# Is $\mathbf{R}^\omega$ in the uniform topology connected?

Let $\mathbf{R}^\omega$ be the set of all (infinite) sequences of real numbers. Then is this space connected in the uniform topology? How to determine this?

The uniform metric $p \colon \mathbf{R}^\omega \times \mathbf{R}^\omega \to \mathbf{R}$ is defined as follows: $$p((x_n),(y_n)) := \sup_{n\in Z^+} \min\{|x_n-y_n|,1\}$$ for sequences $(x_n)$, $(y_n)$ of real numbers.

• Have you looked at arcwise connectedness? Feb 11, 2013 at 12:37
• No, I haven't. What is that? Feb 11, 2013 at 12:38
• It means that given $a\neq b$ in a topological space $X$, we can find $\gamma\colon [0,1]\to X$ continuous so that $\gamma(0)=a$ and $\gamma(1)=b$. Show that this implies connectedness. Feb 11, 2013 at 12:41
• Yes, it does. But how to demonstrate that $\mathbf{R}^\omega$ is arcwise (or in other words path)-connected? How to given a rigorous proof of this fact? Feb 13, 2013 at 10:31
• @DavideGiraudo I encountered this problem in Munkres's Topology (exercise 8 of section 23, 2nd edition) before the notion of arcwise (or path)-connected. Would you mind repeating your ideas and solutions in a more elementary way again? Feb 5, 2014 at 15:42

• The component of $(0,0,0,\ldots)$ is homeomorphic to $\ell_\infty$, and so are all other components... Feb 11, 2013 at 22:13
• Why does any fixed positive radius work? For instance, take $\epsilon = 2$. According to the definition of the metric $\rho(x,y) = \sup \{ \min \{\mid x_n −y_n \mid, 1 \} \}$, I don't think that you can still bound the differences of $x_n$ and $y_n$. What is wrong with my argument? Feb 5, 2014 at 15:34
• @hengxin You're right; I had neglected the cutoff in the metric. So I should have said "any positive radius $<1$". Feb 5, 2014 at 16:48