Proof that a mapping onto $[0,1]$ is continuous.

Question

Let each $(A_i,T_i) = (\{0,2\}, T_{discrete})$ and define $\phi : \prod (A_i, T_i) \rightarrow [0,1]$ with $\phi (<a_1, a_2, ...>) = \sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}}$.

In order to show that this is continuous, let $a =<a_1, ...> \in \prod (A_i, T_i)$ be arbitrary and $V \in T_{[0,1]}$ be arbitrary. Then I need to find a $U$ open in $ \prod (A_i, T_i)$ such that $a \in U \land f(U) \subseteq V$.

If I let $V =$ the open interval $(\sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}} - \epsilon, \sum^{\infty}_{i=1} \frac{a_i}{2^{i+1}} + \epsilon)$, I figure I could let $U = \{a_1\} \times \{a_2\} \times ...\{a_n\} \times A_{n+1} \times ...$, but from here I can't figure out a way to show that $\phi(U) \subseteq V$.

Anyone have any ideas?

Answer 1

HINT: Suppose that $x=\langle x_1,x_2,x_3,\ldots\rangle\in U$, so that $x_k=a_k$ for $k=1,\ldots,n$. Then

$$\begin{align*} \left|\varphi(x)-\varphi(a)\right|&=\left|\sum_{k\ge 1}\frac{a_k}{2^{k+1}}-\sum_{k\ge 1}\frac{x_k}{2^{k+1}}\right|\\ &=\left|\sum_{k\ge n+1}\frac{a_k}{2^{k+1}}-\sum_{k\ge n+1}\frac{x_k}{2^{k+1}}\right|\\ &=\left|\sum_{k\ge n+1}\frac{a_k-x_k}{2^{k+1}}\right|\\ &\le\sum_{k\ge n+1}\frac{|a_k-x_k|}{2^{k+1}}\;, \end{align*}$$

where the last step uses the triangle inequality for absolute values.

• What are the possible values of $|a_k-x_k|$? What is the largest of them?
• If $|a_k-x_k|$ had the largest possible value for every $k\ge n+1$, what would $\sum_{k\ge n+1}\frac{|a_k-x_k|}{2^{k+1}}$ be?
• How can you choose $n$ to make that maximum possible sum less than a given positive $\epsilon$?