Cantor Space is Nowhere Dense

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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 1

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.

$\blacksquare$


Proof 2

Let $S_n$ and $C_n$ be as in the definition of the Cantor set as a limit of a decreasing sequence.

Then the length of every interval in $S_n$ is seen to be $\dfrac 1 {3^n} = 3^{-n}$.


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

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

Let $n \in \N$ such that $3^{-n} < b - a$, so that the length of every interval in $S_n$ is $3^{-n} < b - a$.


Therefore, as the intervals in $S_n$ do not overlap, no interval of length $b - a$ is contained in $C_n = \displaystyle \bigcup S_n$.

Consequently, as $\mathcal C \subseteq C_n$, no interval of length $b - a$ is contained in $\mathcal C$.

Since the interval $\left({a \,.\,.\, b}\right)$ was of arbitrary length, there do not exist any open intervals in $\mathcal C$.

Hence the result, by definition of nowhere dense.

$\blacksquare$


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