Show an increasing function with an interval as a range is continuous My professor gave us this problem out of his book:
Show, if f: [a, b] --> R is an increasing function and the range of f is an interval, then f is continuous
I'm not sure if I'm understanding correctly though. Wouldn't "the range of f is an interval" just mean that there is some A=[f(a),f(b)]? Why would that mean that f is continuous? Couldn't there be a hole in that range?
Any help would be appreciated!
 A: For this problem, the range has to be understood to include only those values that the function maps to.  So, for example, if $f(x)=\begin {cases} x& 0 \le x \le 1 \\x+1& 1 \lt x \le 2 \end {cases}$ the range would be $[0,1] \cup (2,3]$, not $[0,3]$.  Otherwise, you are correct, there could be a hole.
A: To prove that $f$ is continuous on $[a,b]$ it suffices to show that the preimage of any closed set in $[f(a),f(b)]$ is closed.  It suffices to prove this for closed intervals, since any closed set in $[f(a),f(b)]$ is the union of subintervals and points.  
So let $I$ be a subinterval in $[f(a),f(b)]$. Then by surjectivity $I=[f(s),f(t)]$ for some $s$ and $t$ in $[a,b]$.  We claim that the preimage of $[f(s),f(t)]$ is $[s,t]$.  To see this let $c\in [f(a),f(b)]$.  Then $c=f(r)$ for some $r\in [a,b]$ (again by surjectivity).  We now must show that $r\in [s,t]$.  But this follows since $f$ is increasing. I.e, if $r<s$ or $r>t$ then we get the respective contradictions $f(r)<f(s)$ and $f(r)>f(t)$. We're done since $[s,t]$ is obviously closed.
