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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!

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 $xinDelta$ and $epsilon>0$ be arbitrary. Choose $ninBbb N$ 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 $Isubseteq B_epsilon(x)$.

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 $x$. $Deltacap I_1$ is non-empty for the same reason that $Delta$ is non-empty (why is that?), so let $yinDeltacap I_1$. Then $yinDeltacapbig(B_epsilon(x)setminus{x}big)$, and since $epsilon>0$ was arbitrary, $x$ is a limit point of $Delta$. Finally, $xinDelta$ 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 $a,binDeltacap B_epsilon(x)$, and since $x$ cannot be equal to both $a$ and $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.

Let $xin Delta$, For any $epsilon>0$, consider $(x-epsilon,x+epsilon)$ Using archemedean propery, $exists Nin mathbb N: frac{1}{3^N}<epsilon$. Let $frac{M}{3^N}=max {frac{m}{3^N}:frac{m}{3^N}<x space text{and} space min mathbb N}$. So $xin [frac{M}{3^N},frac{M+1}{3^N}]subset (x-epsilon,x+epsilon).$ 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 $xin [frac{M}{3^N},frac{3M+1}{3^{N+1}}].$ there exists $cin Delta$: $cin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$. Hence $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ So, $x$ is a limit point of $Delta$.

Similarly

Case 2 suppose $xin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$, 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.