3

I recently read a fact that surprised me:

The power set of the natural numbers is equinumerous with $\mathbb{R}$.

In other words, $|\mathcal{P}(\mathbb{N})| = |\mathbb{R}|$.

I don't intuitively see why this is true. Can anyone explain why?

Edit: duplicate has been found, please vote to close.

Andrés E. Caicedo
  • 79,201
Newb
  • 17,672
  • 13
  • 67
  • 114
  • $|\Bbb R|\geqslant \aleph_1$ is all you can say. – Pedro Jan 31 '14 at 05:04
  • 1
    http://math.stackexchange.com/questions/553526/the-set-of-real-numbers – bradhd Jan 31 '14 at 05:06
  • Brad found a similar thread; I've voted to close mine. – Newb Jan 31 '14 at 05:08
  • 3
    @AustinMohr The other thread was particularly difficult to find. I'd rather leave this open, and let it be a duplicate linking to the other thread. Thus, the other thread becomes easier to find for other users. – Newb Jan 31 '14 at 05:35

2 Answers2

2

Hint: think of binary numbers between $0$ and $1$. What subset of $\mathbf N$ might you associate to the binary number $0.10101010101...$, say?

(There is a minor issue arising from the fact that rational numbers have two different binary expansions $(1=0.1111111...)$, but away from this countable set, this doesn't pose any problem.)

Bruno Joyal
  • 54,711
1

If you know that $\mathbb{Q}\simeq\mathbb{N}$ then you know $\mathcal{P}(\mathbb{Q})\simeq\mathcal{P}(\mathbb{N})$. There is one construction of $\mathbb{R}$ which defines them as a subset of $\mathcal{P}(\mathbb{Q})$. Thus there is an injection $\mathbb{R}\hookrightarrow\mathcal{P}(\mathbb{Q})$---inclusion. Here is an injection from $\mathcal{P}(\mathbb{N})$ to $\mathbb{R}$:

Let $S\subseteq\mathbb{N}$ and define $f(S)=\sum_{k\in S} 2^{-k}$. This is an injection. Now that $\mathbb{R}\simeq\mathcal{P}(\mathbb{N})$ follows from Schroder-Bernstein.

  • ...What construction of $\mathbb{R}$ defines the reals as a subset of $\mathcal{P}\mathbb{Q}$? I've only seen the ones by Dedekind Cuts and Cauchy Sequences. – Newb Jan 31 '14 at 06:06
  • the dedekind cuts is that construction –  Jan 31 '14 at 06:13