Problem: Consider $\Bbb R^\omega$ with the uniform topology. Show that $\mathbf x = (x_1,x_2,\dots)$ and $\mathbf y = (y_1,y_2,\dots)$ lie in the same component of $\Bbb R^\omega $ if and only if the sequence $\mathbf x -\mathbf y = (x_1-y_1,x_2-y_2,\dots)$ is bounded.
I can prove it from left: assume that $|x_i-y_i| ≤ M$ for all $i∈\omega$. Consider $f(t)=t\mathbf x +(1-t)\mathbf y$. We can check that $f$ is a continuous function from $[0,1]$ to $\Bbb R^\omega$. Then $\mathbf x$ and $\mathbf y$ can be joined by $f(t)$, so $\mathbf x $ and $\mathbf y$ lie in the same component. But, how to prove it from right?
