1. ## Prove a sup

LET A={a: a>0}, where a is a rea No ,and A is bounded from above.

Define $\sqrt A$= { $\sqrt a:a\in A$}

Prove : $Sup\sqrt A = \sqrt{Sup A}$

2. Here is an outline.
For the set $\alpha=\sup{A}$.
If $a\le\alpha$ then $\sqrt{a}\le\sqrt{\alpha}.$
If $\beta=\sup\sqrt{A}$ and $\beta<\sqrt{\alpha}$ then $\beta^2<\alpha$.
So some element of $A$ is between $\beta^2~\&~\alpha.$
That gives a contradiction. I leave it to you to find it.

3. Originally Posted by Plato
Here is an outline.
For the set $\alpha=\sup{A}$.
If $a\le\alpha$ then $\sqrt{a}\le\sqrt{\alpha}.$
If $\beta=\sup\sqrt{A}$ and $\beta<\sqrt{\alpha}$ then $\beta^2<\alpha$.
So some element of $A$ is between $\beta^2~\&~\alpha.$
That gives a contradiction. I leave it to you to find it.
Does your method work for proving : SupAB = SupASupB , where A={a: a>0},B={b:b>0} e.t.c e.t.c??

4. Originally Posted by alexandros
Does your method work for proving : SupAB = SupASupB , where A={a: a>0},B={b:b>0} e.t.c e.t.c??
You know that I am just not inclined to answer your questions.
You should either post some of your own work on problems or explain what you do not understand about the question. But instead you seem to demand a solution without any effort on your part.
Is that being unfair to you?