# Proving equinumerosity of intervals of reals

I've seen some other posts on this topic, but I was wondering if my particular method below works as a proof for the following:

For any two real numbers $\alpha$ and $\beta$, with $\alpha < \beta$, construct a bijection which proves the equinumerosity $\left[\alpha, \beta\right) \approx \left[\alpha, \beta\right] \approx \mathbb{R}.$

**Note: My proof below comes after a proof that $(0,1) \approx \mathbb{R}$, so I use this fact without proving it again.

Proof.

For simplicity take $\alpha = 0, \beta = 1$, though the following argument works for any $\alpha, \beta \in \mathbb{R}$.

We can find a bijection between $[0,1]$ and $[0,1)$ as follows:

$\hspace{1cm}$ Let $J$ be the function defined by:

$$J(x) = \begin{cases} 1/2^{j+1}, &\text{if } x = 1/2^j \:\:\text{for } j \in \left\{0,1,2,\dots\right\} \\[5pt] x, & otherwise \end{cases}$$

To prove that $J$ is a bijection we show that $J$ is both injective and surjective.

$\hspace{1cm}$ First, let $x, x' \in [0,1]$.

$\hspace{1cm}$ Then then are two cases.

$\hspace{1cm}$ Either $J(x) = J(x') \implies x = x' \: \text{directly}$, or we have

\begin{align*} J(x) = J(x') &\implies \frac{1}{2^{j+1}} = \frac{1}{2^{k+1}} \: \text{for }j,k \in \{0,1,\dots\} \\ &\implies j = k \\ &\implies x = x' \\ \end{align*}

$\hspace{1cm}$ Therefore, the function $J$ is injective.

$\hspace{1cm}$ Now, let $y \in [0,1)$.

$\hspace{1cm}$ For any such $y$, we can find an $x \in [0,1]$ with $J(x) = y$.

$\hspace{1cm}$ Either we have $J(x) = y \implies y = x$ directly, or we have \begin{align*}J(x) = y &\implies y = \frac{1}{2^{j+1}} \\ &\implies x = \frac{1}{2^j} \end{align*}

$\hspace{1cm}$ and so we can find an $x$ such that $J(x) = y$.

$\hspace{1cm}$ Therefore $J$ is surjective.

This shows that the function $J$ is indeed a bijection, from which it follows that $[0,1) \approx [0,1]$.

Now, we have already seen that $(0,1) \approx \mathbb{R}$. So to show that $[0,1)$ and $[0,1]$ are also equinumerous to $\mathbb{R}$ it suffices to show that $(0,1) \approx [0,1]$.

Consider the following two mappings:

1. $f: (0,1) \to [0,1],\,\: x \mapsto x$, which is clearly an injection.

2. $g: [0,1] \to (0,1),\,\: x \mapsto \frac{1}{2}\left(x + \frac{1}{2}\right)$, which is also an injection since it maps all of $[0,1]$ to a subset of $(0,1)$.

So, we have injections $(0,1) \to [0,1]$ and $[0,1] \to (0,1)$.

By Cantor-Schroeder-Bernstein, this is sufficient to show that there must also exist a bijection.

It follows that $(0,1)$ and $[0,1]$ are equinumerous.

This shows the intended result that $[0,1] \approx [0,1) \approx \mathbb{R}$.

End proof.

Thanks in advance for any comments. Much appreciated!

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Isn't it easier to go $[\alpha, \beta) \approx \mathbb{R^+}$ by $\phi(x) = (x - \alpha) / (x - \beta)$, and then use $[\alpha, \beta) \le [\alpha, \beta]$ and $[\alpha, \beta] \approx [\alpha, \beta / 2 ] \le [\alpha, \beta)$ and then apply <en.wikipedia.org/wiki/…;? Try to use previous results to shorten your proof! – vonbrand Feb 3 '13 at 19:03
Yes, thank you! I like it. – WilloW Feb 3 '13 at 19:14
Appreciate it the help. – WilloW Feb 3 '13 at 19:20
you are obviously right. I plead too little coffee. – vonbrand Feb 3 '13 at 19:23
Wait, actually I might be wrong. If we define the domain of $\phi$ to be only $[\alpha, \beta)$ then we do we still include the asymptotic behavior when approaching $\beta$ from the right? – WilloW Feb 3 '13 at 19:26

Your proofs seem fine to me. I would just perhaps stress the (relatively easy) fact that $J(x)=J(x')$ cannot happen for $x=\frac1{2^k}$ and $x'$ which is not of this form.
It is worth mentioning that your argument is nicely summarized in Did's answer here: If you can find a copy of $\mathbb N$ in your set, then you can add one point.
If you want to use Cantor-Bernstein theorem (as opposed to writing down an explicit formula for the bijection) and if you already know $|(0,1)|=|\mathbb R|$ then you get from $$|\mathbb R|=|(0,1)|\le|[0,1)|\le|[0,1]|\le|\mathbb R|$$ that all sets in the above chain of inequalities have the same cardinality.