I'm sorry, I'm not sure I understand your question. Can you clarify?

[knowLittle]

Set Theory -- Uncountable Sets

Homework Statement

Prove or disprove.
There is no set A such that $2^A$ is denumberable.

The Attempt at a Solution

A set is denumerable if $|A| = |N|$

My book shows that the statement is true.
If A is denumerable, then since $|2^A| > |A|, 2^A$ is not denumerable.


I don't understand why they state that 2^A > A? I thought that in denumerable sets you can't really say that there is one set greater than other? I thought that there is denumerable or uncountable and that's it.
But, there is not a denumerable set greater than another denumerable set.

Help please.

 

[haruspex]

I don't understand why they state that 2^A > A? I thought that in denumerable sets you can't really say that there is one set greater than other?

That's true, but 2^A is not countable - that's the point.

I thought that there is denumerable or uncountable and that's it.

Well, no - there are infinitely many orders of infinity. If B is uncountable, there is still no bijection between B and its power set, so |2^B| is a higher order of infinity.
Your book appears to be using the general theorem that |2^A|>|A|. I suggest you locate that in your book and study it.

 

[HallsofIvy]

Some textbooks use the term "denumerable" to mean "finite or countable".

If A is a finite set then 2^A is also finite so "denumerable".

If, however, you are using "denumerable" to mean "countably infinite" then the statement is true.

 

[knowLittle]

So, for a countably(denumerable) infinite A, $|2^A|$ is an uncountably infinite?

 

[micromass]

So, for a countably(denumerable) infinite A, $|2^A|$ is an uncountably infinite?

Yes.

 

[HallsofIvy]

So, for a countably(denumerable) infinite A, $|2^A|$ is an uncountably infinite?

Since "infinite" is an adjective, the word "an" should not be there!

 

[knowLittle]

Since "infinite" is an adjective, the word "an" should not be there!

Yes.
:/