# Is $\Bbb{R}^\omega$ (with uniform topology) locally path connected?

Okay. Let $A \subseteq \Bbb{R}^\omega$ be the set of all bounded sequences in $\Bbb{R}$. The problem I am working on is trying to show that $x \in \Bbb{R}^\omega$ is lies in the same component of $0$ if and only if $x$ is a bounded sequence, where $\Bbb{R}^\omega$ is endowed with the uniform topology; or, in other words, that $C(0)=A$, where $C(0)$ is the connected component of $0$. Here is how I set about doing this. I mistakenly thought that $f : \Bbb{R} \to \Bbb{R}^\omega$ defined by $f(t) = xt + (1-t)y$ is continuous, where $x,y \in \Bbb{R}^\omega$. I then proceeded to use this to show that the $\epsilon$-balls are path connected, which would show that $\Bbb{R}^\omega$ is locally path connected and therefore the components and path components coincide. Of course, if it were the case that $f$ is continuous, $\Bbb{R}^\omega$ would be a path-connected and therefore connected space; but it isn't as $A$ is clopen. My first question is, is $\Bbb{R}^\omega$ locally path connected?

I did, however, show that $f$ is continuous when the codomain is restricted to $A$ (see General Topology Chat); and from this I was able to conclude that $A$ is path connected. Martin Sleziak and I were able to show that $A$ is in fact a path component of $0$. But seeing as $\Bbb{R}^\omega$ being locally path connected is up in the air at this point, I cannot conclude that $A$ is also a component of $0$. How do I fix this?

EDIT:

As I said in my comment below, most of the concepts mentioned in Alex Ravsky's answer haven't yet been introduced in my book. But I think I gathered from his answer, as well as Brian Scott's which he linked below, the essential ideas.

Claim: Let $\{P_i\}$ be the path components of some topological space $X$. If the $P_i$ are open in $X$ and are locally path connected, then $X$ is also locally path connected (LPC).

Proof: If $P_i$ is LPC, then its (subspace?) topology is by a basis $\mathcal{B}_i$ that is entirely comprised of path connected sets. Since $P_i$ is open in $X$, each $B \in \mathcal{B}_i$ is open in $X$, and, moreover, they are path connected in $X$. If $x \in X$, then $x \in P_i$ for some $i$; and so there exists a $B \in \mathcal{B}_i$ such that $x \in B \subseteq P_i \subseteq X$. This means that $\mathcal{B} = \bigcup \mathcal{B}_i$ is a basis for $X$'s topology entirely comprised of path connected sets, thereby showing that $X$ is LPC.

Now, since A is open (in fact, it is clopen) and locally path connected, and all other path components of $\Bbb{R}^\omega$ are homeomorphic to $A$, then $\Bbb{R}^\omega$ must also be locally path connected, in which case the components and path components coincide. Martin and I proved that $A$ is a path component of $0$, so it must also a component of $0$.

How does this sound?

• – Alex Ravsky Nov 9 '17 at 10:03

You complicated a simple question. Since $A$ is a linear topological (even a normed) space, it is path connected and locally path connected. Since $\Bbb R^\omega$ is a (disjoint) topological sum of cosets with respect to its clopen subgroup $A$, it is locally path connencted too. Since the set $A$ contains $0$ and is path-connected, $A\subset C(0)$. Since the set $A$ is clopen, $C(0)\subset A$.