Cantor Space is Nowhere Dense/Proof 1

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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, $\mathcal C$ is closed.

So from Closed Set equals its Closure:

$\mathcal C^- = \mathcal C$

where $\mathcal C^-$ denotes the closure of $\mathcal C$.


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

Then $I = \left({a \,.\,.\, b}\right)$ is an open interval of $\left[{0 \,.\,.\, 1}\right]$.

Let $\epsilon = b - a$.

Clearly $\epsilon > 0$.

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



So there exists an open interval of $\left[{0 \,.\,.\, 1}\right]$ which has been deleted from $\left[{0 \,.\,.\, 1}\right]$ during the process of creating $\mathcal C$.

Thus no open interval of $\left[{0 \,.\,.\, 1}\right]$ is disjoint from all the open intervals deleted from $\left[{0 \,.\,.\, 1}\right]$.

So any open interval of $\left[{0 \,.\,.\, 1}\right]$ can not be a subset of $\mathcal C = \mathcal C^-$.

Hence the result, by definition of nowhere dense.

$\blacksquare$


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