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claim

Let $\aleph_0 = card(\Bbb N)$ and $\mathfrak c = card(\Bbb R)$ then

$2^{\aleph_0}=\mathfrak c$

proof

$2^{\aleph_0}=\mathfrak c \iff \exists \text{ bijection } f: \Bbb N \times \{0,1\}\rightarrow \Bbb R$

How to show the existence of bijection for the given sets? What is the most widely used approach?

Henrik supports the community
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delog
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    But $\aleph_1$ need not be the cardinality of $\Bbb R$ – Hagen von Eitzen Jun 05 '17 at 12:02
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    There is no bijection from $\mathbb{N}\times {0,1}$ to $\mathbb{R}$. – Mundron Schmidt Jun 05 '17 at 12:02
  • @HagenvonEitzen sorry actually my textbook define card(R) as some kind of greek? Hebrew? notation seemingly resemble to "c" but I can't tell what it is so I just borrowed the aleph notation. please edit the OP if you know what this notation refer to – delog Jun 05 '17 at 12:07
  • OK there is a problem with defining $\aleph_1=card(\mathbb{R})$ since there is already a definition for $\aleph_1$.Try using the Cauchy sequences with binary representation. – mathma Jun 05 '17 at 12:09
  • @delog: Perhaps fraktur? $\mathfrak{c}$ gives $\mathfrak{c}$. – Andrew D. Hwang Jun 05 '17 at 12:09
  • @AndrewD.Hwang that one. thx – delog Jun 05 '17 at 12:12
  • @MundronSchmidt sorry but my textbook, peculiarly, it just gives me the definition of "equal" of cardinals as existence of bijection. This book asks the reader first advance the logic with this then it denotes will define the definition of cardinal later. Since this question is put before the exact definition of cardinal, I had denoted that I need bijection between above two.. – delog Jun 05 '17 at 12:18
  • you must show that exists a bijection from the powerset of $\Bbb N$ to $\Bbb R$. – Masacroso Jun 05 '17 at 12:40

1 Answers1

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$2^{\aleph_0}=\aleph_1$ is the Continuum Hypothesis which is famously independent of ZFC so you will find it difficult to prove.

Added following comments.

Look at the Beth Nummbers

Look at the link within that to Cardinality of the continuum where you will find the proof that you want.

$\mathfrak{c}$ is a common symbol for the cardinality of the continuum. Hence $\mathfrak{c} = \beth_1 = 2^{\aleph_0}$ but maybe not $\aleph_1$.

badjohn
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    Well, no. The notation the asker defines is $\aleph_1 := |\mathbb{R}|$ – MathematicsStudent1122 Jun 05 '17 at 12:05
  • He should look at beth numbers: https://en.wikipedia.org/wiki/Beth_number. – badjohn Jun 05 '17 at 12:06
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    @MathematicsStudent1122 sorry actually my textbook define card(R) as some kind of greek? Hebrew? notation seemingly resemble to "c" but I can't tell what it is so I just borrowed the aleph notation. – delog Jun 05 '17 at 12:06
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    @mathma No, it's a matter of notation. All this answer does is pick on notation. What OP is asking is to prove $2^{\aleph_0} = \mathfrak{c}$ which is trivial. – MathematicsStudent1122 Jun 05 '17 at 12:08
  • As I just said, look at $\beth$ (beth) numbers which seem to be what you want. $\aleph$ (aleph) is the first letter of the Hebrew alphabet and $\beth$ (beth) is the second. Note the similarity to alpha and beta. – badjohn Jun 05 '17 at 12:08
  • @MathematicsStudent1122 OK got it I posted a comment upstairs. The thing is there is already a definition of $\aleph_1$ in the literature and it is not the cardinality of $\mathbb{R}$ – mathma Jun 05 '17 at 12:12
  • @MathematicsStudent1122 I hope that by following up the beth numbers, he will find what he wants. The Wikipedia link lists the real numbers and the power set of the natural numbers as examples of sets with cardinality $\beth_1$. – badjohn Jun 05 '17 at 12:12
  • @mathma edited op with \mathfrak c – delog Jun 05 '17 at 12:14
  • @badjohn edited op with \mathfrak c. Is it still the same to the continumm hypothesis? – delog Jun 05 '17 at 12:14
  • this is totally unrelated to the question. – Masacroso Jun 05 '17 at 12:39
  • @Masacroso I gave a link that contained the proof that he wanted. – badjohn Jun 05 '17 at 12:41