Is every function from $R^\omega$ to $R^\omega$ with uniform topology continuous?

Question

So I was studying topology and I came across the next theorem:

A function $f: X \to Y$ is continous iff for every $\epsilon>0$ there is $\delta>0$ such that if $d_x(x,y)<\delta$ then $d_y(f(x), f(y))<\epsilon$.

Since every distance in the uniform topology is at most 1, the theorem will always be true for any function from $R^\omega$ to $R^\omega$ with the uniform topology, am I wrong?

Answer 1

Counter-example: $f(x_i)=(y_i)$ where $y_i=1$ if $x_i \geq 0$ and $y_i=0$ if $x_i < 0$. This function is not continuous because $(0,0,...,0,\frac {(-1)^{n}} n,\frac {(-1)^{n+1}} {n+1},... ) \to (0,0,...)$ but $d((0,0,...,0,\frac {(-1)^{n}} n,\frac {(-1)^{n+1}} {n+1},... ), f(0,0,...))=1$ for all $n$. [Here $(0,0,...,0,\frac {(-1)^{n}} n,\frac {(-1)^{n+1}} {n+1},... )$ starts with $n-1$ zeros].