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Suppose $f$ satisfies the intermediate value property, i.e. if $f(a)<c<f(b)$, then there exists $a<x<b$ such that $f(x)=c$ and for every rational $r$, $S_r$ such that $f(x)=r$ is a closed set. Prove $f$ is continuous.

This looks pretty daunting. I am guessing it uses some sort of sequential continuity argument, but I am somewhat lost. Hints would be most appreciated.

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up vote 3 down vote accepted

suppose $f$ has at least one discontinuity point $x$, i.e. $$ (\exists\epsilon>0)(\forall\delta>0)(\exists x'\in(x-\delta,x+\delta))\quad|f(x')-f(x)|\ge\epsilon\text; $$ in particular, for each $n\in\mathbb N$ we can choose $x_n$ so that $$ |x_n-x|<\frac1{n+1}\quad\&\quad|f(x_n)-f(x)|\ge\epsilon\text; $$ choose $$ r_+\in(f(x),f(x)+\epsilon)\cap\mathbb Q \\ r_-\in(f(x)-\epsilon,f(x))\cap\mathbb Q $$ then, by the IVP, there exists some $y_n$ between $x_n$ and $x$ such that $$ f(y_n)=r_+\quad\text{or}\quad f(y_n)=r_- $$ (depending whether $f(x_n)>f(x)$ or $f(x_n)<f(x)$); in particular, $$ y_n\rightarrow x\text; $$ suppose the '$+$' case occurs infinitely often (otherwise consider the '$-$' case): the corresponding subsequence lies in $f^{-1}(r_+)$ (closed by assumption), thus $$ f(x)=r_+\text, $$ in contradiction to the choice of $r_+$. this shows that no discontinuity point exists, i.e. $f$ is continuous.

EDIT my previous attempt (below) was wrong, disregard

consider the preimage of $I\cap\mathbb Q$, where $I=[a,b]$, and show that its closure is the preimage of $I$.

for any $x\in \overline {f^{-1}(I\cap\mathbb Q)}$ take a sequence $\{x_n\}$ in $f^{-1}(I\cap\mathbb Q)$ converging to $x$. then either infinitely many $x_n$'s lie in one of the $f^{-1}(r)$'s, or each $f^{-1}(r)$ only contains finitely many $x_n$'s (so the sequence intersects infinitely many $f^{-1}(r)$'s).

  • in the first case, the subsequence lying in $f^{-1}(r)$ (closed) converges in this set, so $x\in f^{-1}(r)$.

  • in the second case, construct a sequence $\{r_m\}$ and show $$f(x)=\lim_{m\rightarrow\infty}r_m$$

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I saw it as a comment, thinking... :) – Lost1 Apr 9 '13 at 13:42
It is a union of lots of closed sets, I could not see why that helps us. – Lost1 Apr 9 '13 at 13:43
we have $a\leq f(x_n)\leq b$, so it must have a convergent subsequence. Let $x_{n_k}$ denote this subsequence. I am not sure how to show this $f(x_{n_k})$ converges to $f(x)$. I can see why this is true, if for any arbitary $N$, there exists $n,m> N$ such $f(x_m)<f(x)<f(x_n)$ thus we can appeal to intermdeiate value property. – Lost1 Apr 9 '13 at 14:21
every sequence has a monotone subsequence. one side of your last inequality is then enough because then (e.g. assuming $\{x_n\}$ monotone decreasing) $f(x)<f(x_{n+k})<f(x_n)$ – suissidle Apr 10 '13 at 5:55
Yes I realised like this, but I didn't see how that worked. we only shown $f(x_{n})$ is bounded below by $f(x)$. How does that mean it actually converges to $f(x)$. Also, suppose, we now proved the closure of pre-image is the the interval $[a,b]$, why does that mean we are done? – Lost1 Apr 10 '13 at 10:22

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