Cantor set prove question solution verfication

Question

Question: Let C be the Cantor set contained in $[0, 1]$, as constructed in class. Let A be any (nonempty) connected subset of C. Prove that A consists of exactly one point. You may use the key theorems that we have proved about connected sets, and the definition of the Cantor set as constructed (it is the intersection of the sets Cn, which are unions of $2^n$ separated intervals each of width $1/3^n$).

My proof: Let $C⊆[0,1]$ be the Cantor set. For every x,y∈C with x≠y there exists by the construction of C some z∈[0,1] with z∉C and $x<z<y$. Then both $(−1,z)∩C=[−1,z]∩C $and $(z,2)∩C=[z,2]∩C$ are clopen in C with respect to the subspace topology.

So every two points of C are respectively contained in two disjoint clopen subsets of C. Because every connected component of C is contained in a clopen subset of C it follows that no two distinct points of C are contained in the same connected component. So C is totally disconnected, i.e. all connected components are singletons.

Can anyone help me check my proof is right or wrong? thank you so much

Answer 1

You are assuming that the construction of $C$ ensures that between any two points of $C$ there is a point not in $C$. If you actually proved that fact during the construction of $C$, then you can use it as you did, the problem is trivial, and your proof is correct. If not, however, you are essentially assuming the result that you’re trying to prove, and you have more work to do.

HINT: Suppose that $x,y\in C$, and $x<y$. For each $n\in\Bbb N$ let $I_n(x)$ be the interval of $C_n$ containing $x$, and let $I_n(y)$ be the interval of $C_n$ containing $y$.

• Show that $\bigcap_{n\in\Bbb N}I_n(x)=\{x\}$ and $\bigcap_{n\in\Bbb N}I_n(y)=\{y\}$.
• Conclude that there is an $n\in\Bbb N$ such that $I_n(x)\cap I_n(y)=\varnothing$.
• Let $I_n(x)=[a,b]$ and $I_n(y)=[c,d]$. Show that $b<c$.

Now you can pick any $z\in(b,c)$ and complete the argument just as you did originally.