3
$\begingroup$

Previous questions established, for example, that a continuous bijection $f:(0,1)\to [0,1]$ does not exist, but the proof relied on continuity. Clearly, with continuity similar proofs can be invoked to show there is no bijection $f:(-\infty,\infty)\to[0,1]$.

Similarly, one can show there is no order-preserving bijection: Suppose there were, then there is an $x\in(-\infty,\infty)$ such that $f(x)=1$. But there exists some $y\in(-\infty,\infty)$ such that $y>x$, and for a bijection this requires $f(y)>f(x)=1$, which is a contradiction.

But what if the bijection need not be continuous? Is there a bijection, or can you prove there isn't?

$\endgroup$
  • $\begingroup$ Why do you tag this topology if there is no continuity involved? $\endgroup$ – Stefan Hamcke Dec 6 '13 at 18:33
  • $\begingroup$ If you don't take topology into account, then "compact" becomes meaningless... $\endgroup$ – Najib Idrissi Dec 6 '13 at 18:33
  • 1
    $\begingroup$ But note that an order-preserving bijection will be continuous and open - one can show that $f[(a,b)]=(f(a),f(b)).$ $\endgroup$ – Stefan Hamcke Dec 6 '13 at 18:36
  • $\begingroup$ Thanks, I suspected order-preservation implies continuity if the domain has no "gaps". $\endgroup$ – Nameless Dec 6 '13 at 18:48
  • $\begingroup$ @Nameless : this question or equivalent has probably been asked dozens of times on this forum. I don't feel like finding the duplicates right now. The easiest way to show a bijection exists is not by constructing one but by using the Canotr-Schroeder-Bernstein Theorem. $\endgroup$ – Stefan Smith Dec 6 '13 at 19:48
6
$\begingroup$

Without continuity there is a bijection.

Let $x_1=\frac{1}{2}, x_2=\frac{1}{3},.., x_n=\frac{1}{n+1},...$. Define $f(x_1)=1, f(x_2)=0, f(x_3)=x_1,..., f(x_n)=x_{n-2},..$ and $f(x)=x$ for all $x$ not in the sequence.

Then $f$ is a bijection from $(0,1)$ to $[0,1]$.

Now use the fact that $\frac{1}{2}+\frac{1}{\pi}\arctan(x)$ is a bijection from $(-\infty, \infty)$ to $(0,1)$.

$\endgroup$
  • 1
    $\begingroup$ That is a bijection from $(0,1)$ to $[0,1]$. You still need to compose it with something like $\frac12+\frac{x}{2\sqrt{1+x^2}}$ to get a bijection from $(-\infty,\infty)$ to $[0,1]$. $\endgroup$ – robjohn Dec 6 '13 at 18:43
3
$\begingroup$

If you don't demand continuity then compactness is irrelevant. The question becomes a set-theoretic one and therefore the analytical/topological properties of the sets in question are irrelevant. In particular, there is a bijection between the two sets you mention, as can be shown using standard techniques of elementary set theory.

$\endgroup$
3
$\begingroup$

Hint: Try the Schroeder-Bernstein theorem.

$\endgroup$

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.