# Cantor Space is Nowhere Dense/Proof 1

## Theorem

Let $T = \left({\mathcal C, \tau_d}\right)$ be the Cantor space.

Then $T$ is nowhere dense in $\left[{0 \,.\,.\, 1}\right]$.

## Proof

From Cantor Set is Closed in Real Number Space, $\CC$ is closed.

So from Closed Set equals its Closure:

- $\CC^- = \CC$

where $\CC^-$ denotes the closure of $\CC$.

Let $0 \le a < b \le 1$.

Then $I = \openint a b$ is an open interval of $\closedint 0 1$.

Let $\epsilon = b - a$.

Clearly $\epsilon > 0$.

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

So there exists an open interval of $\closedint 0 1$ which has been deleted from $\closedint 0 1$ during the process of creating $\CC$.

Thus no open interval of $\closedint 0 1$ is disjoint from *all* the open intervals deleted from $\closedint 0 1$.

So an open interval of $\closedint 0 1$ can not be a subset of $\CC = \CC^-$.

Hence the result, by definition of nowhere dense.

$\blacksquare$

## 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: $4$