real analysis - Constructing A Space Filling Curve that fills the Unit Square

I'm reading Neal Carothers' Real Analysis, and he constructs a curve defined over $[0,1]$ that fills the unit square as follows:

Let $f$ be a real-valued function over $[0,1]$ such that $f$ is $0$ over $[0,\dfrac{1}{3}]$, $3t-1$ over $(\dfrac{1}{3},\dfrac{2}{3})$, and $1$ over $[\dfrac{2}{3},1]$. Now extend $f$ over $\mathbb{R}$ by specifying that $f$ is even and $2$-periodic.

Here is the equation I don't understand: $$f(3^kt)=f(0.(2a_k)(2a_{k+1})(2a_{k+2})...)=a_k$$ if $t$ is a number in the Cantor Set and $0.(2a_0)(2a_{1})(2a_{2})...$ is the base $3$ decimal expansion of $t$, where $(2a_i)$ is the $(i+1)^{th}$ digit in the expansion. Since it's base $3$ and $t$ is in the Cantor Set, each $a_i$ is either $0$ or $1$.

I tried to prove the equation by writing the decimal expansion as a series multiplied by $3^k$ but I can't seem to make the math come out. In the book, Carothers prefaces the equation with "since $f$ is periodic with period $2$;" I don't know what to make of this hint since $t$ is fixed.

Any help at all would be much appreciated!

Answer

We're in base $3$, so multiplying by $3^k$ shifts the decimal $k$ digits to the right.

So f($3^kt)=((2a_0)(2a_1)\cdots(2a_{k-1}).(2a_k)(2a_{k+1})\cdots)$. We also note that $f$ has period $2$, and that $(2a_0)(2a_1)\cdots(2a_{k-1})$ is an even number and is hence a multiple of the period of $f$.

So we are left with

$f(3^kt)=f(0.(2a_k)(2a_{k+1})\cdots)$

Then if $a_k=0,$ we will have that $0.(0)(2a_{k+1})\cdots=0$ because (in base $3$) it will be within the interval $[0,\frac{1}{3}]$ and $f$ is equal to $0$ there (since that's how $f$ was defined). On the other hand if $a_k=1$, then we will have that $0.2(a_{k+1})\cdots=1$ (in base $3$) since $f$ was defined to be equal to $1$ in the interval $[\frac{2}{3},1]$.

So if $a_k=0$ then $f(3^kt)=a_k=0$, but if $a_k=1$,then $f(3^kt)=a_k=1$, so either way we have that $f(3^kt)=a_k.$