Verify the properties of the power set of a set

Question

Let $A$ be any set .Let $P(A)$ be the power set of $A$.Then which are correct?

• $P(A)=\emptyset $ for some $A$.
• $P(A)= $ is a finite set for some $A$.
• $P(A)= $ is a countable set for some $A$.

• $P(A)=$ uncountable for some $A$.

  1. is false because $P(A)\neq \emptyset $ for any set at best it can be $\{\emptyset\}$.

2.true,Take $A=\{a,b\}$ then $P(A)$ is finite

3.True ,Take $A=\{a,b\}$ then $P(A)$ is finite and hence countable.

4.true take $A=\Bbb N$

But 3 is given as false.i am not getting it.Please help

Answer 1

The word "countable" is used in two different ways:

• Sometimes it means "can be injected into $\mathbb{N}$". (This is what I'll mean by "countable" for the rest of this answer.) Under this definition, $\{a, b\}$ is countable.

• Other times it means "can be bijected to $\mathbb{N}$". (That is, countable and infinite.) Under this definition, $\{a, b\}$ is not countable.

The former definition is in my experience by far the most common - however, many texts still use the latter. I suspect that your book is using the latter definition.

So what (c) is really asking is: "Is there a set whose powerset is countable and infinite?" The answer is indeed "no". For the proof, here's a hint: if $P(A)$ is countable and infinite, show that $A$ is countable and infinite; this yields a contradiction (do you see why?).

Technically, if you assume the Axiom of Choice, showing just "if $P(A)$ is countable and infinite then $A$ is infinite" is enough to get a contradiction; this implicitly uses the statement "every infinite set contains a countable infinite subset," which is not provable from ZF (= set theory without choice) alone. However, the argument using the hint I gave above doesn't use any choice.

Answer 2

If a set is finite we do not say that it is countable, rather we say that the set is atmost countable. Thus, it is not correct to say that $P(A)$ is countable if $P(A)$ is finite.

The statement '$P(A)$ is countable set for some set $A$' is false. To see why, first note that any set $A$ is either at most countable or uncountable.

If $A$ is at most countable, then there are two cases, either $A$ is finite or $A$ is countable. If $A$ is finite, then $P(A)$ is also finite, in fact we have $|P(A)|=2^{|A|}$. Now, suppose that $A$ is countable. Since $A$ is countable, the sets $A$ and $\mathbb{N}$ have the same cardinality. But the set $P(\mathbb{N})$, sometimes denoted by $2^\mathbb{N}$, is uncountable, and thus $P(A)$ is also uncountable.

If $A$ is uncountable, then $P(A)$ cannot be countable. Therefore the statement 'P(A) is countable set for some set A' is false.