# Set Theory - Uncountable Sets

1. May 11, 2014

### knowLittle

Set Theory -- Uncountable Sets

1. The problem statement, all variables and given/known data
Prove or disprove.
There is no set A such that $2^A$ is denumberable.

3. 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.

2. May 11, 2014

### haruspex

That's true, but 2A is not countable - that's the point.
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 |2B| is a higher order of infinity.
Your book appears to be using the general theorem that |2A|>|A|. I suggest you locate that in your book and study it.

3. May 12, 2014

### HallsofIvy

Staff Emeritus
Some text books 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.

4. May 12, 2014

### knowLittle

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

5. May 12, 2014

### micromass

Staff Emeritus
Yes.

6. May 12, 2014

### HallsofIvy

Staff Emeritus
Since "infinite" is an adjective, the word "an" should not be there!:tongue:

7. May 12, 2014

Yes.
:/