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We know (Show that the countable product of metric spaces is metrizable) that the following is true:

Given a countable collection of metric spaces $\{(X_n,\rho_n)\}_{n=1}^{\infty}$. Form the Cartesian Product of these sets $X=\displaystyle\prod_{n=1}^{\infty}X_n$, and define $\rho:X\times X\rightarrow\mathbb{R}$ by

$$\rho(x,y)=\displaystyle\sum_{n=1}^{\infty}\frac{\rho_n(x_n,y_n)}{2^n[1+\rho_n(x_n,y_n)]}.$$

Show that $\rho$ is a metric on $X$ whose induced topology is equivalent to the product topology on $X$.

Is the proof still correct if I define instead $$\rho(x,y)=\displaystyle\sum_{n=1}^{\infty}b^n\frac{\rho_n(x_n,y_n)}{[1+\rho_n(x_n,y_n)]}.$$ where $b\in (0,1)$ ?

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Rnst
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1 Answers1

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Short answer is yes. Long answer is: take the proof for $b = 1/2$, replace $1/2$ by $b$ everywhere, check that proof still holds.

Circonflexe
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