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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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    @Belgi: Open rays to the left and right of $z$. Another notation is $(-\infty,z)\cap A$ and $(z,\infty)\cap A$.2012-11-04

2 Answers 2

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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.

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    @Belgi: Sorry, that last bit wasn’t very clear. I meant that if there are open $U,V\subseteq C$ that disconnecte $X$, then $U\cap X$ and $V\cap X$ are non-empty separated subsets of $X$, and $X$ is their union, so $X$ is disconnected no matter what space it’s embedded in.2012-11-04
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@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.