they are actually referring to two separate but similar problems.

In the first S is bounded from above and x = sup S

In the second S is bounded from below and x = inf S

If you have any idea what sup and inf are this should make sense.

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- February 23rd 2014, 07:27 PM #1

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## sup S (inf S) -- What Does This Mean?

I have a homework problem due tomorrow.

Let S be a nonempty set of real numbers that is bounded from above (below) and let x=sup S (inf S). Prove that x either belongs to S or is an accumulation point (limit point) of S.

This is all the entirety of the problem.

I am sure that I can solve it if I understand what it means but what is the meaning?

- February 23rd 2014, 09:08 PM #2

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## Re: sup S (inf S) -- What Does This Mean?

they are actually referring to two separate but similar problems.

In the first S is bounded from above and x = sup S

In the second S is bounded from below and x = inf S

If you have any idea what sup and inf are this should make sense.

- February 24th 2014, 07:22 AM #3

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- February 24th 2014, 09:18 AM #4

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## Re: sup S (inf S) -- What Does This Mean?

Sup(S) means: the supremum (least upper bound) of the set S.

Inf(S) means the infimum (greatest lower bound) of the set S.

It is an axiom of the real number system that any non-empty set bounded above has a least upper bound, and similarly that every non-empty set bounded from below has a greatest lower bound (lub and glb are two other common abbreviations). This property is called the "completeness property" of the real numbers...it is a continuum, and doesn't have any "holes".

For example, the least upper bound of the open interval (a,b) which is the set {x in R: a < x < b} is the real number b. Similarly the greatest lower bound is the real number a.

If all we were concerned about was INTERVALS, this problem would be easy. But consider this set:

S = {x in (0,1): the decimal expansion consists of only 1's and 0's}

This is kind of a "weird" set, because it isn't "connected". It turns out that sup(S) = 1/9, which is in S, and that inf(S) = 0, which is not in S.