I have a question regarding Cantor set given to me as a homework question (well, part of it):

a. Prove that the only connected components of Cantor set are the singletons $\{x\}$ where $x\in C$

b. Prove that $C$ is metrizable

I am having some problems with this exercise:

My thoughts about $a$:

I know that in general path connectedness and connectedness are not equivalent, but I know that$\mathbb{R}$ is path connected, I want to say something like that since if $\gamma(t):C\to C$ is continues then $\gamma(t)\equiv x$ for some $x\in C$ then I have it that the connected components of $C$ can be only the singltons.

But I lack any justification - connectedness and path connectedness are not the same thing - but maybe since $\mathbb{R}$ is path connected we can justify somehow that if $C$ had any connected component then it is also path connected ? another thing that confuses me is that the open sets relative to $C$ and relative to $\mathbb{R}$ are not the same so I am also having a problem working with the definition of when a space is called connected

My thoughts about b:

Myabe there is something that I don't understand - but isn't $C$ metrizable since its a subspace of a $[0,1]$ with the topology that comes from the standard metric on $\mathbb{R}$ ?

I would appreciate any explanations and help with this exercise!

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    $\begingroup$ Regarding part a, try proving for any $x, y \in C$, there exists $z \in [0, 1] - C$ that lies between $x$ and $y$. This means that $C$ is totally disconnected, i.e., the only connected components are singletons. $\endgroup$ – Karatug Ozan Bircan Nov 4 '12 at 21:13
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    $\begingroup$ @KaratugOzanBircan I can prove this result but I don't see how the statement follows, can you please explain ? $\endgroup$ – Belgi Nov 4 '12 at 21:15
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    $\begingroup$ Regarding @Karatug’s comment: if $A$ is any subset of $C$ containing both $x$ and $y$, $(\leftarrow,z)\cap A$ and $(z,\to)\cap A$ form a separation of $A$ with $x$ in one and $y$ in the other, so $A$ can’t be connected. $\endgroup$ – Brian M. Scott Nov 4 '12 at 21:22
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    $\begingroup$ @BrianM.Scott - Can you please explain your notaion ? what is it that you intersect with $A$ ? $\endgroup$ – Belgi Nov 4 '12 at 21:27
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    $\begingroup$ @Belgi: Open rays to the left and right of $z$. Another notation is $(-\infty,z)\cap A$ and $(z,\infty)\cap A$. $\endgroup$ – Brian M. Scott Nov 4 '12 at 21:28

For (a), are you aware that the only connected subsets of $\mathbb{R}$ are intervals? If so, you could prove that $C$ doesn't contain any real intervals other than those of the form $[x, x] = \{x\}$.

Recall: A "real interval" is a set of real numbers $I$ such that for any $a, b, x \in \mathbb{R}$ with $a < x < b$, if $a, b \in I$ then $x \in I$.

For (b), this sounds good. If you are familiar with the concept, you can go even further and say it's a "complete" metric space, since $C$ is closed.

  • $\begingroup$ I am a bit confused, the open sets in $C$ are the open sets in $\mathbb{R}$ intersected with $C$. So the connected subsets of $\mathbb{R}$ (that I do know that they are the intervals) are not the same as $C$, Why a connected set in $C$ have to be an interval (hence a singleton) ? $\endgroup$ – Belgi Nov 4 '12 at 21:23
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    $\begingroup$ @Belgi: Connectedness is a topological property. Since $C$ inherits its topology from $\Bbb R$, every subset of $C$ has the same relative topology in $C$ that it has in $\Bbb R$, so if it’s connected in one, it’s connected in the other. $\endgroup$ – Brian M. Scott Nov 4 '12 at 21:27
  • $\begingroup$ @BrianM.Scott - I think I understand it in one direction, if $X$ is connected in $C$ but not in $\mathbb{R}$ then $X$ have a separation $U\cup V$ and since $X\subseteq C$ then $U,V\subseteq C$ so $\mbox{U{\cup}V =(U{\cap}C){\cup}(V{\cap}C)}$ is a separation in $C$. Am I correct ? I don't understand the other direction (that is irrelevant for the question, but still interesting) $\endgroup$ – Belgi Nov 4 '12 at 21:34
  • $\begingroup$ @Belgi: More simply, if $U$ and $V$ are open sets $\Bbb R$ such that $U\cap X\ne\varnothing\ne U\cap X$ and $X\subseteq U\cup V$, then $U\cap C$ and $V\cap C$ are a separation of $X$ in $C$. To go the other way, it’s easier to work with the separated sets characterization of connectedness (nr. 5 here): $U\cap X$ and $V\cap X$ are separated sets in $X$ no matter what space $X$ may be embedded in. $\endgroup$ – Brian M. Scott Nov 4 '12 at 21:46
  • $\begingroup$ Thanks for your answer, so the first thing you proved it that if there is a separation in $\mathbb{R}$ then there is one in $C$. I didn't understand the other direction - what are $U,V$ ? is their intersection with $X$ open in $\mathbb{R}$ ? $\endgroup$ – Belgi Nov 4 '12 at 21:50

@B.D’s answer is fine, and I’ve voted it up. But here is an explicit strategy that you may use. Let $S\subset C$ be a subset with at least two points. Now show that $S$ is not connected.


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