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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.


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 <…;? 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

You have already received some comments on your proof. I am posting a CW-answer so that this question does not remain unanswered. And, of course, if other users feel like adding some useful comments, they are invited to.

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.

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