Question

Let A be a set with a partial order R. If \( C \subset B \subset A \) and sup(C) and sup(B) exist then \( \text { sup(C) }\le \text { sup(B) } \)

Answer 1

Is this a good enough proof?
sup(B) is an upper bound for B. Therefore sup(B) is an upper bound for C. Thus \( \text { sup(C) } \le \text { sup(B) } \)

Answer 2

Yes

Answer 3

Depends on how rigorous your professor requires your proofs to be. That's the basic logic of it though.

Answer 4

i kinda found it like it was lacking some information

Answer 5

the definition of sup is the least upper-bound. So your proof is correct.

Answer 6

Consider the element x=sup(C) in C. C is a subset of B, therefore x is in B. Since x is in B, x <= Sup(B).

Answer 7

By definition of Supremum.

Answer 8

But x might not be in C

Answer 9

C=[0,1[ sup C =1 and 1 does not belong to C

Answer 10

Right. :c

Answer 11

I guess my original proof is good enough :D

Answer 12

Proof by contradiction would be easy.
Assume sup(C)> Sup(B)
There exists an x in C such that sup(C)>x>sup(B). x is an element of C and C is a subset of B, therefore x is an element of B. However, x >sup(B), so x is not an element of B.
Reductio ad absurdum.

Answer 13

rather, there exists an x such that sup(C)>=x>sup(B)

Answer 14

Ohhhh i like thisssss

Answer 15

Don't like it too much. Elias is about to poke a big hole in it, I expect. lol

Answer 16

idk I think its clearer