Prove supS <= infT

  • #1

Homework Statement


Let S and T be non-empty subsets of R, and suppose that for all s [tex]\in[/tex] S and t [tex]\in[/tex] T, we have s [tex]\leq[/tex] t.

Prove that supS [tex]\leq[/tex] infT.

Homework Equations


N/A


The Attempt at a Solution



Since s [tex]\in[/tex] S [tex]\Rightarrow[/tex] s [tex]\in[/tex] T, supT is an upper bound for S.
Since supS is the least upper bound, supS [tex]\leq[/tex] supT.


How does this look to you? Feedback is appreciated.
 

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,103
274
[tex]supS \leq supT[/tex] doesn't seem that useful as a start to be honest (supT and infT aren't very close to each other in general). To show that [tex] supS \leq infT[/tex], can you show that infT is an upper bound of S?
 

Related Threads on Prove supS <= infT

Replies
1
Views
943
Replies
3
Views
912
  • Last Post
Replies
7
Views
896
Replies
5
Views
2K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
3K
Top