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(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=$2^{\aleph_o}$?
$\Bbb R$ 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 $\mathfrak{c} \leq {\aleph_0}^{\aleph_0} = 2^{\aleph_0}$.
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.
