Make a bijection between $\mathcal{P}(A)$ and $\{0,1\}×\{0,1\}×\{0,1\}×...×\{0,1\}$ ($k$ times)

Question

If A is a finite set, with $|A|=k$, how i can make a bijection between $\mathcal{P}(A)$ and $\{0,1\}×\{0,1\}×\{0,1\}×...×\{0,1\}$ ($k$ times)? I defined a indicator function:

Let $X\in \mathcal{P}(A)$ and $g_X(a)=\begin{cases}1\,,a\in X\\0,\,a\in A-X\end{cases}$

Let $B=\{0,1\}×\{0,1\}×\{0,1\}×...×\{0,1\}$ ($k$ times) to save writing

But, my problem is idk how I can relate $f:P(A)\to B$ with $g_X$ so that $f$ is bijective. I think it makes no sense to define $f(X) = g_X$ because $g_X=0$ or $1$ and $f(X)=$ something like $(0,0,1,0,1,...)$

Sorry if the question is silly or if it's duplicated, but I am a newbie at this.

Answer 1

Since $|A|=k$, we can find a bijection between $A$ and $\{1, \ldots, k\}$, hence we can index the elements of $A$ where $a_1, \ldots, a_k$

Let $f(X) = b$ where $b \in B$ where $b_i=1$ if $a_i \in X$ that is $$b_i = g_X(a_i)$$

Answer 2

You have set $A$ of size $k$, what is $P(A)$? it is the set of all subsets of $A$, right? meaning that every $e\in P(A)$ defines a partition of $A$ to a set $e$ of selected elements from $A$ and set $A/e$ of unselected elements from $A$.

Now, notice that we can look on $B = {0,1}^k$ as a binary string of length $k$, because every string of length $k$ over the characters 0,1 is uniquely-defining an element in $B$ (define a bijection if you don't see it).

Now, we get to the problem itself, which is to show the bijection between $P(A)$ and $B$. And you had the right direction, so I will just explain what happens in it:

We want to define function $h$ that it's argument is in $P(A)$ and the image is in $B$. (formally: $h:P(A)\to B$). Let's assume that the elements in $A$ are indexed $a_1,..,a_k$. So now, it is clear that a natural mapping is that for each subset $X\in P(A)$ we want to map to the binary string that represents it, so that $B_i = 1 \iff a_i \in X$ and otherwise $B_i =0$. right?

More formally, we can take the indicator you defined and say that:

$h(X) = \times_{i=1}^{k}g_X(a_k)$