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

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

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

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- Dec 6th 2010, 03:59 PMalexandrosProve a sup
LET A={a: a>0}, where a is a rea No ,and A is bounded from above.

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

Prove :$\displaystyle Sup\sqrt A = \sqrt{Sup A}$ - Dec 6th 2010, 04:24 PMPlato
Here is an outline.

For the set $\displaystyle \alpha=\sup{A}$.

If $\displaystyle a\le\alpha$ then $\displaystyle \sqrt{a}\le\sqrt{\alpha}.$

If $\displaystyle \beta=\sup\sqrt{A}$ and $\displaystyle \beta<\sqrt{\alpha}$ then $\displaystyle \beta^2<\alpha$.

So some element of $\displaystyle A$ is between $\displaystyle \beta^2~\&~\alpha.$

That gives a contradiction. I leave it to you to find it. - Dec 6th 2010, 05:55 PMalexandros
- Dec 6th 2010, 06:01 PMPlato
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?

If so, please explain yourself. - Dec 6th 2010, 06:21 PMalexandros