Talk:Intermediate Value Theorem

I have a question : In this proof and on which it is based on (Image of Interval by Continuous Function is Interval) why are we not using the fact that all connected sets on the real line are intervals (closed, open or half-closed or half-open etc.). The proof becomes much simpler because of it. --Hamood Khan @ 03:30, 16 July, 2013 (PST: Pakistan Standard Time).


 * Because this is how it appears in the text from which this is sourced. --prime mover (talk) 22:44, 15 July 2013 (UTC)

As a second suggestion (rather recommendation) I would request that the theorem be proven in its full generality, i.e., for all real, continuous functions of several varaibles $f: D \to \R$, where $D \subset \R^n$ is a connected set, the image of $f$ is also connected and hence an interval (because all connected subsets of $\R$ are intervals). --Hamood Khan @ 03:45, 16 July, 2013 (PST: Pakistan Standard Time)


 * The page as it appears stays as it is. If you want to write a general version then feel free. --prime mover (talk) 22:44, 15 July 2013 (UTC)

Theorem is false as stated
Continuity of f is required on [a..b]. See https://math.stackexchange.com/questions/1615108 --barto (talk) 08:21, 29 April 2017 (EDT)


 * Yes, it doesn't bode well for our reputation if they're talking about us behind our backs on other websites. I understand the whole world takes it as a truism that is utter rubbish, but I never realised just how rubbish we were.


 * How nice it would be if such mistakes were raised directly with us, rather than going to StackExchange.


 * Should we give up?


 * I think this calls for an errata page to be raised concerning the initial statement of this theorem in Binmore. --prime mover (talk) 09:21, 29 April 2017 (EDT)


 * Never give up. We have to confront reality. There's much left to do, but the fact that is very unique gives hope; see my comment on this question: https://math.stackexchange.com/questions/2183803 --barto (talk) 09:30, 29 April 2017 (EDT)


 * Your support is appreciated. Every little helps. --prime mover (talk) 11:31, 29 April 2017 (EDT)


 * The problem stemmed with me, misinterpreting the theorem and restating it (inaccurately) back in 2009. The version I have posted up should now be correct, as it ensures that the function is continuous on the interval containing the endpoints. --prime mover (talk) 17:23, 29 April 2017 (EDT)


 * $S$ and $I$ make it unnecessarily complicated. Compare with the version of 17:47, April 29, 2017 ‎(I made this for a reason): it says exactly the same, but the newer one is less transparent. --barto (talk) 18:28, 29 April 2017 (EDT)


 * Not as general though. The original statement says that $f$ is a (real) function which is continuous on an interval such that $a$ and $b$ are within that interval. This is how it is presented in Binmore (I had somehow managed to misread it) and indeed how it had been originally presented in the initial exposition by "Black". The reason for $S$ is to make it so that it is clear it applies to a function defined on any subset of $\R$ -- as the initial statement did not give any domain for $f$, and I thought it might be important.


 * The version you provide expresses a function which is defined only on $[a..b]$.


 * Whatever. Analysis is the suckiest most boring heap of mammalian solid effluent in the fornicating universe. --prime mover (talk) 18:40, 29 April 2017 (EDT)


 * Yes, precisely. What this does is allowing the function to be defined on a larger set. But when applying a theorem like this, we tend to (silently) apply it to the restriction of f to the interval in question anyway, so there's no need to make this theorem statement more complicated if it's just for that. This is also what we do constantly in abstract algebra when dealing with (linear) mappings and subspaces.


 * Imagine we'd start each theorem in analysis along the lines: "Let S be a set. Let T be a subset which has a topology. Let U be a subset on which f is continuous. Let V be a subset of U such that etc etc..." only to avoid the word 'restriction'. --barto (talk) 02:41, 30 April 2017 (EDT)