Open neighborhoods in the set of $K=\prod_1^{\infty}\{0,1\}$

Question

The problem given to me goes as follows:

Define $K=\prod_1^{\infty}\{0,1\}$, in the product topology. Let $S=s_n$ be a sequence of nonnegative real numbers such that $\sum_1^{\infty}s_i=1$. Define a map such that $f_s:K\rightarrow[0,1]$ by $k\mapsto \sum_1^{\infty}s_ik_i$. Show that $f_S$ is continuous.

I was thinking that the easiest way to show this would be to take an open neighborhood $V$ in $[0,1]$ (so it is of the form $(a,b)$) and show that $f_S^{-1}(V)$ is open.

But I am having troubles understanding what open sets in $K$ look like, can someone give me an example?

In addition if you think there is an easier way to show this, I would be welcome to suggestions.

Answer 1

Suppose that $x=\langle x_i:i\in\Bbb Z^+\rangle\in K$. For $n\in\Bbb Z^+$ let

$$B_n(x)=\{\langle y_i:i\in\Bbb Z^+\rangle\in K:y_i=x_i\text{ for }i=1,\ldots,n\}\;;$$

$B_n(x)$ is a basic open set in the product topology, and it’s not hard to check that $\{B_n(x):n\in\Bbb Z^+\}$ is a nbhd base at $x$ in $K$. You can use this to show that if a sequence in $K$ converges to $x$, its image under $f$ converges to $f(x)$ and hence that $f$ is continuous.

Answer 2

I personally find this easier to visualize by using the neighbourhood version of continuity.

Fix $x\in K$ and $\varepsilon>0$. By convergence, there is some $n$ such that $\sum_{i>n} s_i <\varepsilon$. Let $U=\bigcap_{i\le n}\{ y\in K: y_i=x_i\}$. This set is open in the product topology and, given any $y\in U$, $$ |f(x)-f(y)|= \sum_{i>n} s_i <\varepsilon $$