Intermediate Value Theorem question

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The discussion centers on proving that a continuous function f on the interval [a,b], where f(a) < f(b) and f(s) ≠ f(t) for all s ≠ t, is strictly increasing. Participants clarify that the provided counterexample violates the condition that f(s) ≠ f(t), which is crucial for the proof. The Intermediate Value Theorem (IVT) is referenced to support the argument that there exists a value c in [a,b] such that f(c) equals any value between f(a) and f(b). The conversation also touches on minor language corrections regarding the terms "prove" and "proof." Overall, the proof's validity hinges on adhering to the specified conditions of the function.
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Consider a continuous function f in [a,b] and f(a) &lt; f(b). Suppose that \forall s \neq t in [a,b], f(s) \neq f(t). Proof that f is strictly increasing function in [a,b].

Homework Equations



I.V.T: If f is continuous in [a,b] and \gamma is a real in [f(a),f(b)], then there'll be at least one c in [a,b] such that f(c) = \gamma.

The Attempt at a Solution



This exercise is very strange to me. Besides I can apply the I.V.T to show that for any sub interval in [a,b] there will be an intermediate value in f(a), f(b), I can easily draw and counter example of what it pretends:
https://www.dropbox.com/s/dtj28xo4ilaai4z/pf.eps?dl=0

I am missing something important?

Thanks in advance!
 
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Portuga said:
Consider a continuous function f in [a,b] and f(a) &lt; f(b). Suppose that \forall s \neq t in [a,b], f(s) \neq f(t). Proof that f is strictly increasing function in [a,b].

Homework Equations



I.V.T: If f is continuous in [a,b] and \gamma is a real in [f(a),f(b)], then there'll be at least one c in [a,b] such that f(c) = \gamma.

The Attempt at a Solution



This exercise is very strange to me. Besides I can apply the I.V.T to show that for any sub interval in [a,b] there will be an intermediate value in f(a), f(b), I can easily draw and counter example of what it pretends:
https://www.dropbox.com/s/dtj28xo4ilaai4z/pf.eps?dl=0

I am missing something important?

Thanks in advance!

Your example violates the hypotheses of the claim: your function ##f(x)## has ##f(s) = f(t)## for several pairs ##(s,t)## with ##s \neq t##.
 
Picture doesn't make it. Blank screen. Can you copy/paste it in the post ?
 
BvU said:
Picture doesn't make it. Blank screen. Can you copy/paste it in the post ?
upload_2017-7-5_11-24-35.png
 
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I did the first one in libreoffice, so I made the mistake pointed by Ray.
 
Portuga said:
Portuga said:
Suppose that \forall s \neq t in [a,b], f(s) \neq f(t).
Your drawing violates the assumption that ##f(s) \neq f(t)##
Portuga said:
Proof that f is strictly increasing function in [a,b].

Minor point. The verb is "to prove". The noun is "proof".
 
Thank you! Now I got the point! Sorry for my poor English! Thank you very much. Now it's clear for me!
 
Portuga said:
Sorry for my poor English!
No need for an apology. Lots of native speakers of English also get this wrong (prove vs. proof), sometimes spelling "prove" as "proove."
 
Figure from link in OP:
upload_2017-7-8_11-40-23.png
 

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