Prove that $\exists$ $f:(a,b)\to [a,b]$ such that $f$ is Bijective.

Question

Let $a,b \in \Bbb{R}$ such that $a<b$ Then Prove that $\:\exists\:$ a continuous function $f:(a,b)\to [a,b]$ such that $f$ is Bijective.

I am not sure how to prove formally, But i tested the claim for $f(x)=x$. It is of course injective but not surjective since codomain is different from range.

EDIT: As per the link given in the comment section, this is what i understood. First let us assume $f:(a,b)\to [a,b]$ is onto. So $\exists$ $c,d \in (a,b)$ such that $f(c)=a,f(d)=b$. Without loss of generality let $c<d$.

Now Since $f$ is continuous, by IVT we have: $$f\left([c,d]\right)=[a,b]$$ But $(a,b)-[c,d]\ne \phi$ So $f$ cannot be injective.

Hence if $f$ is Surjective, $f$ cannot be injective. This implies that: If $f$ is Injective, $f$ cannot be surjective.

The best example for $f$ to be continuous and injective is $f(x)=x$. But I am unable to find example for $f$ to be continuous and surjective.

You cannot prove it because it is wrong. See for example https://math.stackexchange.com/q/42308/42969. — Martin R, Feb 05 '21 at 05:59
Continuous bijection between these two intervals will be a homeomorphism. — neelkanth, Feb 05 '21 at 06:28
Continuous and surjective is simple: Set $f(x) =a$ for $a < x \le a+\epsilon$, $f(x) = b$ for $b-\epsilon \le x < b$, and as a suitable linear function on $[a+\epsilon, b- \epsilon]$. — Martin R, Feb 05 '21 at 09:05
How about $f:(0,1)\to [0,1]$, $f(x)=\cos^2(2\pi x)$ is also continuous and surjective — Ekaveera Gouribhatla, Feb 05 '21 at 11:39

Prove that $\exists$ $f:(a,b)\to [a,b]$ such that $f$ is Bijective.

0 Answers0