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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!

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    $\begingroup$ Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point? $\endgroup$ – Somabha Mukherjee Jan 4 '15 at 13:22
  • $\begingroup$ @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. :) $\endgroup$ – Kolmin Jan 4 '15 at 14:24
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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 $b$, $\Delta\cap\big(B_\epsilon(x)\setminus\{x\}\big)\ne\varnothing$.

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  • $\begingroup$ 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? $\endgroup$ – Kolmin Jan 5 '15 at 10:43
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    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Jan 5 '15 at 10:57
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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 $\frac{M}{3^N}=\max \{\frac{m}{3^N}:\frac{m}{3^N}<x \space \text{and} \space m\in \mathbb N\}$. 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.

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