(w_j)$

[Indeed: Let $\lambda\in B_\epsilon^{d_j}(w_j)=\{\lambda\in X_j|d_j(\lambda,w_j)<\epsilon\}$. Since $B_\epsilon^{d}(w)\subseteq W$ we have, by construction of $d$, that $\sum_{i=1}^m d_j(u_j, w_j)<\epsilon$. In particular $d_j(u_j, w_j)<\epsilon$ and $pr_j(W)\supseteq B_\epsilon^{d_j}(w_j)$]

Hence $\color{red}{f^{-1}(W)\supseteq f_j^{-1}(B_\epsilon^{d_j}(w_j))}$

[And I fail to proof this, what really bugs me...]

The rest of the proof is clear.

For $f(y)=w$ it is $y\in f_j^{-1}(B_\epsilon^{d_j}(w_j))$, so

$y\in\underbrace{\bigcap_{i=1}^m f_j^{-1}(B_\epsilon^{d_j}(w_j))}_{\text{open}}\subseteq f^{-1}(W)$ so $f^{-1}(W)$ is open.

Can you help me out? Thanks in advance.

Answer 1

The implication doesn't work. Consider, for example, $$f : \Bbb{R} \to \Bbb{R} \times \Bbb{R} : x \mapsto (x, 2\cos(x)).$$ I'm equipping $\Bbb{R}$ with its usual metric, and according to the construction, we have $d$ is the metric derived from the $1$-norm.

Let $W = B^d_1(0, 2)$, and in particular, take $w = (0, 2)$ and $\varepsilon = 1$. I also want to consider $j = 2$. We have $w_2 = 2$ and $f_2 = 2\cos$.

Now, $B^{\Bbb{R}}_1(2) = (1, 3)$, the interval, not the point in $\Bbb{R}^2$. Then, $$f_2^{-1}(B_1^{\Bbb{R}}(2)) = f_2^{-1}(1, 3) = \cdots \cup \left(-2\pi - \frac{\pi}{3}, -2\pi + \frac{\pi}{3}\right) \cup \left(-\frac{\pi}{3}, \frac{\pi}{3}\right) \cup \left(2\pi - \frac{\pi}{3}, 2\pi + \frac{\pi}{3}\right) \cup \cdots$$ In particular, this set is unbounded.

On the other hand, $f^{-1}(B^d_1(0, 2))$ definitely is bounded. If $x \in f^{-1}(B^d_1(0, 2))$, then $$(x, 2\cos(x)) = f(x) \in B^d_1(0, 2) \implies |x - 0| + |2 - 2\cos(x)| < 1 \implies |x| < 1.$$ Thus, $f^{-1}(B^d_1(0, 2)) \subseteq B_1^{\Bbb{R}}(0)$, which makes it bounded by definition.

While the theorem is true, this proof is invalid.

Answer 2

Here is the proof on the implication II)$\Rightarrow$ I):

Let $f_1,\dotso, f_m$ be continuous.

We have to show, that $f$ is continuous. Let $W\subseteq X$ be open. We show, that $f^{-1}(W)$ is open.

Since $W$ is open, there is for $w\in W$ a $\epsilon >0$ such that $B_\epsilon^d(w)\subseteq X$.

Now holds: $\underbrace{B_{\epsilon/m}^{d_1}(w_1)\times\dotso\times B_{\epsilon/m}^{d_m}(w_m)}_{=: U}\subseteq B_\epsilon^d(w)$

Indeed:

Let $u\in U$. Then it is $d(w,u)=\sum_{i=1}^m d_i(u_i,w_i)<\sum_{i=1}^m \epsilon/m=\epsilon$.

Hence $u\in B_\epsilon^d(w)$.

Now we show, that $\underbrace{\bigcap_{i=1}^m f_i^{-1}(B_{\epsilon/m}^{d_i}(w_i))}_{\text{open}}\subseteq f^{-1}(W)$.

Indeed:

Let $y\in \bigcap_{i=1}^m f_i^{-1}(B_{\epsilon/m}^{d_i}(w_i)$. Then $y\in f_i^{-1}(B_{\epsilon/m}^{d_i}(w_i))$ for every $1\leq i\leq m$.

So $y\in\{y\in Y|f_i(y)\in B_{\epsilon/m}^{d_i}(w_i)\}$ for every $1\leq i\leq m$.

We have $f_i(y)=\operatorname{pr}_i(f(y))$ for every $1\leq i\leq m$, hence $f(y)\in U$.

We conclude $y\in f^{-1}(U)\subseteq f^{-1}(B_{\epsilon}^d(w))\subseteq f^{-1}(W)$.

As desired $f^{-1}(W)$ is open.

Does everything seem to be in order here?

Thanks in advance.