2
$\begingroup$

Let $a < b$ be real numbers, and let $f:[a,b]\to\textbf{R}$ be a function which is both continuous and strictly monotone increasing. Then $f$ is a bijection from $[a,b]$ to $[f(a),f(b)]$, and the inverse $f^{-1}:[f(a),f(b)]\to[a,b]$ is also continuous and strictly monotone increasing.

My solution

To start with, let us prove that $f$ is bijective.

Since $f$ is strictly monotone increasing, it is injective.

Indeed, suppose that $x \neq y$. Then either $x > y$ or $x < y$. In the first case, $f(x) > f(y)$ and in the second case $f(x) < f(y)$. In both case, $f(x)\neq f(y)$. Thus $f$ is injective.

We do also have that $f$ is surjective. Indeed, given $y\in[f(a),f(b)]$, due to the intermediate value theorem there corresponds a $c\in[a,b]$ such that $f(c) = y$. Thus $f$ is surjective.

Let us now prove that $f^{-1}$ is strictly increasing. Let $f(a) > f(b)$. Then $a > b$, otherwise we would have $f(a) < f(b)$, which contradicts our assumption. Consequently, $f$ is strictly increasing.

Now it remains to prove that $f^{-1}$ is continuous, but I get stuck.

Can someone prove this last part using only the $\varepsilon-\delta$ definition or its sequential characterization?

Any comments on the previous attempts are welcome as well.

$\endgroup$
1
  • $\begingroup$ $f$ maps open intervals to open intervals. $\endgroup$
    – Ningxin
    May 16, 2020 at 22:11

2 Answers 2

1
$\begingroup$

Let $c=f(a), d=f(b) $ and $g=f^{-1}$. You can directly prove that $g$ is continuous on $[c, d] $. Let $p\in(c, d) $ so that $g(p) \in(a, b) $. Thus there exists a positive number $\epsilon_0$ such that $(g(p) - \epsilon_0,g(p)+\epsilon_0)\subseteq (a, b) $ (in particular you can take $\epsilon _0=\min(g(p)-a,b-g(p))$).

Consider an arbitrary $\epsilon >0$ and let $\epsilon'=\min(\epsilon, \epsilon_0)$. Then we have $(g(p) - \epsilon ', g(p) +\epsilon') \subseteq (a, b) $. It follows that $r=f(g(p)-\epsilon'),s=f(g(p)+\epsilon ') $ both lie in $(c, d) $ and $r<p<s$. Let us write $\delta=\min(p-r,s-p) $ so that $(p-\delta, p+\delta) \subseteq (r, s) $ and therefore $$g((p-\delta, p+\delta)) \subseteq g((r, s)) \subseteq (g(r), g(s)) =(g(p) - \epsilon', g(p)+\epsilon') \subseteq (g(p) - \epsilon, g(p) +\epsilon) $$ And this proves that $g$ is continuous at $p$. The proof can be easily modified/adapted for the case when $p=c$ or $p=d$.

The proof uses the fact that both $f, g$ are strictly increasing on their domains and you should be able to figure out where this has been used in above proof.

$\endgroup$
-1
$\begingroup$

Take any $x_n \rightarrow x$ and $f$ continuous, then $f(x_n) \rightarrow f(x)$. But then $f^{-1}(f(x_n)) = x_n \rightarrow x$ by definition. On the other hand, if $y_n = f(x_n)$ and $y = f(x)$, that means $f^{-1}(y_n) \rightarrow f^{-1}(y)$, so by the sequential criterion for continuity, $f^{-1}$ is continuous.

$\endgroup$
1
  • $\begingroup$ Here you have an arbitrary sequence $x_n\to x$. But by definition of sequential continuity you should start with an arbitrary sequence $y_n\to y$ and define $x_n=f^{-1}(y_n)$. $\endgroup$
    – Paramanand Singh
    May 17, 2020 at 0:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .