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wsalem
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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.
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.