Is this function, $f$, a homomorphism under addition and subtraction? Allow me to construct $f$
$\textbf{Lemma}$: For all $x \in [0,1]$, there exists a sequence $\{b_k\}_{k=1}^\infty$ such that
$$\forall k \in \mathbb{N}, b_k \in \{0,1\}$$
And
$$x = \sum_{k=1}^\infty \frac{b_k}{2^k}$$
$\textbf{Proof}$: For any number $x \in [0,1]$, it has a base 2 form (binary) consisting of only $0$ before the Radix point. For example,
$$1.0_{10} = 0.11\bar 1_2$$
$$0.5_{10} = 0.10\bar 0_2$$
$$0.0_{10} = 0.00\bar 0_2$$
Consider the sequence $\{b_k\}_{k=1}^\infty$, where $b_k$ is the value at the $k$th location past the Radix point of $x$'s base 2 form.
Because $b_k$ is a bit in a base 2 number, for all $k \in \mathbb{N}$, $b_k \in \{0, 1\}$. In addition, because of the method used to convert base 2 numbers to base 10, and $x \in [0,1]$
$$x = \sum_{k=1}^\infty \frac{b_k}{2^k}$$
And our lemma is true.
Using this Lemma, we construct a sequence of functions
$$\{b_k(x)\}_{k=1}^\infty:\ [0,1] \rightarrow \{0, 1\}$$
Such that for some $x \in [0,1]$ and some corresponding $\{b_k\}_{k=1}^\infty$ as defined in our lemma, $b_k(x) = b_k$.
Notice that for all $x \in [0,1]$,
$$x = \sum_{k=1}^\infty \frac{b_k(x)}{2^k}$$
We will now use this function $b_k(x)$ to construct $f:\ [0,1] \rightarrow [0,1]$
Define
$$f(x) := \sum_{k=1}^\infty \frac{2b_k(x)}{3^k}$$
Is this function a homomorphism under addition and subtraction? That is for $x,y \in [0,1]$, does
$$(x+y) \in [0,1] \Rightarrow f(x + y) = f(x) + f(y)$$
$$(x-y) \in [0,1] \Rightarrow f(x - y) = f(x) - f(y)$$
The purpose of this is to show that $f$ is continuous. If $f$ is homomorphic, it is not too hard to prove. I have tried to prove it is continuous without this assumption but keep getting stuck.
The function intrigued me and can be thought of as first converting $x$ to binary, replacing the 1's with 2's, and then converting it from ternary to decimal.

This map is not additive (even when restricted to cases it is well-defined).
Consider $x = y  = 1/3$.  Then
$$
x = y = 1/3 = 0.\overline{01}_2
\\
x+y = 2/3 = 0.\overline{10}_2
\\
f(x) = f(y) = 0.\overline{02}_3 = 1/4
\\
f(x+y) = 0.\overline{20}_3 = 3/4
$$
and $f(x+y) \ne f(x)+f(y)$.

Is $f$ continuous?  Yes, it is continuous at the points with unique binary expansion.  For counterexample we have to look at a point with two different binary expansions.
$$
1/2 = 0.1\overline{0}_2 = 0.0\overline{1}_2 .
$$
So conider a sequence $a_n$,
$$
a_{2n} = 0.10^n1\overline{0}_2,\\
a_{2n+1} = 0.1^n\overline{0}_2 .
$$
where $0^n$ means $n$ zeros in a row, $1^n$ means $n$ ones in a row.
Then $a_n$ converges to $1/2$, with terms alternately above and below $1/2$.
But $f(a_n)$ does not converge.  The even terms $f(a_{2n})$ converge
to $2/3$ from above.  The odd terms $f(a_{2n+1})$ converge
to $1/3$ from below.
