# C(reals) = C(P(naturals))?

1. Nov 26, 2003

### pjmunki

C(reals) = C(P(naturals))??

hello,
I am studying Cantor's set theory at present, but am a little confused as to why he concludes that the cardinality of the set of real numbers is equal to the cardinal number of the power set of naturals (2^aleph0).

Thanks.

2. Nov 26, 2003

### NateTG

The elements of 2N can be treated as sequences of 1's and 0's where a sequence corresponding to a subset A of N has a 1 in the n th position if n is in A.

You should have no trouble seeing that this mapping is bijective.

Now, if you look at the real numbers on [0,1] base 2, you get numbers like:
0.10101011101000110...
which are also sequences of ones and zeros. So there's a natural mapping.

Unfortunately there is a problem because
0.011111111111111111111111....=
0.100000000000000000000000....
in the reals.

But that only occurs a countable number of (N) times. So we can certainly construct a bijection to [0,1] + N.

So we have |2N| = |[0,1] + N|

but you should already know that |[0,1] + N|=|[0,1]|=|R|
(Cantor certainly did)

So by substitution we get:
|2N |=|R|

Last edited: Nov 26, 2003