Can we use the Cantor function to prove that Cantor set is uncountable?

Question

Since the cantor function is an onto function from the cantor set to $[0,1]$, the cantor set must be uncountable.

Is this statement right? I think there is a simpler proof for it than what I have seen.

Answer 1

Since you have $\mathcal C\subset[0,1]$, it also follows that $|\mathcal C|\leq\left|[0,1]\right|$. Now since you also have $\mathcal c(\mathcal C)=[0,1]$, it follows that $|\mathcal C|\geq\left|[0,1]\right|$. Conclusion: the two sets have the same cardinality!

Whatever solution you end up providing, I think it still needs to take into account something equivalent to the base 3 vs. base 2 way of thinking, but a different way of thinking about it could be:

• Create alternative sets, say $\mathcal D_s$, that are defined by repeatedly removing the middle part in proportion $1:s$ of the interval $[0,1]$ and the subintervals this creates.
• Now we have $\mathcal D_{1/3}=\mathcal C$ and $\mathcal D_0=[0,1]$
• Now we can continually morph the first set into the second by letting $s:1/3\to 0$, and every point will be smoothly maintained in each set.

This illustrates the principle for the fourth generation of the cantor set: