# R=2^\alepha 0 vs Continuum hypothesis! A result in a taste of topology

1. Jan 24, 2009

### wsalem

R=2^\alepha 0 vs Continuum hypothesis! A result in "a taste of topology"

A year ago or so I read a proof in A Taste Of Topology, Runde that the cardinality of the continuum equals the cardinality of the powerset of the natural numbers. But a few hours ago I found Hurkyl making that statement $$|\mathbb{R}| = |\mathcal{P}(\mathbb{N})|$$ is undecidable in ZFC, I almost put in this proof but I realized this is the continuum hypothesis (would have been embarrassing!), so there must be something wrong, either in the proof of Runde or a misunderstanding (on by behalf) of the statement! Interestingly though, of all the reviews written on the book, there isn't a single mention of that statement!

Proposition $$c = 2^{\aleph_0}$$.
Here $$\aleph_0$$ denotes the cardinality of $$\mathbf{N}$$ and $$c$$ the cardinality of $$R$$
The proof uses Cantor-Bernstein theorem, basically if $$2^{\aleph_0} \leq c$$ and $$2^{\aleph_0} \geq c$$ holds then $$2^{\aleph_0} = c$$

Direction: $$2^{\aleph_0} \leq c$$.

Given $$S \subset N$$, define $$(\sigma_{n}(S))^{\infty}_{n=1}$$ by letting $$\sigma_{n}(S) = 1$$ if $$n \in S$$ and $$\sigma_{n}(S) = 2$$ if $$n \notin S$$, and let $$r(S) := \sum_{n=1}^{\infty} \frac{\sigma_{n}(S)}{10^n}}$$
Then $$P(\mathbf{N}) \rightarrow (0,1)$$ defined by $$S \rightarrow r(S)$$ is injective

Direction: $$2^{\aleph_0} \geq c$$

For the converse inequality, we use the fact that every $$r \in (0, 1)$$ not only has a decimal expansion, but also a binary one: $$r := \sum_{n=1}^{\infty} \frac{\sigma_{n}(r)}{2^n}}$$ with $$\sigma_{n}(r) \in \{0,1\}$$ for $$n \in \mathbf{N}$$.
Hence, every number in $$(0, 1)$$ can be represented by a string of zeros and ones.
This representation, however, is not unique: for example, both 1000 ... and 0111 ... represent the number $$\frac{1}{2}$$.
This, however, is the only way ambiguity can occur. Hence, whenever $$r \in (0,1)$$ has a period $$\overline{1}$$, we convene to pick its nonperiodic binary expansion. In this fashion, we assign, to each $$r \in (0, 1)$$, a unique sequence $$(\sigma_{n}(r))^{\infty}_{n}=1$$ in $$\{0, 1\}$$.

The map $$(0, 1) \rightarrow P(N)$$, $$r \rightarrow \{n \in N : \sigma_{n}(r) = 1\}$$ is then injective.

2. Jan 24, 2009

### CompuChip

Re: R=2^\alepha 0 vs Continuum hypothesis! A result in "a taste of topology"

I think by definition of the cardinal numbers $\aleph_n$, $|\mathbb{R}| = |\mathcal P(\mathbb{N})|$; the continuum hypothesis just asks whether there exists a set A such that |A| lies strictly between those (equivalently, whether $2^{\aleph_0} = \aleph_1$).

3. Jan 24, 2009

### wsalem

Re: R=2^\alepha 0 vs Continuum hypothesis! A result in "a taste of topology"

That would make perfect sense then. I think I was in a hurry comparing the result to the continuum hypothesis. So in other words, that would make the function Hurkyl defined in https://www.physicsforums.com/showpost.php?p=2046993&postcount=7 decidable and f(0)=1!

Last edited: Jan 24, 2009
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