# Pointwise Convergence on an Infinite Intersection of Function Images

This is the most interesting real analysis question I have run into thus far (that I understand):

For all $n\in\mathbb{N}$, let $f_n(x)$ and $f(x)$ be one-to-one continuous functions such that $$B=\bigcap_{n=1}^{\infty}f_n(A)$$ is a nonempty interval. If $f_n(x)$ converges pointwise to $f(x)$ on $A$, then $f_n^{-1}(x)$ converges pointwise to $f^{-1}(x)$ on $B$.

Do you think this is a valid assertion, or are other conditions necessary for it to hold water?

Edit 1: Because $f_n\to f$, is it safe to assume that $f:A\to B$? If so, then could I conclude that $f$ is bijective on $A$ and therefore $f^{-1}:B\to A$? I still feel that this is not rigorous enough, however.

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It seems intuitive to me, but continuity is definitely a keypoint here. I am not sure it can be done in such generality though, but I'll give it a try when I won't be half asleep trying to TeX on my cellphone. –  Patrick Da Silva Mar 16 '12 at 4:20
@PatrickDaSilva, I was thinking the same, and I was also thinking that the $f_n$ may need to be differentiable (for reasons that I have yet to fully go over). By the way, I take off my hat to you for checking M.SE on your phone! –  Josué Molina Mar 16 '12 at 4:31
After thinking about it I think this question is really hard and deserves more than 15 minutes to prove. Maybe I'll give it a thought some time later. –  Patrick Da Silva Mar 16 '12 at 17:05
Here $A$ is an arbitrary set of reals, or an interval ? –  Ewan Delanoy Mar 18 '12 at 7:00
I can show that this is true when $A$ is a finite union of intervals. It suffices to treat the case when $A$ is a single interval.
Patrick da Silva is absolutely right : continuity is definitely a keypoint here. If $A$ is an interval, then a continuous function on $A$ must be monotonous. So $f$ (and each $f_n$ also) is decreasing or increasing. We may assume that $f$ is increasing. Then for large enough $n$, each $f_n$ will be increasing also (indeed, let $a<b$ in $A$, then $f(a)<f(b)$, so $f_n(a)<f_n(b)$ for all large enough $n$).
Now assume by contradiction that $(f_n^{-1}(y))$ does not converge to $f^{-1}(y)$ for some $y\in B$. This means that some subsequence stays away from $f^{-1}(y)$. Passing to a subsequence if necessary, we may assume that there is a $\varepsilon >0$ such that $f_n^{-1}(y) \in A \setminus [f^{-1}(y)-\varepsilon,f^{-1}(y)+\varepsilon]$ for all $n$. Passing to a subsequence again, we may further assume that $f_n^{-1}(y)$ is always on the same side of $f^{-1}(y)$, with e.g. $f_n^{-1}(y) \geq f^{-1}(y)+\varepsilon$. Now $f^{-1}(y)+\varepsilon \in A$ since $A$ is an interval. Since $f_n$ is increasing, we deduce $y=f_n(f_n^{-1}(y)) \geq f_n(f^{-1}(y)+\varepsilon)$. Now the right-hand side converges to $f(f^{-1}(y)+\varepsilon)$. Passing to the limit, one obtains $y \geq f(f^{-1}(y)+\varepsilon)$. And since $f^{-1}$ is increasing, we deduce $f^{-1}(y) \geq f^{-1}(y)+\varepsilon$, a contradiction.