I thought this question was a simple restatement of the extreme value theorem, but I got it wrong.

My attempt:

True. Let $$I:=[a,b]$$ Since $f$ is continuous on $I$, $\forall\epsilon>0,\exists\delta>0:\forall c\in I,0<\vert x-c\vert<\delta\Rightarrow \vert f(x)-f(c)\vert<\epsilon$

Since $I$ is closed and bounded by $a$ and $b$, by the Extreme Value Theorem, $f$ is bounded on $I$. Since $f$ is bounded, it has a least upper bound and a greatest lower bound. Let $$M=\sup_{x\in I}f(x) \text{ and }m=\inf_{x\in I}f(x)$$ Also by the Extreme Value Theorem, we know $\exists x_M,x_m\in I$ such that $$f(x_M)=M,f(x_m)=m$$ We know $J:=f(I)$. Then $$J=[m,M]$$

  • $\begingroup$ That's a fine proof. $\endgroup$ – Bernard Nov 3 '15 at 22:12
  • $\begingroup$ Why does $J$ contain $(m+M)/2$, for example? $\endgroup$ – John Dawkins Nov 3 '15 at 22:14

Your proof is incomplete: what the extreme value theorem says is that $m$ and $M$ (minimum and maximum of $f$ over $[a,b]$ exist and so $f(I)\subseteq[m,M]$.

However you can conclude by using the intermediate value theorem.


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