Question

Let A,B be sets
s = supA, t=supB
Prove that s + t is upperBound of A + B

Answer 1

Lol I got it I think:

\[(\forall a \in A, \forall b \in B)[ a \le s, b \le t] \rightarrow (a+b \le s + t)\]
then
\[A + B = \left\{ a+b, \forall a \in A, \forall b \in B \right\}\]
then s+t is upperbound for A + B

Answer 2

Next, let u be an element from {upper bounds of A + B}
Prove
\[\forall a \in A, t \le u - a\]

Answer 3

why did you mention me in chat?

Answer 4

@Gdeinward ye so a few years ago (I was naughty) I also gave direct answers to people and I didn't follow the rules. Eventually the people just kept telling me to give answers to them but they were very easy to do, and over time I felt like they were not doing them at all. So now I stopped giving answer to them and I feel like I spoiled them too cause they developed a habit of just posting a super-easy question without bothering to google it. This was my experience.

Answer 5

I mean I thought you could give direct answers, but as long as you went in-depth and explained it was fine. But boy was I wrong.

Answer 6

So my proof would be:

let u = s + t
then u is an element of {upperbounds of A + B} (as proven in the 1nd post)
since s = supA, forall a in A, a <= s
then a <= s + t - t
then a <= u - t
then t <= u - a