Talk:Extreme value theorem

I propose the following as an informal introduction:

The extreme value theorem in mathematics is a way of stating that a smooth line has one or more highest and lowest points where no other point is higher than the highest point and no other point is lower than the lowest point. 213.55.136.205 01:28, 14 Dec 2004 (UTC)

Perhaps we should add to the first line " [...] if a real-valued non-constant function f is continuous in the closed and bounded interval [a,b], then f must attain a maximum and a minimum", if only for clarity. It is true that we can have a max(f)=min(f) (as is the case for constant functions), but the question "what if f is constant" is one that I hear in every class I have taught this in. I think the confusion lies in a misunderstanding that the minimum is necessarily smaller than the maximum, perhaps we ought to clarify this? — Preceding unsigned comment added by Okran (talk • contribs) 14:27, 29 November 2014 (UTC)

The Boundedness Theorem

I was readig over this and remembered reading Uniform boundedness principle, I'm not sure it's the same theorem. Webhat 08:02, 13 May 2006 (UTC)

Boundedness is not weaker than having a max/min.

I disagree with the statement that the boundedness theorem is "A weaker version of this[Extreme Value] theorem". Every closed and bounded interval has a maximium and a minimum value, so the Extreme Value Theorem is a consequence of Boundedness--it is because the interval is bounded that we know it has a max/min. So, they are actually the same theorem, but Boundedness is more general--which (in my opinion) makes it stronger, not weaker: if an interval is closed and bounded we know that it has a max/min, but there are other properties of closed and bounded intervals in addition to just having a max/min.

I wish you people would just occasionally write a paragraph about what this sort of thing actually means. I came to this page from Laffer curve, hoping that it would be illuminating. But it isn't. I'd need to spend a while rereading this before I could grasp what it's actually saying. Much of our statistical and economics output is like this. Grace Note 00:54, 19 October 2007 (UTC)

Agreed About: "Boundedness is not weaker"

I absolutely agree with that statement. I believe that we should not allude to the fact that the boundedness theorem is a substitute or another weaker version of the EVT. I suggest we delete the word "weaker" and just mention that it's related.(Helios2k6 (talk) 00:52, 7 November 2008 (UTC))

Non standard proof

Is N a "infinite hyperinteger" (first occurence in the section) or a finite (later in the section N is called finite)?

OK, I may understand that the first N is indeed infinite, but it is fixed once for all. Then the second N is a generic finite integer, that should not be called the same way, or more caution is needed in the wording. Also, I would remove the parenthesis around the corresponding sentence "(Note that..."), as I understand that it announces the use of "transfer". Bdmy (talk) 10:54, 21 December 2008 (UTC)

I am not necessarily against hyper-arguments, but my feeling is that the section as it stands is rather counterproductive if the intention is to advocate for non-standard. I feel that some sentence like: "one shows here the power of the tools on a VERY simple result in Analysis, where of course the old method is shorter." One could try to make the reader feel why of course the proof is basically the same as old dichotomy. Bdmy (talk) 09:42, 21 December 2008 (UTC)

The parenthetical remark was supposed to motivate the application of the transfer principle. If you feel that it is confusing instead, perhaps we should remove it. I am surprised you say the old method is shorter. I was thinking on the contrary that the NSA method is more elegant and shorter. The basic idea is that the somewhat cumbersome manipulations with sequences are replaced by the choice of a single hyperinteger. Of course the latter are defined in terms of sequences; see Epilogue to Keisler's textbook. That's in my mind the connection between the two. What I just wrote seems a bit too wordy for inclusion in the article, though. I tried it at uniform continuity but it got deleted a couple of times already, so I am ready to give up on this somewhat editorializing comment. Katzmik (talk) 16:49, 21 December 2008 (UTC)

To the contrary, I feel that removing the parenthesis is more clearly a motivation before applying the transfer principle, but I would just rename the finite generic integer, perhaps just n instead of N that is fixed. Also I should agree that the classical proof is not really shorter, but we know it by heart after years of teaching. I don't agree with cumbersome though. Bdmy (talk) 17:19, 21 December 2008 (UTC)

An alternative would be to mention the standard setting inside the parenthesis, which is what I did in reaction to your justified criticism. Either way is OK as far as I am concerned. Katzmik (talk) 17:23, 21 December 2008 (UTC)

Proof in the standard setting.

What we present is really much more complicated then it needs to be. For example, Spivak gives the following proof in his calculus text. (And there may be slicker proofs, this was just the first I looked at.)

Suppose ƒ is continuous on [a,b]. First we prove ƒ is bounded. Let A={x : a≤ x ≤ b and f is bounded on [a,x]}. The set A is non-empty because of the continuity of ƒ at a. The set A is bounded above by b. Let β=sup A. Suppose β<b, since ƒ is continuous at β it is bounded on an interval around β. Contradiction, hence β=b, which implies that ƒ is bounded on [a,x] for every x < b. But ƒ is continuous at b, and hence f is bounded on [a,b].

Now we prove that f achieves its maximum.

Let B={ƒ(x): a≤ x ≤ b}. B is clearly non-empty, and is bounded above by previous argument. Let α=sup B. Let g(x)=1/(α-ƒ(x)). Suppose there is no x in [a,b] so that f(x)=α. Then g(x) is continuous in [a,b] and hence bounded. On the other hand the values of ƒ are arbitrarily close to α, and hence g(x) is unbounded. Contradiction. So there must be some x in [a,b] so that f(x)=α.

The other comment about this proof is that if you find avoids sequences, if you like that sort of thing ;). Thenub314 (talk) 19:19, 21 December 2008 (UTC)

I don't totally agree with you, I find that the subsequence proof is (for my personal taste) the most intuitive. But it could be interesting to present in this basic article several points of view (I mean: include also this other classical proof you mention). Personally, I tend to think that some proofs by least upper bound have an appearance of "black box" (like non-standard proofs?).

Another interesting point, but I am not expert enough to deal with this, would be to point the fact that this theorem is not "calculable", and not admitted by Intuitionists. Bdmy (talk) 19:56, 21 December 2008 (UTC)

I find it more intuitive as well, but somehow I the one above involved less machinery. Somehow the result is so intuitive I am not sure the proofs help make it any clearer. It is an interesting question, what to say about intuitionists. There are there not quite identical constructivist schools of thought. I know for a fact this fails in two of them, but I would need to check it out for the third (though I strongly suspect it does). Thenub314 (talk) 20:57, 21 December 2008 (UTC)

I don't know I am very undecided what we should have in the way of proofs. The manual of style makes the point that Wikipedia is not a textbook (the wikibooks project aims at that) and proofs should be included when, in our best judgment they make the article clearer. I always try to keep in mind I write for a fictitious 10 year old (my image of myself back then) who wants to know more so he can write his school report. That being said, the result seems so clear everything that is interesting about it lies in its proof. So several proofs may not be as hazardous as I initially thought... Thenub314 (talk) 22:51, 21 December 2008 (UTC)

Missing step in the non-standard proof.

When the steps were filled in from Keisler's book there was a small gap that was not filled in. That is, we need to know that $f(x_{i_{0}})$ is finite in order to be able to apply the standard part function. This to my mind requires boundedness theorem. Glancing back at Keisler he does not cover the boundedness theorem, and does not include a complete proof of the extreme value theorem. Does anyone have suggestions for a basic introductory text that proves these facts that we could reference? Thenub314 (talk) 18:52, 7 June 2011 (UTC)

Are examples misleading?

Examples in "Functions to which theorem does not apply" seem to mislead. They discuss about function not being bounded from above where as the theorem assumptions are only on the interval. And f(x) = tan x, x \in [0,pi/2] though the function is not bounded from above, interval is closed and compact and f(pi/2) = \infty is the maximum. Here, I am considering range of f to be extended real numbers. Vinay h (talk) 03:39, 23 June 2012 (UTC)

Extension to multi-dimensional spaces, metric spaces and arbitrary topological spaces

Can someone please describe the extension of the extreme value theorem to multi-dimensional functions, arbitrary metric spaces (if it exists) and general topological spaces? Thanks! --Erel Segal (talk) 17:31, 25 August 2014 (UTC)

We have "This implies the following generalization of the extreme value theorem: a continuous real-valued function on a nonempty compact space is bounded above and attains its supremum." "Bounded" and "extreme value" are problematic if the codomain of the function is not a linearly ordered space, and the result is unlikely to be true unless the codomain is complete. In other words, the codomain should be the reals. — Arthur Rubin (talk) 21:36, 26 August 2014 (UTC)

Proof of the Generalized Case

The proof of the generalized case is very simple, short and instructive. I remember it but unfortunately, I don't remember any source reference, so I can't include it in the article. Here's the proof I mean:

Let $V,\ W$ be topological spaces, $f:V\to W$ a continuous function and $K\subseteq V$ compact. Let $\bigcup {\mathcal {C}}\supseteq f(K)$ be an arbitrary open cover of $f(K)$. Because f is continuous, $\bigcup \left\{f^{-1}(C):C\in {\mathcal {C}}\right\}\supseteq K$ is an open cover of $K$. By compactness of $K$, we can select a finite ${\mathcal {F}}\subseteq {\mathcal {C}}$ such that $\bigcup \left\{f^{-1}(C):C\in {\mathcal {F}}\right\}\supseteq K$ is still an open cover of $K$. But then $\bigcup \left\{C:C\in {\mathcal {F}}\right\}\supseteq \bigcup \left\{f(f^{-1}(C)):C\in {\mathcal {F}}\right\}=f\left(\bigcup \left\{f^{-1}(C):C\in {\mathcal {F}}\right\}\right)\supseteq f(K)$ is a finite cover of $f(K)$. Because ${\mathcal {F}}\subseteq {\mathcal {C}}$, it is an open cover. As $\bigcup {\mathcal {C}}\supseteq f(K)$ was arbitrary, we can conclude that $f(K)$ is compact. #

Herein, $f^{-1}$ refers to the preimage as explained in Continuous_function#Continuous_functions_between_topological_spaces. --Carsten Milkau (talk) 12:30, 13 July 2021 (UTC)