# Infinite sets?

1. Jan 29, 2012

### cragar

I was reading in George Gamow's book and he talks about drawing a one-to-one correspondence between the natural numbers and their squares and how these sets have the same size.
But could I take each natural number to $\aleph_{0}$
like $1^{\aleph_0}, 2^{\aleph_0}$ and draw a one-to-one correspondence between these 2 sets? This is probably ill defined or wrong, but just wondering.

2. Jan 29, 2012

### Integral

Staff Emeritus
$\aleph_{0}$ is the cardinality of the natural numbers. I do not believe that you can use it as if it were real.

Any set which can be but into 1 to 1 correspondecne with the natural numbers has cardinality of $\aleph_{0}$

3. Jan 29, 2012

### Deveno

in set theory, the notation BA usually means the set of all possible mappings from A to B. the reason for this notation, is that if |A| and |B| are finite:

|BA| = |B||A|.

note that there is only one possible mapping A→1, everything goes to the only possible image.

however, the set of mappings A→{0,1} is equivalent to the power set of A (a in A maps to 1 if a is in a given subset S of A, and 0 if it is not).

in particular: $2^{\aleph_0}$ is uncountable, but $1^{\aleph_0}$ is FINITE (it has cardinality 1).

4. Jan 29, 2012

### cragar

ok interesting, can I ask a side question. If I multiply all the real numbers together do I get one. Multiply all of them except 0. because what ever large number, its reciprocal exists.

5. Jan 29, 2012

### Deveno

that depends on "how you multiply them".

infinite products (like infinite sums) only make sense when they converge. if you always pair a real number with its inverse BEFORE you multiply anything else, you will get 1. but if you multiply all the real numbers > 1 FIRST, you'll get a divergent product, and from there, you could wind up with ANYTHING.

see, our notion of "pairing" only works with countable things (at least only works well). there's a LOT of real numbers. there's just as many real numbers between 0 and 1, as there are in the entire set. so "how" you "group" them, is going to affect any potential "product", drastically. in other words, you can re-arrange the groupings, to come up with any desired "product" you like, because there's so many of them, you can always grab a "few infinite more" to compensate.

you can't even say if the "end product" will be positive, or negative (without being very careful about the "grouping").

in the face of uncountably infinite sets, our ordinary intuitions about what happens if we do something with "all" of them, just break down.

you see, each real number is actually an infinite set of rational numbers. it's really a wonder we can add and multiply them at all, because of the difficulties involved.

6. Jan 30, 2012

### SteveL27

Absolutely. Bearing in mind the meaning of $n^{\aleph_0}$. It denotes the set of functions from the set $\aleph_0$ to the set n.

There is certainly a 1-1 correspondence between the natural numbers {1, 2, 3, ...} and the set {$1^{\aleph_0}, 2^{\aleph_0}$, ... }

In fact even though each of the sets $n^{\aleph_0}$ is an uncountable set (can you prove that? Try it for $2^{\aleph_0}$ with the definition I gave) there are only countably many such sets ... one for each n. You are correct.

7. Jan 30, 2012

### cragar

I was using $2^{\aleph_0}$ as a very large number.
why would a really large number be uncountable?

8. Jan 30, 2012

### SteveL27

$2^{\aleph_0}$ has a very specific meaning in set theory. It's the set of all functions from ${\aleph_0}$ to 2.

Since ${\aleph_0}$ is the set {0, 1, 2, 3, 4, ...}; and the set 2 = {0, 1}; a function from ${\aleph_0}$ to 2 is just a binary sequence. The collection of all binary sequences is uncountable.

That's the only definition of $2^{\aleph_0}$. It is of course "a really large number," but its not a natural number. It's a transfinite number.

9. Jan 30, 2012

### cragar

ok, now that I think of it, isn't $2^{\aleph_0}$ and
$10^{\aleph_0}$ the same thing, so this function would not be one-to-one and onto.

10. Jan 30, 2012

### micromass

Staff Emeritus
It can be proven that $2^{\aleph_0}$ and $10^{\aleph_0}$ are indeed equal.

Much of the confusion in this thread comes from the fact that you don't know what $\aleph_0$ really IS. Thinking of it as a really large number is very wrong.

If you have the time, read "Introduction to Set theory" by Hrbacek and Jech.
Or if you don't have that much time: at least read our FAQ on infinity: https://www.physicsforums.com/showthread.php?t=507003 [Broken]
There are 3 parts of the FAQ.

Last edited by a moderator: May 5, 2017
11. Jan 30, 2012

### cragar

how should I think of $\aleph_0}$

12. Jan 30, 2012

### micromass

Staff Emeritus

13. Jan 30, 2012

### SteveL27

They have equal cardinalities, so if you interpret them as cardinal numbers they are equal.

But as sets they are not equal. For example the function that maps every natural number to 3 is an element of $10^{\aleph_0}$ but not $2^{\aleph_0}$.

There's some ambiguity in the notation $n^{\aleph_0}$. You can interpret it as a set of functions; or you can interpret it as the cardinality of that particular set of functions.

The map that assigns to n the set $n^{\aleph_0}$ is injective; but the map that assigns to n the cardinality $n^{\aleph_0}$ is not injective.

Perhaps I should have been more clear earlier about the distinction.

14. Jan 30, 2012

### micromass

Staff Emeritus
There is no ambiguity. If you write $\aleph_0$, then you obviously mean this as a cardinal number. So obviously cardinal arithmetic applies.
If you want to discuss sets, then you write $2^\mathbb{N}$ and not $2^{\aleph_0}$.
If you want to discuss ordinals, then you write $2^\omega$.

There is no ambiguity.

15. Jan 30, 2012

### Char. Limit

So what is $2^\omega$ then?

16. Jan 30, 2012

### micromass

Staff Emeritus
17. Jan 30, 2012

### SteveL27

Interesting point. So the notation $A^B$ may mean two different things. If A and B happen to be cardinals, then it means the cardinality of the set of functions from B to A. If A and B happen to not both be cardinals, then it means the set of functions from B to A.

I'd never thought of it that way before; and I've never seen an explicit discussion of this overloading of the set exponentiation operator.

In your opinion is this a generally understood convention? Or is there a formal definition somewhere that set exponentiation has two different meanings depending on whether the arguments are cardinals?

Hmmm ... now it occurs to me that if the arguments are ordinals, then the meaning IS different and I'm aware of that difference ... so I shouldn't have too much trouble accepting the difference between sets of functions and cardinal arithmetic.

Last edited: Jan 30, 2012
18. Jan 30, 2012

### micromass

Staff Emeritus
I think it's just a convention that most people are aware of. It's usually clear from the context which convention is used.

19. Jan 30, 2012

### jgens

If you want to discuss the set of all mappings $\aleph_1 \rightarrow \{0,1\}$, then how would you suggest to do this using your convention? Your convention rules out $2^{\aleph_1}$ so this seems problematic.

20. Jan 30, 2012

### micromass

Staff Emeritus
Let A be a set of cardinality $\aleph_1$. Look at $2^A$.

That wasn't very hard.