Problems on the Cardinality of the Sets

Question

Problem 1

For any set $A$, let $B^A$ be the set of all functions mapping $A$ into the set $B=\{0,1\}$. Show that $\left \lvert B^A \right \rvert=\left \lvert\mathscr {P}(A)\right \rvert$ where $\left\lvert S \right \rvert$ means the cardinality of $S$, for any set $S$ and $\mathscr {P}(A)$ denotes the Power Set of $A$.

Problem 2

1. Show that (using the same notation as the above problem) $\mathscr {P}(A)$ has too many elements to be put in an one to one correspondence with $A$.

2. Explain why this intuitively means that there are an infinite number of infinite cardinal numbers.

3. Is the set of everything a logically acceptable concept? Why or why not?

I have only proved the problems for finite $\left \lvert A \right \rvert$. For a proof of Problem 1 when $\left \lvert A \right \rvert$ is finite, we note that for each element $a \in A$ there are only two choices from the set $B$. Hence the total number of such functions will be $2^\left \lvert A \right \rvert=\left \lvert \mathscr {P}(A) \right \rvert$.

But if $\left \lvert A \right \rvert$ is not finite then I don't know how to proceed because the argument that has been given for finite $\left \lvert A \right \rvert$ may not apply to the non-finite $\left \lvert A \right \rvert$.

Similar problem is with Problem 2. For the first part if $\left \lvert A \right \rvert$ is finite then we note that for a bijection to exist between $A$ and $\mathscr P(A)$ then $\left \lvert \mathscr {P}(A) \right \rvert$ must be equal to $\left \lvert A \right \rvert$. But since for finite $\left \lvert A \right \rvert$, it is an integer and for all $\left \lvert A \right \rvert \geq1$ we have $\left \lvert A \right \rvert<2^{\left \lvert A \right \rvert}$, we conclude that for finite $\left \lvert A \right \rvert$ there cannot exist a bijection between $A$ and $\mathscr P(A)$.

But here again the problem is when $\left \lvert A \right \rvert$ is non-finite.

For third part of Problem 2, is the answer simply is that since everything is not a well-defined concept, the set of everything is not a logically acceptable concept, or is there some more subtle reasoning?

How can the problem be tackled?

Answer 1

In general, when we consider the cardinality, we use the concept of a 'bijection'.

So, in problem 1, if you want to show such equality in general case, you should construct explicit or implicit form of bijection between $\mathcal{P}(A)$ and $B^{A}$

In problem 2, you should argue by contradiction. Suppose there exists a bijection between two sets, then some contradiction happen.

Suppose $g:A \rightarrow 2^A$, then consider $B=\{x\in A| x\notin g(x)\}$ and note that $B\in 2^A$.

So that, there is $y\in A$ such that $g(y)=B$.

Either $y\in B$ or $y\notin B$, you can easily observe what is contradiction.

Such method used above is called 'Diagonalization'.