Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

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 $$f(x)=\lim_{m\rightarrow\infty}r_m$$

share|improve this answer
    
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
show 9 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.