Cantor Space is Totally Separated

Theorem

Let $T = \struct {\CC, \tau_d}$ be the Cantor space.

Then $T$ is totally separated.

Proof

Let $a, b \in \CC$ such that $a < b$.

Then $b - a = \epsilon$.

Consider $n \in \N$ such that $3^{-n} < \epsilon$.

Work In Progress In particular: To be proved that between $a + 3^{-\paren {n + 1} }$ and $a + 2 \times 3^{-\paren {n + 1} }$ there exists an interval $I$ which is excluded from $\CC$. Or something like that. Then we pick some $r \in I$ and continue as follows. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by completing it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{WIP}} from the code.

So $\exists r \in \R: a < r < b, r \notin \CC$.

Let $A = \CC \cap \hointr 0 r$ and $B = \CC \cap \hointl r 1$.

Thus $A \mid B$ is a separation of $\CC$ such that $a \in A, b \in B$.

Such a separation can be found for any two distinct $a, b \in \CC$.

Hence the result by definition of totally separated.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $29$. The Cantor Set: $6$