Cantor Set is Uncountable/Proof 1
Theorem
The Cantor set $\CC$ is uncountable.
Proof
From the definition as a ternary representation, $\CC$ consists of all the elements of $\closedint 0 1$ which can be written without using a $1$.
So let $x \in \CC$. Then in base $3$ notation, we have (as $0 \le x \le 1$):
- $\ds x = \sum_{i \mathop = 1}^\infty r_j 3^{-j}$
From the definition of the Cantor set, we have $\forall j: r_j \in \set {0, 2}$.
Furthermore, from Representation of Ternary Expansions, the $r_j$ are unique.
Now define the following function:
- $\ds f: \CC \to \closedint 0 1, \quad \map f {\sum_{i \mathop = 1}^\infty r_j 3^{-j} } = \sum_{i \mathop = 1}^\infty \frac {r_j} 2 2^{-j}$
Observe that $\forall j: \dfrac {r_j} 2 \in \set {0, 1}$.
That the right hand side expression is in fact an element of $\closedint 0 1$ now follows from binary notation.
Furthermore by Existence of Base-N Representation, any element $y$ of $\closedint 0 1$ may be written in the following form (where $\forall j: b_j \in \set {0, 1}$):
- $\ds y = \sum_{i \mathop = 1}^\infty b_j 2^{-j}$
Obviously, $y = \map f x$, where $x \in \CC$ is defined as follows:
- $\ds x = \sum_{i \mathop = 1}^\infty 2 b_j 3^{-j}$
It follows that $f$ is surjective.
From Closed Interval in Reals is Uncountable, the closed interval $\closedint 0 1$ is uncountable.
The result follows.
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$\blacksquare$
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: $5$