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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!!!

Asaf Karagila
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aloyce temba
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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 01 '11 at 15:51
  • @Henno: You have given an answer here: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2007;task=show_msg;msg=2733.0001.0001 –  Jun 01 '11 at 15:53
  • @Henno: I was going to post that link as an answer :) –  Jun 01 '11 at 15:53
  • @Chandru: This fits perfectly into topology, analysis does not really deal with isolated points. – Asaf Karagila Jun 01 '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 01 '11 at 16:00
  • @Chandru: What do you mean by every point of the Cantor set is a limit point of itself? Even if you're trying to say every point of the Cantor set is a limit point of the others, I don't really see how that helps, because that's precisely what's asked here. – t.b. Jun 01 '11 at 16:05
  • @Asaf: I don't think so. I have learn't about Isolated points in analysis, also :) –  Jun 01 '11 at 16:06
  • @Chandru: In all my experience the reason they taught us about isolated points in analysis was usually some very introductory level topology in order to define some basic topological properties of the space ($\mathbb R^n$ usually). My point remains, this is perfectly fitting under just [topology]. – Asaf Karagila Jun 01 '11 at 16:09
  • @aloyce: your accounts have been merged. Please register your account so that you can log in more easily in the future. – Qiaochu Yuan Jun 01 '11 at 16:26
  • @Asaf: May be. I actually feel, that topology and analysis are so related that it's tough to categorize them. –  Jun 01 '11 at 16:37

2 Answers2

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$C_i$ 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.

Thomas Andrews
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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.

André Nicolas
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