# Cantor's s(w)et

1. Sep 17, 2010

### Hippasos

Cantor's s(w)et :)

Hi!

I don't know if this makes any sense, but after reading the book "Pi in the Sky"
this summer (It got really exceptionally hot here in high latitudes) I was let wondering the following:

Let C be the number of the continuum of numbers (with decimals).

Let N be the number of all natural numbers.

When counting decimals in the numbers which belongs to C,
is there numbers which have C-number decimals
more than numbers which have N-number decimals ?

Thanks!

Last edited: Sep 17, 2010
2. Sep 17, 2010

### mathman

Re: Cantor's s(w)et :)

What do you mean by C-number decimals and N-number decimals?

3. Sep 18, 2010

### Hippasos

Re: Cantor's s(w)et :)

I apologize for my mediocre English...
Thinking another way to rephrase this.

4. Sep 19, 2010

### HallsofIvy

Re: Cantor's s(w)et :)

Unfortunately, I can't quite make out what you are asking here. You say "Let C be the number of the continuum of numbers" but then you ask about "numbers which belong to C" so apparently what you really mean is "Let C be the continuum of numbers (the set of such numbers)" and "Let N be the set of all natural numbers". Now what do you mean by "counting decimals"? Are you asking "is the Cardinality of C greater than the Cardinality of N"? If so, the answer is definitely yes- that was shown by Cantor's "diagonal" method. We "count" cardinality by setting up a "one to one correspondence", that is an assignment to such that each object in one set is assigned to one and only one object in the other set. If such is possible, then the two sets have the same Cardinality. Cantor showed that, however you try to set up such an assignment, there are numbers in C that are "left over"- that is, in this sense, there are "more" numbers in C than there are in N and so C has greater cardinality than N.

5. Sep 20, 2010

### Hippasos

Re: Cantor's s(w)et :)

Hi!

I think I got it right this time... :D

Let C be the number of the continuum of numbers (with decimals).

Let N be the number of all natural numbers.

When counting decimals in the numbers which belongs to the continuum,
is there numbers which have C decimals
more than numbers which have N decimals ?

6. Sep 20, 2010

### CRGreathouse

Re: Cantor's s(w)et :)

I think that C is $$\mathbb{R}$$, the set of real numbers. In that case I interpret the question as: Is $$|\mathbb{R}|>|\mathbb{N}|$$? The answer, of course, is yes: this is Cantor's theorem.