How to show Cantor set is dense-in-itself?

Question

Here is my attempt to the proof. Suppose $x\in C$. Then we can find another element $x_n\in C$ such that $|x-x_n|<\frac{1}{3^n}<\epsilon$ for $n\in\mathbb{N}$ by letting $x_n$ to be one of the endpoint of construction of Cantor set at stage $n$ . This eventually forms a sequence of $\{x_n\}$ such that this sequence converges to $x$ as $n\rightarrow\infty$. Therefore, $x$ is not isolated. Since $x$ is arbitrary, it follows by definition that Cantor set is dense-in-itself.

Answer 1

That's the way to prove it. Simply, since it is a subset of $\mathbb{R}$, the topology we shall connsider in the Cantor set is the subspace topology, i.e. we will say a subset $U$ of the Cantor set $C$ is open if, and only if there exists some open $V$ in $\mathbb{R}$ such that $U=V\cap\mathbb{R}$

In this context, $C$ is a dense space if for any $x\in C$ and any open neighborhood of $x$ in $C$, it will always contain points of $C$ other than $x$, or in other words, $U\setminus\{x\}\cap C\not=\varnothing$ for every open neighborhood $U$ of $x$ in C.

So, compare this with what you have proved. For a fixed $x\in C$, the elements $|x-y|<\epsilon$ with $y\in C$, $\epsilon>0$ form a basic neighborhood system of $x$ in the subspace topology, so in fact, to prove that the Cantor set is dense, you only have to check that any basic open neighborhood of any point $x\in C$ contains points of $C$ other than $x$.

In the end, your answer is correct, because by definition, if a sequence $(x_n)_{n\in\mathbb{N}}\subseteq C$ converges to a point $x\in C$, every basic open neighborhood of $x$ contains points of $C$ other than $x$; a certain tail $(x_n)_{n\geq N}$ for some $N\in\mathbb{N}$.

I hope my comment provides some further insight about why your answer is correct.

Answer 2

Method $1.$ Show that the Cantor set $C$ is the set of those $x\in [0,1]$ that can be represented in ternary (Base-$3$) without using the digit $1$. E.g. in ternary, $1=0.\overline 2.$

Now if $\epsilon >0$ and $x\in C$ and in ternary $x=0.x_1x_2x_3...$ where each $x_j\in \{0,2\},$ then take $n$ large enough that $2\cdot 3^{-n}<\epsilon.$ And let $y_j=x_j$ when $j\ne n,$ while $y_n=2-x_n.$ Then the ternary sequence $0.y_1y_2y_3...$ represents a number $y\in C.$

And $0<|x-y|=2\cdot 3^{-n}<\epsilon.$

Method $2$. Consider the set $J$ of pair-wise disjoint open intervals thet are "removed" from $[0,1]$ to form $C.$ That is, $C=[0,1]\setminus (\cup J).$ The "removal" is often described as a sequence of stages: At Stage 1, one interval $(1/3,2/3)$ is removed. At Stage 2, two intervals $(1/9,2/9),(7/9,8/9)$ are removed. At Stage $n$ there are $2^{n-1}$ intervals removed.

Show by induction on the "Stage number" $n$ that no two members of $J$ share a common end-point. And now:

$(i).$ Suppose by contradiction that $x$ is an isolated member of $C$ and $0<x<1.$

Then for some $r>0$ the intervals $(-r+x,x)$ and $(x,x+r)$ are subsets of $\cup J.$ And since the members of $J$ are pairwise-disjoint open intervals, there must exist positive $r',r''$ such that $(-r+x,x)\subseteq (-r',x)=j'\in J$ and $(x,x+r)\subseteq (x,x+r'')=j''\in J.$ But then $x$ is a common end-point of $j'$ and $j'',$ a contradiction.

$(ii).$ The members of $J$ are pairwise-disjoint so the end-points of all the members of $J$ are all in $C.$ And for every $n\in \Bbb N$ the points $3^{-n}$ and $1-3^{-n}$ are end-points of some members of $J.$ So neither $0$ nor $1$ is an isolated point of $C$.