Prove that the set S = {x: a<x<b } is open in R (=Real Nos ) and not in [math]R^2[/math]

why don't you give it a hot yourself this time and we'll show you how to build on it eh?

 

you need to give us something for us to help you.

O.K

 

S is open in R iff ,for all ,xεS there exists an ε>ο such that for all , y:

 

[math]y\in B(x,\epsilon)\Longrightarrow y\in S[/math].

 

Here the ball B(x,ε) = {z : |z-x|<ε }.

 

So in this case we are seeking an ε>0

 

ΑΝD:

 

S is not open in [math]R^2[/math] ,iff there exists an ,xεS such that for all ε>0 ,there exists a pair of [math] (x_{1},y_{1})[/math] belonging to the ball

 

B((x,0),ε) and not belonging to S.

 

Here the ball B((x,0),ε) = {(z,w) : [math]\sqrt{(z-x)^2 + w^2}<\epsilon[/math]}.

 

So in this case we are looking for a pair of [math](x_{1},y_{1})[/math].

 

Any suggestions??

Well that shows you know the definitions. But that's not really the point

 

I'll show you how to proof that (0,1) is open in [math]\mathbb{R}[/math]. Firstly, take [math]x\in (0,1)[/math]. Then, we want to show that there exists an [math]\epsilon[/math] such that [math]B(x,\epsilon) \subset (0,1)[/math].

 

So, draw a line and label 0 and 1, then put x somewhere on that line between the two points. To make a ball of radius [math]\epsilon[/math] fit in there, you need [math]\epsilon[/math] to be smaller than the distance from 0 to x, or x to 1; whichever is the smallest. And just to be on the safe side, we can divide that distance by two to make sure the ball still fits.

 

So let [math]\epsilon = \tfrac{1}{2}\min\{ x, 1-x \}[/math], and you're done. Of course you will want to formalise this argument a bit, but the key argument is there.

 

In terms of proving it isn't open in [math]\mathbb{R}^2[/math], since you have the extra dimension to play with, this should be easy. It's like saying; can I fit a circle onto a straight line. Of course the answer is no.

O.K

For the first problem taking ε= min{ x,1-x } it works ,but for the 2nd

 

problem ,wcich you consider as an easy problem ,what would you suggest for the pair [math](x_{1},y_{1})[/math]

Okay. So... I don't get it At this point, the problem is really trivial and you should be able to get it with everything i've given you, or at least make some kind of stab at guessing the solution. I'm going to give you a final hint: take a point in (0,1). Say, for simplicity, a half. Then take ANY ball of radius [math]\epsilon[/math]. Is there a point which is in a ball of radius [math]\epsilon[/math] and centred at the origin, but not contained within (0,1)? It will probably help if you draw a picture

Okay. So... I don't get it At this point, the problem is really trivial and you should be able to get it with everything i've given you, or at least make some kind of stab at guessing the solution. I'm going to give you a final hint: take a point in (0,1). Say, for simplicity, a half. Then take ANY ball of radius [math]\epsilon[/math]. Is there a point which is in a ball of radius [math]\epsilon[/math] and centred at the origin, but not contained within (0,1)? It will probably help if you draw a picture

 

Please be so kind as to read my posts again .

 

I am not asking for a geometrical picture where intutively we can say that ,since we cannot fit a disc into a line ,then S is not open in [math]R^2[/math].

 

WE want an analytical proof .

 

For example, we can easily see that a continuous function in a certain interval ,with f(a)>0 and f(b)<0 for two valeus a,b the function will at least cross the x-axis .

 

But to prove that analytically is quite a job

With all due respect, I have read your posts quite clearly. I understand your desire for a rigorous proof. But metric spaces and topology in general takes such a basis in geometry that you can derive a great deal of intuition from pictures, especially at simple levels such as proving sets are open.

 

My purpose here is not to give you the answer, because you won't learn anything from that. It is to provide enough information to give you a good shot at proving it yourself.