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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!

K B
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    The range is an interval so it couldn't possibly have a hole since then it wouldn't be an interval. – Alex R. Nov 21 '13 at 22:19
  • By range, your prof means image, i.e. $f$ is meant to be surjective on an interval. – Doc Nov 21 '13 at 22:19

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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.

Ross Millikan
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  • Thank you! So how would that definition of the range help prove that the function is continuous if the range only has the values that f maps to? It doesn't seem like you can use that to show that f is in fact continuous – K B Nov 21 '13 at 22:29
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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.

Doc
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