# Is there a bijection from $A = ]0,1[$ to $B = A \cup \{1,2,3,4\}$?

Is there a bijection from $A = ]0,1[$ to $B = A \cup \{1,2,3,4\}$? If there is, give example.

I had this question on exam few days ago, and I have been googling for days, but I can't find a solution. I think (probably wrong) that there is not because $A$ is subset of $B$, but I am totally unsure.

Usually on exams we get function and test it this time, we got this, and no one knows how to solve it.

• What does your notation $\langle 0,1\rangle$ mean? – Bungo Jul 7 '16 at 0:12
• It means all numbers between 0 and 1, excluding 0 and 1 – Vulisha Jul 7 '16 at 0:13
• OK, the answer is, there is a bijection between $A$ and $B$. Let $(a_n)_{n=1}^{\infty}$ be any sequence of distinct elements of $A$ (e.g. an enumeration of the rationals contained in $A$), and define $f(a_1) = 1$, $f(a_2) = 2$, $f(a_3) = 3$, $f(a_4) = 4$, then for all $n > 4$ define $f(a_n) = a_{n-4}$. For elements $x\in A$ which are not represented in the sequence $(a_n)$, define $f(x) = x$. – Bungo Jul 7 '16 at 0:15
• Oh, it looks so simple now, it looks like one of easiest problems(since usually we have to solve it on full page :P ) , but I would have never thought of it since all I practiced did not look nearly abstract like this example. Well Bungo, thank you very much sir! – Vulisha Jul 7 '16 at 0:39
• @Vulisha: you could write up the answer suggested by Bungo, then (after an enforced delay) accept it. That way the question doesn't stay open. We encourage self answers if you find the answer after you post the question. – Ross Millikan Jul 7 '16 at 1:07

Let $(a_n)_{n=1}^{\infty}$ be any sequence of distinct elements of $A$ (e.g. an enumeration of the rationals contained in $A$), and define $f(a_1) = 1$, $f(a_2) = 2$, $f(a_3) = 3$, $f(a_4) = 4$, then for all $n > 4$ define $f(a_n) = a_{n-4}$. For elements $x\in A$ which are not represented in the sequence $(a_n)$, define $f(x) = x$.