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I need to prove $((0,1)\cap\mathbb{Q})\cup((2,3)\cap\mathbb{Q}) $ is isomorphic to $(0,10)\cap\mathbb{Q}$. I have tried by looking for a bijective relation between sets and Yi think my best try has been a piecewise function:

\begin{equation} f(x)= \begin{cases} x, & 0\lt x\lt 1 \\ 9x-17, & 2\lt x\lt 3 \end{cases} \end{equation} The problem is $1$ belongs to $(0,10)$ but it is not on the image of $f$ so it doesn't work as a bijection. I'd really appreciate some help on this problem. Thanks Edit: I need an order-preserving bijection.

Bernard
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CoolJedi132
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    This question (https://math.stackexchange.com/questions/28568/bijection-between-an-open-and-a-closed-interval) may help you – Tito Eliatron Nov 08 '20 at 18:51
  • Do you just need a bijection, or do you want an order-preserving bijection? – Hagen von Eitzen Nov 08 '20 at 18:52
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    Some hints: ignore intersecting with $\mathbb{Q}$ for the time being. Find a bijection between $(0,1)$ and $[0,1)$ first. Then find a bijection between $[0,1)$ and $[0,5)$ and one between $(2,3)$ and $(5,10)$. – Integrand Nov 08 '20 at 18:52
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    @HagenvonEitzen I need an order-preserving bijection, sry. – CoolJedi132 Nov 08 '20 at 18:55

2 Answers2

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First, find an irrational "hole" in $(0,10)$, let's say $Z=\pi$.

Can you now construct an order-preserving bijection of $(0,1)\cap\mathbb Q$ to $(0,Z)\cap\mathbb Q$? For example:

  • Make a (strictly) monotonically increasing sequence of rational numbers $a_i\in(0,1)$ such that $\lim_{n\to\infty}a_i=1$, e.g. $0.9, 0.99, 0.999,\ldots$.
  • Also, make another (strictly) monotonically increasing sequence of rational numbers $b_i\in(0,Z)$ such that $\lim_{n\to\infty}b_n=Z$, e.g. $3, 3.1, 3.14, 3.141\ldots$.
  • Now make the bijection piecewise-linear on: $(0,a_1]\to(0,b_1], [a_1,a_2]\to[b_1,b_2], [a_2,a_3]\to[b_2,b_3]$ etc.

Altogether, these linear mappings will provide an order-preserving bijection of $(0,1)\cap\mathbb Q$ to $(0,Z)\cap\mathbb Q$.

Similarly, construct an order-preserving bijection of $(2,3)\cap\mathbb Q$ to $(Z,10)\cap\mathbb Q$. (You will need strictly monotonically decreasing sequences of rational numbers converging to $2$ and $Z$ "from above".)

Finally, glue those two bijections together.

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Let $a_n=1-\frac1n$ and partition $(0,1)$ as $$(0,1)=\bigcup_{n=1}^\infty (a_n,a_{n+1}].$$ Likewise, let $b_n=2+\frac1n$ and partition $$(2,3)=\bigcup_{n=1}^\infty [b_{n+1},b_n).$$

Next, fix a real number $\alpha$ between $0$ and $10$. Let $\{c_n\}_n$ be a strictly increasing sequence of rational numbers that converge to $\alpha$ and starts with $c_1=0$. Similarly, let $\{d_n\}_n$ be a strictly decreasing sequence of rational numbers that converge to $\alpha$ and starts with $d_1=10$.

Note that $$ (0,10)=\bigcup_{n=1}^\infty (c_n,c_{n+1}]\;\cup\{\alpha\}\cup \;\bigcup_{n=1}^\infty [d_{n+1},d_n)$$

Now we can readily biject $(a_n,a_{n+1}]\to (c_n,c_{n+1}]$. This preserves rationality as all $a_n$ and all $c_n$ are rational. We can do the same with $[b_{n+1},b_n)\to [d_{n+1},d_n)$. Ultimately, this gives us an order-preseving bijection $$\bigl((0,1)\cap \Bbb Q\bigr) \cup \bigl((2,3)\cap \Bbb Q\bigr)\to \bigl((0,10)\cap \Bbb Q\bigr)\setminus\{\alpha\}.$$ But we can ignore that spurious "$\setminus\{\alpha\}$" if we pick $\alpha$ to be irrational!

Hagen von Eitzen
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