# Proving the continuity of the inverse of a continuous, stricly monotonic function

Please tell me if the following is correct.

We have a continuous, strictly monotonic, increasing function on some closed and bounded interval $$I$$, and $$x_0\in I$$. Let $$g(y)$$ be its inverse, and $$f(x_0)=y_0$$. I want to show that $$|g(y)-g(y_0)|<\epsilon\implies|y-y_0|<\delta$$. \begin{align*} |g(y)-g(y_0)|<\epsilon&\Leftrightarrow x_0-\epsilon If I consider only the $$y$$s that are extremely close to $$y_0$$, then I think that I can set $$\delta = \min(y_0-f(x_0-\epsilon), f(x_0+\epsilon)-y_0)$$. For the case where the function is decreasing, I just flip the inequality symbols from line one to two, and take $$\delta=\min(f(x_0-\epsilon)-y_0,y_0-f(x_0+\epsilon))$$.

• You need to mention $f$ is strictly increasing, otherwise you can't guarantee the function is bijetive – user732848 Aug 6 '20 at 16:41
• right, I mixed up monotonic and "strictly". – Luyw Aug 6 '20 at 16:44
• @Shamim is the approach correct, though? – Luyw Aug 6 '20 at 16:45
• Indeed but a bit formalization would be enough. You can just use the continuity of $f$, that will do the job – user732848 Aug 6 '20 at 16:46
• I also assume you meant 'closed and bounded' interval? Because if not, it doesn't make sense – user732848 Aug 6 '20 at 16:52

I don't prove that $$f$$ is bijective, as it's very easy. Just injectivity requires the monotonicity and surjectivity requires IVT. We know since $$f$$ is continuous on $$[a,b]$$ then for any $$\varepsilon >0$$ one can find $$\delta >0$$ such that $$|x-c|<\delta$$ implies $$|f(x)-f(c)|<\varepsilon$$. Now there is a unique $$y,y_0$$ such that $$|x-c|=|g(y)-g(y_0)|$$ and this solves the problem. (Just change the job of epsilon and delta in your approach)
• Note that I define $y=f(x), y_0=f(c)$ – user732848 Aug 6 '20 at 16:55