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Let S be a connected subset of the reals. Then all the continuous functions on S satisfies the intermediate value theorem.

Is this correct? I am sure it is not when we use the version of IVT from Wikipedia with f(a) and f(b). But when I use the below version, I got a bit confused. Thank you very much! l

scsnm
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  • What does it mean for a continuous function on $[0,1)$ to satisfy the IVT? This has to be said. Otherwise the task is not doable. – amsmath Oct 24 '19 at 02:34
  • I think this is why I am considering the max min value version of IVT? – scsnm Oct 24 '19 at 02:37
  • But for $\frac 1x$ on $(0,1]$ there is no max. – amsmath Oct 24 '19 at 02:37
  • i am not even sure if these two versions are equivalent! – scsnm Oct 24 '19 at 02:37
  • They are....... – amsmath Oct 24 '19 at 02:38
  • What exactly are you confused about? Do you understand what the quoted theorem is saying? It implies your version since if $f(a) \leq y\leq b$ then of course $m\leq f(a)\leq y\leq f(b)\leq M$, so there is $c\in[a,b]$ with $f(c)=y$, right? – MPW Oct 24 '19 at 02:38
  • they are ?I didn't find any repetition online of this version. I found it in a textbook. – scsnm Oct 24 '19 at 02:39
  • Maybe the version you need says that given $a, b \in Im(f)$ with $a < b$ then for every $y \in \mathbb{R}$, $a < y < b$, there exists $x \in Dom(f)$ such that $f(x) = y$, does it? – ABP Oct 24 '19 at 02:40
  • The main question is: where did you get the claim from which is to prove? – amsmath Oct 24 '19 at 02:40
  • @ amsmath it's a GRE question. So they are equivalent, you are saying? – scsnm Oct 24 '19 at 02:44
  • Here is a question which may help you: https://math.stackexchange.com/questions/1663112/prove-x-is-connected-iff-for-every-continuous-fx-rightarrow-mathbb-r?rq=1 Just replace $X = S$ if you are not familiar with metric spaces. – ABP Oct 24 '19 at 03:04

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