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Consider an infinite sequence of subsets of the interval $[0,1]$ obtained in the following way; set $C_{0}=[0,1]$. $C_{1}$ is obtained by removing the middle open half of $C_{0}$, that is

$$C_{1}= C_{0} -(1/4,3/4)=[0,1/4] \cup [3/4,1]$$

C2 is obtained by removing the middle open half of the interval of C1, that is ,C2=C1-(1/16,3/16)-(13/16,15,16). The set $C=\cap C_{n}$ for nϵ{0,1,2…∞} is called cantor set. Consider C as a subspace of [0,1]. Show that C is closed, compact and contains no isolated points.

I think for $C$ is closed because $C$ is an intersection of closed sets $C_n$ , looking $[0,1]$ is a closed interval. but any closed subset of $\mathbb R$ is compact, hence compactness of $C$ comes because it is a closed subset of $[0,1]$. I have no idea how to show that $C$ contains no isolated point! How may I get to this? Thank you!!!

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Not every closed set is compact, every closed and bounded set is, so closed subsets of $[0,1]$ are indeed compact. – Henno Brandsma Jun 1 '11 at 15:51
@Henno: You have given an answer here:… – user9413 Jun 1 '11 at 15:53
@Henno: I was going to post that link as an answer :) – user9413 Jun 1 '11 at 15:53
@Chandru: This fits perfectly into topology, analysis does not really deal with isolated points. – Asaf Karagila Jun 1 '11 at 16:00
@girdav: note that it's not the Cantor middle thirds set that is described here (but of course the reason is similar). – t.b. Jun 1 '11 at 16:00

$C_i$ is the union of a finite number of disjoint closed intervals of lengths $\left(\frac{1}{4}\right)^i$. The trick is to realize that the endpoints of these interval at step $i$ are in $C$ - they never get removed.

Let $\epsilon>0$. Then choose $n$ so that $\left(\frac{1}{4}\right)^n<\epsilon$.

Assume $x\in C$. Then $x\in C_n$, so $x$ is in a closed interval of length $\left(\frac{1}{4}\right)^n$ contained in $C_n$. Pick $y\neq x$ to be one of the endpoints of that interval. Then $|y-x|\leq\left(\frac{1}{4}\right)^n<\epsilon$. But by the observation above, $y\in C$.

So no point $x\in C$ is isolated.

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The argument below uses base $4$ expansions. We always use the "non-terminating" expansion when a number has two expansions.

The numbers that were removed are the ones that have a $1$ or a $2$ in their base $4$ expansion. So what is left is the numbers that only have $0$'s and/or $3$'s in their base $4$ expansion. Now given such a number $x$, it is possible to produce an infinite sequence $(x_n)$ of such numbers, all different from $x$, but approaching $x$. This is done simply by changing "digits" of $x$ ($0$ to $3$ or $3$ to $0$) further and further along in the base $4$ expansion of $x$.

Once one has a solid intuitive grasp of the process, base $4$ expansions can be stripped from the argument, and we end up with a concise elegant proof such as the one given by Thomas Andrews.

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