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Here the two point set would have the discrete topology.

The product has the product topology.

And the Cantor set has the subspace topology.

I defined my function $f$ to be,

$f: C \rightarrow X$

$\sum_{n=1}^{\infty} \frac{a_n}{2^n} \rightarrow (\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}, ...)$, where $a_n \in \left\{0, 2\right\}$

Then the inverse is...

$f^{-1}: X \rightarrow C$

$x \rightarrow \sum_{n=1}^{\infty} \frac{2 \cdot x_n}{2^n}$, where $x = (x_1, x_2, x_3, ...), x_n \in \left\{0, 1\right\}$

**Checking they are inverses:**

$f(f^{-1}(x)) = f(\sum_{n=1}^{\infty} \frac{2 \cdot x_n}{2^n}) = (\frac{2 \cdot x_1}{2}, \frac{2 \cdot x_2}{2}, \frac{2 \cdot x_3}{2}, ...) = (x_1, x_2, x_3, ...) = x$

$f^{-1}(f(\sum_{n=1}^{\infty} \frac{a_n}{2^n})) = f^{-1}(\frac{a_1}{2}, \frac{a_2}{2}, \frac{a_3}{2}, ...) = \sum_{n=1}^{\infty} \frac{2 \cdot a_n/2}{2^n} = \sum_{n=1}^{\infty} \frac{a_n}{2^n}$

$\implies$ they are inverses.

$\implies$ f is bijective.

All that is left is to show $f$ and $f^{-1}$ are continuous.

I'm having some trouble showing that $f^{-1}$ is continuous.

Consider an open set in $U \subset C$. Then, $U = C \cap (a, b), (a, b) \subset R$.

It seems straight forward that $f(U) = f(C \cap (a, b)$ would be in $X$. How do I show it would be open in $X$?