# A proof that the Cantor set is Perfect

I found in a book a proof that the Cantor Set $\Delta$ is perfect, however I would like to know if "my proof" does the job in the same way.

Theorem: The Cantor Set $\Delta$ is perfect.

Proof: Let $x \in \Delta$ and fix $\epsilon > 0$. Then, we can take a $n_0 = n$ sufficiently large to have $\epsilon > 1/3^{n_0}$. Thus, the interval $[a, b]$ where $x$ lies is a subset of $B_\epsilon > (x)$. Hence, by iterating the construction of the Cantor set for $N > n_0$, we have intervals of length $1/3^N$ all included in $B_\epsilon (x)$, but with only one of those intervals such that $x$ lies within.

The intution behind the proof was that we should prove that for every $x$, if $x \in \Delta$, then for every $\epsilon >0$, $B_\epsilon (x) \setminus \{x\} \cap \Delta \neq \varnothing$.

Now, I do not particularly like my reference to the $[a, b]$ interval that is not mentioned before. Moreover, here – by choosing a closed interval – I am trying to address all at once the case in which $x$ is an endpoint of one of the closed intervals that form $\Delta$. Finally, I did not close the proof with a statement like "Thus, there are infinitely many points that differ from $x$ and that lie within $B_\epsilon (x)$.

In the end, I am not completely sure if this can be considered a proof or not. The intuition is correct (I am kind of positive about it), but I am not sure if I was actually able to write down my intuition in a good way.

As always any feedback is more than welcome.
Thank you!

• Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point? – Somabha Mukherjee Jan 4 '15 at 13:22
• @SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :) – Kolmin Jan 4 '15 at 14:24

Your idea is sound, but you’ve not expressed it clearly enough for you to have a real proof. I’ll write up an argument along the general lines that you have in mind.

Let $$x\in\Delta$$ and $$\epsilon>0$$ be arbitrary. Choose $$n\in\Bbb N$$ large enough so that $$3^{-n}<\epsilon$$. $$C_n$$, the $$n$$-th stage in the standard construction of $$\Delta$$, is the union of $$2^n$$ pairwise disjoint closed intervals, each of length $$3^{-n}$$; let $$I$$ be the one of these intervals containing $$x$$, clearly $$I\subseteq B_\epsilon(x)$$.

Now consider $$C_{n+1}$$: it’s a disjoint union of $$2^{n+1}$$ closed intervals, each of length $$3^{-(n+1)}$$, and exactly two of these intervals, say $$I_0$$ and $$I_1$$, are subsets of $$I$$. Let $$I_0$$ be the one that contains $$x$$. $$\Delta\cap I_1$$ is non-empty for the same reason that $$\Delta$$ is non-empty (why is that?), so let $$y\in\Delta\cap I_1$$. Then $$y\in\Delta\cap\big(B_\epsilon(x)\setminus\{x\}\big)$$, and since $$\epsilon>0$$ was arbitrary, $$x$$ is a limit point of $$\Delta$$. Finally, $$x\in\Delta$$ was arbitrary, and $$\Delta$$ is closed, so $$\Delta$$ is perfect. $$\dashv$$

It isn’t actually necessary to split $$I$$ into $$I_0$$ and $$I_1$$. Let $$I=[a,b]$$; then $$a,b\in\Delta\cap B_\epsilon(x)$$, and since $$x$$ cannot be equal to both $$a$$ and $$b$$, $$\Delta\cap\big(B_\epsilon(x)\setminus\{x\}\big)\ne\varnothing$$.

• First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval? – Kolmin Jan 5 '15 at 10:43
• @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof. – Brian M. Scott Jan 5 '15 at 10:57

Let $$x\in \Delta$$, For any $$\epsilon>0$$, consider $$(x-\epsilon,x+\epsilon)$$ Using archemedean propery, $$\exists N\in \mathbb N: \frac{1}{3^N}<\epsilon$$. Let $$\frac{M}{3^N}=\max \{\frac{m}{3^N}:\frac{m}{3^N}. So $$x\in [\frac{M}{3^N},\frac{M+1}{3^N}]\subset (x-\epsilon,x+\epsilon).$$ Consider the $$N+1$$ the stage of construction Removing $$(\frac{3M+1}{3^{N+1}},\frac{3M+2}{3^{N+1}})$$ from $$[\frac{M}{3^N},\frac{M+1}{3^N}]$$.

Case 1 $$x\in [\frac{M}{3^N},\frac{3M+1}{3^{N+1}}].$$ there exists $$c\in \Delta$$: $$c\in [\frac{3M+2}{3^{N+1}},\frac{M+1}{3^{N}}]$$. Hence $$(x-\epsilon,x+\epsilon)\cap \Delta \setminus \{x\}\neq \emptyset.$$ So, $$x$$ is a limit point of $$\Delta$$.

Similarly

Case 2 suppose $$x\in [\frac{3M+2}{3^{N+1}},\frac{M+1}{3^{N}}]$$, We have $$(x-\epsilon,x+\epsilon)\cap \Delta \setminus \{x\}\neq \emptyset.$$ Hence, $$x$$ is the limit point of $$\Delta$$. So, $$\Delta$$ is a perfect set.

• How can I verify that $x\in [\frac{M}{3^N},\frac{M+1}{3^N}]$ in the third line? Thanks! – himath Apr 28 at 8:40