Talk:Infimum and supremum

The relation at the end was m < x, not m <= x
I fixed it. Priestoferis (talk) 18:28, 16 December 2009 (UTC) kjopj 'ioiunjjj — Preceding unsigned comment added by 202.91.82.69 (talk) 08:39, 4 July 2012 (UTC)

Any bounded nonempty subset has infimum
Hmmm... I can't fix this myself because I'm not 100% sure of the answer, but I wish that the entry made it 100% clear that any bounded nonempty subset of the real numbers has an infimum in the non-extended real numbers.


 * I added this. - Patrick 08:38, 25 Aug 2003 (UTC)

Wrong?
The infimum and supremum of S are related via
 * $$\inf(S) = -\sup(-S)$$.

? Depends on the set. --Abdull 12:04, 2 December 2005 (UTC)
 * That sounds correct assuming S is a set of real numbers and −S is defined as the set {s|−s∈S}, that is the set found by negating the elements of S. Then this is simply stating that inf and sup are symmetric in a sign sense. —BenFrantzDale 12:25, 2 December 2005 (UTC)

Confusion
So is it true that to find an infimum, one must know a set and a subset that you want to find the infimum of? And in what cases would the infimum *not* belong to the said subset? The example given on this page:
 * $$\inf \{ (-1)^n + 1/n \mid n = 1, 2, 3, \dots \} = -1$$

doesn't seem to work, since :$$\{n = 1, 2, 3, \dots \}$$ doesn't seem to be a subset of $$\{(-1)^n + 1/n \} \!\ $$. Can anyone help? Fresheneesz 01:58, 22 March 2006 (UTC)
 * I think the $$\{n = 1, 2, 3, \dots \}$$ is merely specifying that n must be a natural number in $$\{(-1)^n + 1/n \}$$. The infimum specified is the infimum of the set of all $$(-1)^n + 1/n$$ with natural n – 147.188.225.242 10:18, 15 May 2006 (UTC)
 * Yes, to find an infimum of a set you need an "enclosing set" to look within. The notation $$\{ (-1)^n + 1/n \mid n = 1, 2, 3, \dots \} $$ means just one set: "The quantity $$(-1)^n + 1/n$$ evaluated in turn for each of $$n = 1, 2, 3, \dots$$.  Since that example is in the section "Infima of Real Numbers", it's assumed that the "enclosing set" is $$\mathbb{R}$$. —The preceding unsigned comment was added by 65.57.245.11 (talk) 01:02, 6 February 2007 (UTC).

Instead of "that is smaller than all other elements of the subset", shouldn't it be "that is not bigger than all other elements of the subset"

Is it me or is this wrong?
In mathematics the infimum of a subset of some set is the greatest element, not necessarily in the subset

The "greatest element" is defined as a the largest member of the subset, here it says it doesn't have to be .. isn't this a contradiction? —Preceding unsigned comment added by 217.44.0.55 (talk) 14:38, 9 April 2008 (UTC)
 * The statement is correct. "Greatest element" is the largest member of "some set" (consider, T) that is not bigger than all members of the "subset" (S). It is not necessary that this element belong to the subset (S). This is illustrated by several of the Examples. --Loresayer (talk) 02:35, 1 November 2009 (UTC)
 * The wording "not bigger than all members of S" could be interpreted as either:

Am I right?84.81.167.158 (talk) 22:48, 18 May 2020 (UTC)
 * not bigger than any of the members of S [which, I think, is intended here];
 * not bigger than all, but bigger than some, members of S which seems not to comply with the definition given in the article].

It seems that u r right.. 2409:8930:3E13:D0B:CC64:7E55:D425:22C8 (talk) 17:31, 11 March 2021 (UTC)

Supremum really a dual to infimum?
Currently, the article says
 * Infima are in a precise sense dual to the concept of a supremum.

In Wikipedia, the article Duality (order theory) says every partially ordered set P gives rise to a dual partially ordered set. For one subset, I can see that there is exactly one infinum. How does a single infinum lead to a dual "supremum" poset? Thanks, --Abdull (talk) 21:23, 21 February 2010 (UTC)

Yes, they are really dual. Given a poset P and a subset S, we can reverse the ordering and form a dual poset P*, containing S*. (S* is the dual of S, the same subset of elements, but with the ordering reversed as it is in P*). Now, the statement that infimum and supremum are dual just means

inf(S) =  sup(S*) sup(S) =  inf(S*)

Cgwaldman (talk) 23:23, 27 January 2012 (UTC)

Example figure
I think the example in the figure is not very illustrative because it shows a very special case. The one in the supremum page seems slightly better, though I would change the color of the supremum itself. 212.126.224.100 (talk) 11:25, 5 October 2010 (UTC)

The infimium figure is useless to me, as I am red green colorblind. Automaticgiant (talk) 17:06, 18 February 2015 (UTC)

Merge
It seems to me to be a sensible idea to merge infimum and supremum, perhaps as infimum and supremum. The concepts are so related that separate articles are bound to be redundant to some degree. Thoughts? — Anonymous Dissident  Talk 12:35, 9 March 2011 (UTC)


 * I agree, both with the proposal to merge these article and with the proposed merge target. Compare for example the articles greatest element, upper and lower bounds and semilattice which discuss dual notions in a single article. — Tobias Bergemann (talk) 08:46, 21 July 2011 (UTC)


 * Absolutely. —Mark Dominus (talk) 15:26, 21 July 2011 (UTC)

I think a merge would be a great idea, as the two concept are the inverse of one and other and could be explained better together. — Preceding unsigned comment added by 86.44.146.52 (talk) 18:55, 13 January 2012 (UTC)


 * I don't think they should be merged. They are semantically separate meanings should therefore remain separate.  The more we merge these types of things the more of a muddle wiki will become.  Resist this.  — Preceding unsigned comment added by Adel314 (talk • contribs) 08:10, 1 September 2012 (UTC)


 * WAT 188.4.50.246 (talk) 08:58, 4 October 2012 (UTC)

I infinitely and supremely agree! 188.4.50.246 (talk) 08:57, 4 October 2012 (UTC)

These should have been merged a long time ago! Phillip M. Feldman — Preceding unsigned comment added by Pfeldman (talk • contribs) 02:23, 10 February 2013 (UTC)

I see no reason for them to not be merged. — Preceding unsigned comment added by 71.206.9.215 (talk) 17:27, 1 November 2012 (UTC)

Adding Deutsch (de)
I wanted to add the German page, but I encounted the problem described here. Am I/are we supposed to merge Infimum with Supremum in every language? Please help! -- 85.183.206.154 (talk) 16:35, 11 May 2013 (UTC)
 * I would. GenQuest  "Talk to Me" 00:20, 23 November 2013 (UTC)

Question
Talk:Supremum. Rinconsoleao (talk) 12:04, 20 July 2011 (UTC)

Confusion 2.0
The article is confusing and it is not open to a readership of similar disciplines. The uniqueness statement is overkill in the intro and it misdirects the reader to get the meaning of sup and inf. I move the statement into the relative section.

I am still wondering if the partial ordered set T must be always defined. For instance, the leading example in the intro is confusing because R+* is a subset of R but R is not defined as "taking the role of T" for the purpose of getting the infimum. Even if it is trivial, I cannot understand how the level of triviality could be high.

For the sake of clarity (I try), another thing confused me: what if S={x in R | 0 0 there is an x ∈ A with x < p + ε, and x ≥ p for every x ∈ A

I think it should be:

p = inf A if x ≥ p for every x ∈ A

But I'm not knowledgeable enough to assert this.

Matiasmoreno~enwiki (talk) 14:29, 19 June 2017 (UTC) — Preceding unsigned comment added by Matiasmoreno (talk • contribs)


 * Your proposed change only defines a lower bound for the set A and there are many lower bounds. An inf, when it exists, is the greatest of all of these. To capture this concept ("greatest") precisely, we say that the inf has the property that–if you increase it by any amount, it is no longer a lower bound, i.e., there is some element of A that is now smaller than this increased value.--Bill Cherowitzo (talk) 16:47, 19 June 2017 (UTC)

Join and meet
It was just pointed out on the talkpage of Join and Meet that these two articles are about the same topic, albeit from different perspectives. If you have thoughts about this, please include them over there. --JBL (talk) 12:31, 18 May 2018 (UTC)

Example Correct but Modified from Source
The example from Baby Rudin in the section Infima and suprema of real numbers, though obviously true, was stated in Example 1.9b as $$\sup \mathbb{Q}^- = 0$$ (though not as succinctly) rather than $$\sup \mathbb{R}^- = 0$$. I didn't change it both because it seems to me the former implies the latter and it would require moving the example to a different section. It seems to me there are four options. Bhbuehler (talk) 22:21, 27 June 2018 (UTC)
 * 1) It could be left as is. The downside is that would require a somewhat misleading citation.
 * 2) The example could be corrected to that in Rudin but moved to a different section.
 * 3) Alternatively, maybe the example is obvious enough that it doesn't need a citation and the reference to Baby Rudin could be deleted.
 * 4) Someone could clarify it so the citation refers to the correct example from Rudin and then showing some theorem proves the example for the $$\mathbb{R}^-$$ as well. This has the advantage of providing more information but possibly distracting from the point of the section.

Doesn't this statement depend on what kind of elements are in the sets?

 * Infima and suprema do not necessarily exist. Existence of an infimum of a subset S of P can fail if S has no lower bound at all, or if the set of lower bounds does not contain a maximal element. However, if an infimum or supremum does exist, it is unique.

Is the below not a counterexample. (if a rather odd one)

T is the set of all ordered binomials a+b.

S is {0+2, 0+3, 1+1, 1+2, 1+3, 2+0, 2+1, 2+2, 3+0, 3+1} (and is thus a subset of T).

0+2, 1+1, and 2+0 are all infimums because each is ≤ to all elements of S (is a lower bound), and all other lower bounds are ≤ each of them.

Similarly, 1+3, 2+2, and 3+1 are all supremums because each is ≥ to all elements of S (is an upper bound), and all other upper bounds are ≥ each of them.

I'm probably just misunderstanding what this article is about, but, if so, I think that's evidence that the article is confusing. — Preceding unsigned comment added by 98.20.224.66 (talk)


 * In your example, 0+2 is not less than 1+1 and so 0+2 is not a lower bound on the set, and similarly 1+1 is not less than 0+2 and so 1+1 is not a lower bound. (You seem to be confusing a minimal element -- that which has nothing smaller in the set -- for a lower bound -- which requires that all elements of the set be larger than it.)  Insider of T, your set has infimum 0+0. --JBL (talk) 11:34, 19 August 2018 (UTC)

Image versus text
The image shows "A set T of real numbers (hollow and filled circles), a subset S of T (filled circles)". According to the definition in the text, the infuinum and supremum must both be hollow circles, but both are filled circles in the image. The text extracts are as follows:

".. the infimum of a subset S of a partially ordered set T is the greatest element in T that is less than or equal to all elements of S"

and

"The supremum of a subset S of a partially ordered set T is the least element in T that is greater than or equal to all elements of S." — Preceding unsigned comment added by Kallax (talk • contribs) 10:48, 18 November 2019 (UTC)

A little mistake ?
In the properties of in part "Infima and suprema of real numbers" we have the following inequality:

"inf A + B ≤inf A + inf B; similarly for suprema."

It seems "inf" schould be replaced by a sup. If c_0 is the infimum of A+B, it can be approach with a term of the form a + b. The term a is bigger (or equal) than a_0 = inf A and the same for b that schould be bigger than inf B. — Preceding unsigned comment added by 2A02:AA13:6142:D600:351F:A19E:AB51:2EC3 (talk) 13:20, 28 April 2020 (UTC)
 * The inequality should have been equality. I've fixed the article.  –Deacon Vorbis (carbon &bull; videos) 13:46, 28 April 2020 (UTC)
 * It was broken in this edit a few weeks ago, for those who are curious. --JBL (talk) 13:56, 28 April 2020 (UTC)

Nested of-construction and insufficiently specified use of the word "element" cause for confusion
[Confusion part 1] The introductory sentence of the article contains: "the infimum (abbreviated inf; plural infima) of a subset S of a partially ordered set T is the greatest element in T that is less than or equal to all elements of S" and this had me puzzled for a while. My confusion was - at least partly - caused by the nested of-construction and the lack of a comma between "set T" and "is the greatest element". I have been wondering whether either of these interpretations was intended: or: From the context of this talk page, I seem to grasp that the latter is intended. If so, why not writing:
 * the infimum is in S and this infimum is infimum to T;
 * the infimum is in T and this infimum is infimum to S.
 * "In mathematics, the infimum (abbreviated inf; plural infima) of a set S which is a subset to a partially ordered set T, is the greatest element in T that is less than or equal to all elements of S, if such an element exists."?

[Confusion part 2] The same introductory sentence contains: "is the greatest element in T that is less than or equal to all elements of S, if such an element exists." [boldface Redav] What is the intended referent of "(such an) element"? Is it: or: Sometimes, wording is everything ...Redav (talk) 23:16, 18 May 2020 (UTC)
 * the greatest element in T that ...;
 * all elements of S? [so the clause "if such an element exists" would then mean "if S is not empty"]

Existence and uniqueness: example creates confusion
The following passage is confusing.

The issue is with the last, parenthetical statement. The example is intended to demonstrate that "if the set of lower bounds does not contain a greatest element" then the existence of inf "can fail." But this is not the case for the ordered set $$\mathbb{Q}$$. It is known that for the set $$S=\{ x \in \mathbb{Q} : x^2 < 2\}$$ that $$\sup S = \sqrt{2}$$. Of course $$\sqrt{2}\notin\mathbb{Q}$$ but that's different from saying "no supremum."

Perhaps just remove the example or develop a better, more appropriate and illustrative example relevant for a partially ordered set.

--MarkWayne (talk) 21:21, 5 January 2024 (UTC)
 * I think the problem is that the example assumes that $$P = \mathbb{Q}$$, but neglects to say so explicitly, so I added it to the article. The example would be in error in the case of $$P = \mathbb{R}$$. –MadeOfAtoms (talk) 10:36, 6 January 2024 (UTC)