Prove by induction that the function is injective

[evinda]

Hi! (Wave)

The set $\mathbb{R}$ of real numbers is not countable.

Proof:

We define the function $F: \{0,1\}^{\omega} \to \mathbb{R}$ with the formula:

$$(a_n)_{n \in \omega} \in \{0,1\}^{\omega} \mapsto F((a_n)_{n \in \omega})=\sum_{n=0}^{\infty} \frac{2a_n}{3^{n+1}}$$

Show that $F$ is 1-1 and thus if $\mathbb{R}$ is countable then the set $\{0,1\}^{\omega}$ would also be, that is a contradiction.So we pick $(a_n)_{n \in \omega}, (b_n)_{n \in \omega} \in \{0,1\}^{\omega}$ with $F((a_n)_{n \in \omega})=F((b_n)_{n \in \omega}) \Rightarrow \sum_{n=0}^{\infty} \frac{2a_n}{3^{n+1}}=\sum_{n=0}^{\infty} \frac{2b_n}{3^{n+1}}$We will show that it holds using induction.At the base case, we assume that $a_0 \neq b_0$, let $a_0=0, b_0=1$.
So we have $\sum_{n=1}^{\infty} \frac{2a_n}{3^{n+1}}=\frac{2}{3}+\sum_{n=1}^{\infty} \frac{2b_n}{3^{n+1}}$

How could we find a contradiction? (Thinking)

 

[johng1]

Here's a proof that your function is injective:

 

[evinda]

Here's a proof that your function is injective:

I see.. (Nod)
Like that we have shown that the function $F: \{ 0,1 \}^{\omega} \to \mathbb{R}$ with the formula $(a_n)_{n \in \omega} \in \{0,1\}^{\omega} \mapsto F((a_n)_{n \in \omega})=\sum_{n=0}^{\infty} \frac{2a_n}{3^{n+1}}$ is injective.
Now we suppose that $\mathbb{R}$ is countable. That means that there is an injective function $g: \mathbb{R} \to \omega$.
Since $F: \{ 0,1 \}^{\omega} \to \mathbb{R}$ is injective, do we deduce that $g \circ F: \{0,1\}^{\omega} \to \omega$ is injective, that is a contradiction since then that would mean that $\{0,1\}^{\omega}$ is countable, i.e. that $\{0,1\}^{\omega} \sim \omega$ ? (Thinking)