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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.

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    $\begingroup$ Have you looked at arcwise connectedness? $\endgroup$ Commented Feb 11, 2013 at 12:37
  • $\begingroup$ No, I haven't. What is that? $\endgroup$ Commented Feb 11, 2013 at 12:38
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    $\begingroup$ 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. $\endgroup$ Commented Feb 11, 2013 at 12:41
  • $\begingroup$ 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? $\endgroup$ Commented Feb 13, 2013 at 10:31
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    $\begingroup$ @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? $\endgroup$
    – hengxin
    Commented Feb 5, 2014 at 15:42

1 Answer 1

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

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  • $\begingroup$ The component of $(0,0,0,\ldots)$ is homeomorphic to $\ell_\infty$, and so are all other components... $\endgroup$ Commented Feb 11, 2013 at 22:13
  • $\begingroup$ How is the set of bounded sequences both open and closed in this set? $\endgroup$ Commented Feb 13, 2013 at 10:29
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    $\begingroup$ 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.) $\endgroup$ Commented Feb 13, 2013 at 19:59
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    $\begingroup$ 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? $\endgroup$
    – hengxin
    Commented Feb 5, 2014 at 15:34
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    $\begingroup$ @hengxin You're right; I had neglected the cutoff in the metric. So I should have said "any positive radius $<1$". $\endgroup$ Commented Feb 5, 2014 at 16:48

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