# Cardinality of Unit Interval

Let $A,B,C$ be sets and let $Y^{X}$ be the set of all functions $f: X \rightarrow Y$. Define a function $$\Psi:C^{(A \times B)} \rightarrow (C^{B})^{A}$$ by setting for $F: A \times B \rightarrow C$ $$\Psi(F)(a):B \rightarrow C$$ To be the function $(\Psi(f)(a))(b)=F(a,b)$

I've already shown that $\Psi: C^{(A \times B)} \approx (C^{B})^{A}$

My question is, if I let $I=[0,1]=\{t \in \mathbb{R}|0 \leq t \leq 1 \}$ be the closed unit interval. Prove: $$|I|=2^{\aleph_{0}}$$ The hint given is $2 \leq 10 \leq 2^{\aleph_{0}}$ and the preceding result given.

I understand you are somehow supposed to find an injection from the closed unit interval to the set of all sequences containing 0s and 1s by dividing the unit interval in half over and over but I can't work out the small details. In particular, it would be helpful to know which sets to use to take advantage of the previous result.

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Given an $x \in [0.1]$ we may write $x \sum_{n=1}^{\infty} x_n 2^{-n}$ where each $x_n \in \{0,1\}$.
Proof: Fix an $x \in [0,1]$. If $x < \frac{1}{2}$, set $x_1 = 0$. Else set $x_1 = 1$. So $2^{-1}x_1 \le x$, and $|x - 2^{-1}x_1| \le \frac{1}{2}$. Suppose that $x_1, ..., x_k$ have been constructed with $|\sum_{j=1}^{k} x_j 2^{-j} - x| \le \frac{1}{2^k}$ and $x \ge \sum_{j=1}^{k} x_j 2^{-j}$. Now $x' = x - \sum_{j=1}^k x_j 2^{-j} \in [0, \frac{1}{2^k}]$. So we split this intervals into halves, if $x'$ is in the first half set $x_{k+1} = 0$ and if $x'$ is in the second half set $x_{k+1} = 1$. This completes the inductive construction.
We have that $|x - \sum_{j=1}^N x_j 2^{-j}| \le 2^{-N} \to 0$ as $N \to \infty$. So we have a binary representation of $x$.
Now we have that there's an injection from $[0,1]$ to $\{ (x_n)_{n \in \mathbb{N}} : x_n \in \{0,1\} \}$ which can be identified with the collection of all $f: \mathbb{N} \to \{0,1\}$. This is the injection you were looking for.