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In my lecture notes there we have the following proof that the Cantor space $K_{\infty}$ is totally disconnected:

Let $z<z'$ in $K_{\infty}$, then we can find a $l \geq0$ and an interval $I:=(z'', z''')$ with length $3^{-l}$ such that $I \subset (z, z')$ and $I \cap K_{\infty} = \emptyset$. The mapping $y\mapsto 1_{y \leq z''}$ with $z'' \in I$ is therefore continuous from $K_{\infty}$ to $\{0,1\}$. It follows that $z$ and $z'$ do not belong to the same connected component of $K_{\infty}$.

My first question is how we can see that this mapping is continuous?

If it is indeed continuous, we can say that because $z$ and $z'$ will get mapped to $0$ and $1$ respectively and because there is the interval $I$ between $z$ and $z'$ which does not belong to $K_{\infty}$ we can't find a continuos mapping between $z$ and $z'$. At this point, I am not sure if the above argument is correct or not (of course this has also partly to do with the first question).

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The only point at which the function could be discontinuous is $z''$ (there is a jump at that point), but $z''$ does not belong to its domain. Therefore, the functions is continuous.

Since there is a continuous map $f\colon K_\infty\longrightarrow\mathbb R$ such that $f(z)=1$ and that $f(z')=0$, $z$ and $z'$ belong to distinct connected components of $K_\infty$. Since this happens for any two distinct elements of the Cantor set, this set is totally discontinuous.

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