Two metrics $d_1, d_2$ on a set $X$ are called uniformly equivalent, iff for every $\varepsilon > 0$ there exists $\delta_1, \delta_2$ such that
$$
d_1(x,y) < \delta_1 \Rightarrow d_2(x,y) < \varepsilon \quad \textrm{ and } \quad
d_2(x,y) < \delta_2 \Rightarrow d_2(x,y) < \varepsilon.
$$
Let $X_1, X_2, \ldots$ be a sequence of metrizable spaces and let $\rho_i$ be a metric on the space $X_i$ bounded by $1$ for $i = 1,2, \ldots$ (which could always so choosen). Consider the set $X = \prod_{i=1}^{\infty} X_i$ and for every pair of points $x = \{ x_i \}$ and $y = \{ y_i \}$ let
$$
\rho(x,y) := \sum_{i=1}^{\infty} \frac{1}{2^i} \rho_i(x_i, y_i).
$$
Now I want to show:
If for $i = 1, 2, \ldots$ metrics $\rho_i$ and $\sigma_i$ on a set $X_i$ are uniformly equivalent and bounded by $1$, then the metrics $\rho$ and $\sigma$ on $\prod_{i=1}^{\infty} X_i$, as defined above, are also uniformly equivalent.
I don't know how to find an appropriate $\delta_1, \delta_2$ such that the implications hold, cause I don't know if for a sequence something like
$$
\sum a_i < \delta
$$
what could be said about the individuel elements (besides $a_i < \delta$), the factor $1/2^i$ makes it more difficult for me too… any hints on how to solve this?
Best Answer
Our goal is to choose $\delta$ so that $\rho(x,y)<\delta\Rightarrow\sigma(x,y)<\epsilon$.
For any $\epsilon>0$: for each $i$, we may choose $\delta_i$ so that $$ \rho_i(x_i,y_i)<\delta_i \Rightarrow \sigma_i(x_i,y_i)<\epsilon/2 $$ In addition, there must be some $k$ such that $$ \sum_{i=k+1}^\infty \frac1{2^i} = \frac1{2^k}<\epsilon $$ (in particular, we may choose $k\in\mathbb Z$ so that $k>-\log_2(\epsilon)$).
Now, consider the set $\delta_1,\delta_2,\dots,\delta_k$ with $\delta_i$ as described above and let $\delta = 2^{-k}\min\{\delta_1,\delta_2,\dots,\delta_k\}$. What now?