# Is my proof correct- totally disconnected set.

Show that the Cantor set, C, is totally disconnected.

Let $x,y\in C$ and suppose (WLOG) that $x<y$. Since $x,y\in C$, then $x,y\in C_n$ $\forall n\in$ℕ . Then, by construction of C, $\exists$ N$\in$ℕsuch that $|x-y|>\frac{1}{3^N}$ (where $\frac{1}{3^N}$ is the length of a closed interval in $C_N$. This implies that $x$ and $y$ belong to different closed intervals in $C_N$ and also, there must exist an element, $z$, where $x<z<y$, but $z\notin C_N$ and hence, $z\notin C$ (i.e. $z$ was in the open interval "removed" when constructing the Cantor set). Let $A$={$a|a<z, a\in C$} and $B$={$b|b>z, b\in C$}. Then $x\in A$ and $y\in B$. Note that $\bar A\cup B$=$\emptyset$ and $A\cup \bar B$=$\emptyset$, implying that $A$ and $B$ are separated sets. Then, since $A$ is the set of all elements in $C$ less than $z$ and $B$ is the set of all elements in $C$ greater than $z$ (and $z$ is not in $C$), it follows that $C=A\cup B$, and therefore $C$ is totally disconnected.

Is this correctly done? I am concerned about the part where I stated that $A$ and $B$ are separated sets, do I need to prove this point further? Feedback is appreciated.

Thanks.

I think that it’s acceptable as is, except that you have a typo: you wrote $\overline A\cup B$ and $A\cup\overline B$ where you wanted intersections. However, it’s possible to give more justification quite easily, so you might want to do so. Just note that $A=C\cap(\leftarrow,z)$, so $A$ is open in $C$, and $A=C\cap(\leftarrow,z]$, so $A$ is closed in $C$, and $A=C\cap(\leftarrow,z)$, so $A$ is also open in $C$ and therefore $B=C\setminus A$ is closed in $C$. Thus, $\operatorname{cl}_CA=A$ and $\operatorname{cl}_CB=B$, and $\operatorname{cl}_CA\cap\operatorname{cl}_C=A\cap B=\varnothing$.
• @Alti: No, I mean that $A$ itself is closed: it’s the intersection of $C$ with the closed set $(\leftarrow,z]$. And yes, $\operatorname{cl}XA$ is a standard notation for the closure of a set $A$ in a space $X$. – Brian M. Scott Dec 8 '12 at 14:44
• I'm sorry, I don't see how $z$ is included in the set $A$ because my definition of $A$ implies that $z$ is in not $A$... – Alti Dec 8 '12 at 21:00
• @Alti: $z$ isn’t in $A$; that’s the whole point! That’s why $A=C\cap(\leftarrow,z)=C\cap(\leftarrow,z]$ is both open and closed in $C$. – Brian M. Scott Dec 8 '12 at 22:09