Question

Prove or disprove:
\(sup (A\cup B)\geq sup\{sup(A) , sup (B)\}\)
\(sup (A\cup B)= sup\{sup(A) , sup (B)\}\)
Please, help

Answer 1

for b) I have:
\(sup (A\cup B) = max \{sub (A) , sub (B)\}\)= \(sup\{sup(A), sub(B)\}\)
but I don't know a)

Answer 2

@freckles

Answer 3

where did you start?

Answer 4

contradiction proof?

Answer 5

what do you mean? for part a?

Answer 6

part a what did you consider so far?

Answer 7

part b you said you proved it is correct?

Answer 8

I don't see the difference between a, b but they separate them. For part b, I think I am right.

Answer 9

hmm so part b) is valid for any sup(A intersect B)

Answer 10

i mean union

Answer 11

Actually, for part b, the whole proof is
1) \(A\cup B\) is bounded:
let \(x\in A\cup B\) then \(x \in A\) or \(x\in B\)
So that \(x \geq inf (A) \) or \(x\geq inf B\)
that means \(x\geq min\{inf(A), inf (B)\}\)
the same for sup,
so that \( min\{inf(A), inf (B)\}\leq x\leq max\{sup(A), sup(B)\}\) \(\forall x \in A\cup B\)
that shows \(A\cup B\) is bounded.

Answer 12

2) my proof above.

Answer 13

Note: if the statement is false, just give a counterexample.

Answer 14

Well if statement 2 is true, which it is, then wouldn't that imply that 1 is true as well? \(\ge\) is greater than OR equal (exclusive or) , then since one is true , then the whole statement would be true right?

I'm not too sure if the above is totally sufficient. But to show the 2nd one, first we show \(c\) is an upper bound of \(A \cup B\). If \( x \in A ∪ B\), then \(x \in A\) or
\(x \in B\). As a result, , \(x \le \sup A\) or \(x \le \sup B\). That is, \(x \le \sup\{\sup A,\sup B\}\)

To show \(c\) is the least upper bound, suppose \(c_1\) is any upper bound of
\(A \cup B\). Thus, \(x \le c_1\) for all \(x \in A\) and \(x \in B\). Now since\(\sup A\) is theleast upper bound of A, then \(\sup A \le c_1\). In a similar manner, \( \sup B \le c_1\).
Hence, \(\sup\{\sup A,\sup B\} = c \le c_1\).

Hm I admit I'm not 100% sure of this. I don't remeber too well proofs on sup/inf xD

Answer 15

Check my stuff, please: :)
let \(x \in A\cup B\), then \(x \in A\) --> \(x\leq sub A\)
or \(x\in B\) --> \(x\leq sub B\)
which gives \(x\leq max \{sub A, sub B\}\) \(\forall x \in A\cup B\)
that means \(max\{sub A, sub B\}\) is an upper bound of \(A\cup B\)
but \(sup (A\cup B)\) is the least upper bound of it, so that \(sup (A\cup B)\leq max\{sup A, sup B\}\)

Answer 16

hm yeah actually I think that is more accurate than what I wrote

Answer 17

aaaaaaaaaaaaaaaah!! it confuses me. Let think more. hihihi

Answer 18

Oh, I see it. They give me part a) which is a part or part b. After getting part a), I do something more to get part b.
Just break part b into 2 parts. right?

Answer 19

I dunno I feel slightly confused myself xD.. Um. I think though..., I believe that\(\max\{\sup A,\sup B\}=\sup\{\sup A, \sup B\}\) because you just have a 2-element set so the max is basially the sup (or vice-versa), for part b) so, I think that... from the first part of my argument, you can say you have \(x \le \max\{\sup A,\sup B\}\) and in the conclusion, you could say \(\max\{\sup A,\sup B\} = c \le c_1\) . Then you can just say \(\sup(A \cup B) =\max\{\sup A, \sup B\}\) , meaning that \(\sup(A \cup B) =\sup\{\sup A, \sup B\}\).

But I guess the more general statement is that \(\max(\cdot ) \ge \sup (\cdot)\) which would imply statement 1.

? Ah it's been so long since I have done these... oye

Answer 20

Thanks for being patient to me. I do appreciate. I will have it done tonight :)

Answer 21

One more question: why do we study and then forget??

Answer 22

Lol.. I dunno. I guess your brain has a "limit" to memorizing stuff, or I guess if you don't exercise it enough, it forgets. But I barely used sup/inf in my studies afterwards.. so yeah :P

Answer 23

I suppose if you go the pure math way, you'll see it a lot more.. but I went towards stats/probability so I haven't used it as much hehe

Answer 24

Now I know you are stats/ probability expert. I think I can't let you free when I have a problem on that field. hehehe....