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!
Best Answer
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.