Theorem 20.5 proves that
$D(x,y)=\sup({\frac{\bar d(x_i,y_i)}{i}})$
where $\bar d(a,b)=\min{(|a-b|,1})$
Then D is metric induces the product topology on $R^{\omega}$
I do not understand how to show that any open set U in metric topology for any element $x\in U$ there exist V is product topology such that $x\in V\subset U$.
Because in Product topology any open set has all coordinate as whole $X$ but finite elements are finite subset. So how such open set contain in smaller neighbourhood of elements in metric topology.
$U$ is open set in metric topology containing $x$.
As $V$ has first cordinate open set whose radius is $2\varepsilon$ how can it belong to $U$?
Any help will be appreciated
Best Answer
Read the whole proof: Munkres shows that all $y \in V$ obey $D(x,y) < \varepsilon$ (so that it lies in the ball and hence in $U$), and this computation really depends on the first $N$ coordinates of $y$, as the larger ones contribute so little to $D$ (due to the $\frac{1}{i}$ terms) that they can be ignored (and the sup reduces to a finite max).
Also, $y_1 \in (x_1-\varepsilon, x_1+\varepsilon)$, the first coordinate of $V$, ensures that $\bar{d}(y_1,x_1) < \varepsilon$ ; the diameter is $2\varepsilon$ but we only care about the distance to $\underline{x}$ we started with.
The sup metric that is also introduced in the book does not have this property that all large coordinates are irrelevant and Munkres also shows that the topology induced by that metric is actually not the product one. The $\frac{1}{i}$ terms are essential here, or we can also use a series of metrics that are weighted such that last terms are irrelevant as in this question.