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This question already has an answer here:

(I am a 13 year old so when you answer please don't use things that are TOO hard even though I actually can understand quite complex stuff)

I was studying Infinite sets and their cardinality (not in school, but just for fun), and I know that card($\mathbb N$)=$\aleph_o$ and card($\mathbb R$)=c I now know that c=$2^{\aleph_o}$ which may or may not be $\aleph_1$.

Is there any proof that c=$2^{\aleph_o}$?

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marked as duplicate by Lord_Farin, Dando18, Martin Sleziak, Chinnapparaj R, Xander Henderson Dec 12 '18 at 1:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It is possible to map the set of all subsets of the natural numbers to the set of real numbers with a bijection. I don't recall the map right now, so I'll let someone else supply it. $\endgroup$ – abiessu Sep 16 '13 at 3:34
  • $\begingroup$ See mathoverflow.net/questions/56633/… for a thread about finding such a bijection $\endgroup$ – Ian Coley Sep 16 '13 at 3:35
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    $\begingroup$ It's great that you want to learn more, but it seems that you're doing that in an unstructured way. And that can be very dangerous because you will have to make large skips between points. Try picking up a book. $\endgroup$ – Asaf Karagila Sep 17 '13 at 8:56
  • $\begingroup$ @Asaf: Any suggestions? $\endgroup$ – Zaz Jul 8 '15 at 17:06
  • $\begingroup$ @Zaz: Suggestions for what? For studying set theory? Take a course. Preferably with a world renowned set theorist, with a reputation for being a great teacher. $\endgroup$ – Asaf Karagila Jul 8 '15 at 17:17
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$\Bbb R$ contains the Cantor Ternary Set, so $\mathfrak{c} \geq 2^{\aleph_0}$.

On the other hand $\Bbb R$ is the set of all convergent sequences of rational numbers, so $\mathfrak{c} \leq {\aleph_0}^{\aleph_0} = 2^{\aleph_0}$.

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  • $\begingroup$ what is the cantor ternary set? $\endgroup$ – YYC Sep 25 '13 at 11:42
  • $\begingroup$ @Piman, As far as I know, it's also called the Cantor set, the middle-thirds Cantor set, or the Cantor discontinuum. en.wikipedia.org/wiki/Cantor_set $\endgroup$ – Metta World Peace Sep 25 '13 at 12:59
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Just to make notation clear. $c$ is the cardinality of $\mathbb R$, which is the same as the cardinality of $[0,1]$. $2^{\aleph_0}$ is the cardinality of the power set of a countable set (i.e., $2^{\aleph_0}=|\mathcal P (\mathbb N)|$). But $2^{\aleph_0}$ can also be defined as the cardinality of the set $\{f:\mathbb N \to \{0,1\}\}$, the set of all functions from $\mathbb N$ to the two-point set $\{0,1\}$. Equivalently, it is the set of all sequences of $0$'s and $1$'s.

The two meanings of $2^{\aleph_0}$ above coincide as one can show the relevant sets to admit a bijection, so they have the same cardinality. Now, to see that $c=2^{\aleph_0}$ it is convenient to represent real numbers in $[0,1]$ in binary notation. Thus, every real numbers $x\in [0,1]$ you write as $0.x_1x_2x_3\cdots$ where each $x_i$ is a binary digit, thus an element of $\{0,1\}$. There is a little difficulty here, namely some real numbers may have more than one binary expansion. Let's ignore this for a minute. Pretending that every real numbers $x\in [0,1]$ has a unique binary expansion (and clearly every binary expansion as above defines some such real number), this establishes a bijection between $[0,1]$ and $\{f:\mathbb N \to \{0,1\}\}$, mapping a real number to its sequence of digits in binary expansion. It follows that $c=2^{\aleph_0}$.

Now, getting back to reality one needs to take care of the fact that some real numbers admit more than one expansion. This can be done by invoking the Cantor-Schroeder-Bernstein theorem. I won't give the details here (unless you explicitly ask for the details). The point of the argument above is that by slight modification of reality you get a crisp (yet incorrect) proof of the claim, that can be corrected with a bit of technicality. I hope this answers your question.

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