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

Saaqib Mahmood
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    Have you looked at arcwise connectedness? – Davide Giraudo Feb 11 '13 at 12:37
  • No, I haven't. What is that? – Saaqib Mahmood Feb 11 '13 at 12:38
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    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. – Davide Giraudo Feb 11 '13 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? – Saaqib Mahmood Feb 13 '13 at 10:31
  • Take two elements of $\Bbb R^\omega$, say $x$ and $y$. What is the simplest path between $x$ and $y$? Then show that it is continuous for the metric $p$. – Davide Giraudo Feb 13 '13 at 10:41
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    @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? – hengxin Feb 05 '14 at 15:42

1 Answers1

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The set of bounded sequences is both open and closed in this topology, so the space is disconnected.

Andreas Blass
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  • The component of $(0,0,0,\ldots)$ is homeomorphic to $\ell_\infty$, and so are all other components... – Henno Brandsma Feb 11 '13 at 22:13
  • How is the set of bounded sequences both open and closed in this set? – Saaqib Mahmood Feb 13 '13 at 10:29
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    It's open because it contains the ball of radius 1 around each of its points. It's closed because the same is true of its complement (and also because open subgroups of topological groups are always also closed). (There's nothing special about the radius 1; what I said would be equally true for any fixed positive radius.) – Andreas Blass Feb 13 '13 at 19:59
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    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? – hengxin Feb 05 '14 at 15:34
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    @hengxin You're right; I had neglected the cutoff in the metric. So I should have said "any positive radius $<1$". – Andreas Blass Feb 05 '14 at 16:48