1
$\begingroup$

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 $b\in (0,1)$ ?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.