# Metrizability of product spaces, Horst Herrlich, Topology I

In Horst Herrlich's Topology I, page 117, Statement 4.4.12 the author goes on to prove that product space is metrizable if each component ($$\underline X_i$$) of the product space is metrizable and the corresp. index set ($$I$$) is at most countable.

Assuming first the case when $$I = N$$ is infinitely countable, he then goes on to define such a metric (here each $$d_n$$ belongs to corresp. $$X_n$$)

$$d((x_n),(y_n)) = max \ \{ \frac 1 n \ d_n(x_n, y_n) | n \in I \}$$

If $$x = (x_n)$$ is element of the product space and $$r > 0$$, then $$\exists m \in N, \forall n \ge m : \frac 1 n < r$$

He then goes on to state that it follows:

$$S(x,r) = \{ y \in {\displaystyle \prod_{I} X_n} \ | \ d(x,y) < r \ \} = \bigcap \{ p_n^{-1} [S(x_n, n \cdot r)] \ | \ n < m \} \ \ \ [1]$$

Here $$S(x_n, s) = \{ y \in X_n \ | \ d_n(x_n, y) < s \}$$

I really tried to understand but I can't see how [1] just 'follows'. Can someone help in understanding why the equality really holds?

EDIT. Forgot to mention that $$diam \ \underline X_i \le 1$$

• This might help.. Plus if the $X_i$ are non-trivial and Hausdorff then $\prod_i X_i$ is first countable ifff $I$ is countable.. Commented Oct 29, 2020 at 7:00

Suppose that $$y\in\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n; then $$d_n(x_n,y_n) for each $$n, so $$\frac1n d_n(x_n,y_n) for each $$n. If $$n\ge m$$, then $$\frac1n, so $$\frac1n d_n(x_n,y_n)\le\frac1n\cdot1. (You didn’t mention it, but at this point he is clearly assuming that the metrics $$d_n$$ are bounded by $$1$$.) Thus, $$\frac1n d_n(x_n,y_n) for all $$n\in\Bbb N$$, and therefore $$d(x,y)=\max\left\{\frac1n d_n(x_n,y_n):n\in\Bbb N\right\} so $$y\in S(x,r)$$. (The fact that this set has a maximum also depends on the assumption that the metrics $$d_n$$ have a common upper bound.) This shows that

$$\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n

Now suppose that $$y\in S(x,r)$$. Then

$$d(x,y)=\max\left\{\frac1n d_n(x_n,y_n):n\in\Bbb N\right\}

so $$\frac1n d_n(x_n,y_n) for each $$n\in\Bbb N$$, and therefore $$d_n(x_n,y_n) for each $$n\in\Bbb N$$. In particular, $$d_n(x_n,y_n) for each $$n, so $$y_n\in S(x_n,nr)$$ for each $$n, and therefore $$y\in\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n. Thus,

$$S(x,r)\subseteq\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n

so

$$\bigcap\left\{p_n^{-1}[S(x_n,nr)]:n