0
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ What does your notation $\langle 0,1\rangle$ mean? $\endgroup$ – Bungo Jul 7 '16 at 0:12
  • $\begingroup$ It means all numbers between 0 and 1, excluding 0 and 1 $\endgroup$ – Vulisha Jul 7 '16 at 0:13
  • 1
    $\begingroup$ 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 $f(x) = x$. $\endgroup$ – Bungo Jul 7 '16 at 0:15
  • $\begingroup$ 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! $\endgroup$ – Vulisha Jul 7 '16 at 0:39
  • $\begingroup$ @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. $\endgroup$ – Ross Millikan Jul 7 '16 at 1:07
1
$\begingroup$

With the help of Bungo we got the solution. Answer is there IS bijection ,because in sets A and B there is uncountably infinite numbers regardles of additional {1,2,3,4} in B set so we can define function that is one to one and onto (injection and surjection). This is type of " Hilbert's hotel paradox"

Example as Bungo said would be:

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 $f(x) = x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.