How to prove continuity in Baire space?

Question

Let $X=(\omega^\omega,d)$ be Baire space with the metric $d$ defined in assignment $1$. Define a function $G:X\to X$ by letting, for $f\in X$, the function $G(f)$ be defined by: $$(G(f))(n)=\begin{cases}f(n/2),&\text{if }n\text{ is even; and}\\0,&\text{if }n\text{ is odd.}\end{cases}$$ Show that $G$ is continuous at the identity function $f(n)=n$.

Can anyone please tell me how to prove continuity in this case? Thanks a lot

The metric is shown below

Answer 1

I’ll outline the approach.

It may be a little easier to think of the points of $X$ as infinite sequences: $f\in X$ is simply the sequence $\langle f(n):n\in\omega\rangle$. In particular, if $f$ is the identity function, the sequence is $\langle 0,1,2,3,\ldots\rangle$.

• Verify that $G(f)$ is the sequence $\langle 0,0,1,0,2,0,3,0,4,0,\ldots\rangle$. If we let $g=G(f)$, then in function terms $$g(n)=\begin{cases}0,&\text{if }n\text{ is odd}\\n/2,&\text{if }n\text{ is even}\;.\end{cases}$$

A tabular display may be helpful:

$$\begin{array}{rcc} n:&0&1&2&3&4&5&6&7&8\\ f(n):&0&1&2&3&4&5&6&7&8\\ g(n):&0&0&1&0&2&0&3&0&4 \end{array}$$

For $x\in X$ and $n\in\omega$ let $B_n(x)=\{y\in X:d(x,y)<2^{-n}\}$.

• Verify that $\{B_n(x):n\in\omega\}$ is a local base at $x$.

The following observation will be useful:

• Show that $B_n(x)=\{y\in X:y(k)=x(k)\text{ for each }k<n\}$. In other words, $B_n(x)$ is the set of all sequences in $X$ that agree with $x$ for the first $n$ terms.

In order to show that $G$ is continuous at $f$, you must show that for each $\epsilon>0$ there is a $\delta>0$ such that if $d(f,x)<\delta$, then $d(g,G(x))<\epsilon$.

• Show that for each $m\in\omega$ there is an $n\in\omega$ such that $G[B_n(f)]\subseteq B_m(g)$. You should be able to express a suitable $n$ as a simple function of $m$.
• Then explain why this proves that $G$ is continuous at $f$.