Talk:Infimum

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[edit] 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)

[edit] 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 souns 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)

[edit] 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"

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