Prove a sup

Printable View

December 6th 2010, 03:59 PM
alexandros

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 : $Sup\sqrt A = \sqrt{Sup A}$
December 6th 2010, 04:24 PM
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.
December 6th 2010, 05:55 PM
alexandros

Quote:

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??
December 6th 2010, 06:01 PM
Plato

Quote:

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?
If so, please explain yourself.
December 6th 2010, 06:21 PM
alexandros

Quote:

Originally Posted by Plato

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.

Check the thread :Supremum and infemum

All times are GMT -8. The time now is 11:06 AM.