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

• wsalem
In summary, the conversation is discussing the relationship between R=2^\aleph_0 (the cardinality of the real numbers) and the Continuum Hypothesis, which states that there is no set with cardinality between \aleph_0 (the cardinality of the natural numbers) and R. The conversation mentions a proof in the book "A Taste of Topology" that shows R=2^\aleph_0, but also notes that the statement is undecidable in ZFC (Zermelo-Fraenkel set theory with the Axiom of Choice). The conversation then goes on to discuss a proof using the Cantor-Bernstein theorem, and ultimately concludes that the Continuum Hypothesis is
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.

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 $2^{\aleph_0} = \aleph_1$).

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## 1. What is the R=2ℵ0 vs Continuum hypothesis?

The R=2ℵ0 vs Continuum hypothesis is a mathematical statement in topology that compares the cardinality of the set of all real numbers (continuum) to the cardinality of the set of all countable infinite sets (R=2ℵ0). It is a fundamental topic in set theory and has been a subject of much research and debate among mathematicians.

## 2. What does it mean for a set to have a cardinality of 2ℵ0?

A set with a cardinality of 2ℵ0 is a set that has the same number of elements as the set of all countable infinite sets. This means that the set is uncountably infinite and cannot be put into a one-to-one correspondence with the natural numbers.

## 3. What is the significance of the Continuum hypothesis in topology?

The Continuum hypothesis has significant implications in topology as it addresses the question of whether there exists a set with a cardinality between the countable and the uncountable sets. It also has connections to other areas of mathematics such as analysis and logic.

## 4. Has the R=2ℵ0 vs Continuum hypothesis been proven?

No, the R=2ℵ0 vs Continuum hypothesis has not been proven. In fact, it has been shown to be independent of the standard axioms of set theory, meaning that it cannot be proved or disproved using these axioms alone. This has led to much discussion and research on alternative axioms that may help resolve the hypothesis.

## 5. What are some applications of the R=2ℵ0 vs Continuum hypothesis?

The R=2ℵ0 vs Continuum hypothesis has implications in various areas of mathematics, including topology, set theory, and logic. It also has connections to computer science and has been used to study the complexity of algorithms. Additionally, it has sparked new research and developments in alternative axioms and foundations of mathematics.

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