# 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? Jan 4, 2015 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. :) Jan 4, 2015 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? Jan 5, 2015 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. Jan 5, 2015 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! Apr 28, 2020 at 8:40