Proof that given the info below, sup S <= inf T

  • Thread starter Thread starter thedoctor818
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
SUMMARY

The discussion centers on proving that the supremum of set S is less than or equal to the infimum of set T, given that S and T are nonempty subsets of the real numbers with the condition that every element s in S is less than or equal to every element t in T. Participants clarify that while the title states "sup S ≤ inf T," there is confusion regarding the implications of this relationship, particularly when considering examples like S = (0,1) and T = (2,3). The conclusion drawn is that sup S may equal inf T, but it is not necessarily true that sup S is strictly less than inf T.

PREREQUISITES
  • Understanding of supremum and infimum in real analysis
  • Familiarity with the properties of bounded sets
  • Knowledge of proof techniques, including proof by contradiction
  • Basic concepts of set theory and real number subsets
NEXT STEPS
  • Study the properties of supremum and infimum in real analysis
  • Learn about proof techniques, particularly proof by contradiction
  • Explore examples of bounded and unbounded sets in real analysis
  • Investigate the implications of set relationships in real number subsets
USEFUL FOR

Mathematics students, particularly those studying real analysis, educators teaching set theory concepts, and anyone interested in understanding the properties of bounded sets and their implications in proofs.

thedoctor818
Messages
1
Reaction score
0

Homework Statement


Let S and T be nonempty subsets of [tex]\mathbb{R} \backepsilon s \leq t \forall s \in S \wedge t \in T.[/tex] A) Observe that S is bounded above and that T is bounded below. B) Prove that [tex]sup S = inf T.[/tex]


Homework Equations





The Attempt at a Solution


Let [tex]s_0 = sup S. \text{ Then } s \leq s_o \forall s \in S \wedge s \leq t \forall s \in S \Rightarrow s_0 \leq t.[/tex]
Let [tex]t_0 = inf S. \text{ Then } t_0 \leq t \forall t \in T \wedge s \leq t \forall t \in T \Rightarrow s \leq t_0.[/tex]
 
Physics news on Phys.org
I am probbably missing something, but it doesn't seem that it is even true. COnsider
S=(0,1) and T=(2,3). Then for all s in S s=<t for all t in T, but clearly supS=/=infT. Are you sure the problem is not asking you to show : supS=<infT? If this is what the problem is asking, then try a proof by contradiction. Namely, assume that infT<supS, and derive a contradiction.
 
Last edited:
Yes. The title says "sup S[itex]\le[/itex] inf T" but in your post you say "sup S< sup T" which does not follow from your hypotheses. They might be equal. If sup S> sup T, then let [itex]\delta[/itex]= Sup S- sup T. There must be some member of S closer to S than that and so must be some member of S larger than sup T.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
1K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
1
Views
2K