Question

hmm

Answer 1

It's not true every subset of R is bounded above.

Answer 2

But the rest of what you're proposing makes sense.

Answer 3

Can you prove that max(a,b) is an upper bound for A U B ?

Answer 4

One way to argue is by contraction. Suppose max(a,b) were not an upper bound for AUB. That is, there exists \( x \in A \cup B \) such that
\[ x > max(a,b) \]
Well, as \( max(a,b) \geq a \) and \( max(a,b) \geq b \), that must mean that
\[ x > a \ \ \ \ \hbox{and} \ \ \ \ x > b \]
But as x is a member of A or B, it must be bounded above by at least one of a or b. Contradiction.

Answer 5

I think I got the R thinkg confused with, every subset of R that is upperbounded has a supremum.

Answer 6

yes

Answer 7

o great, that makes perfect sense thank you.

Answer 8

so crazy how hard it is to say the simplest thing.

Answer 9

not hard but ....