Proving Openness of Set S in R^2

[poutsos.A]

How do we prove that the following set is open??


S= R^2- {(x,y) : y=x^2, xεR}

WHERE R is for real Nos

 

[HallsofIvy]

How about using the definition of "open set", which, in a metric space, is that every point of the set is an interior point. Let (x,y) be a point in S which means it is a pair of real numbers such that y is NOT equal to x^2. Then show that there exist a distance r such that all points within r of (x,y) are also in S.

One way to do that is to find the shortest distance from (x,y) to the curve y= x^2 and take half that distance as r.

Frankly, it is hard to see how you could be expected to do a problem like this if you could not see how to prove that the limit of a constant sequence is that constant.

 

[poutsos.A]

Thanx, but how do we calculate that distance d(x,y) =r??

 

[NoMoreExams]

Thanx, but how do we calculate that distance d(x,y) =r??

distance formula?

 

[HallsofIvy]

Thanx, but how do we calculate that distance d(x,y) =r??

As long as you know there is such a non-zero distance, you don't need to know what it is!

 

[poutsos.A]

From the definition of the open set we have:

S is open iff for all xεS THERE exists r>0 such that Disc(x,r) \subseteq S

........or(equivalently)..........


S is open iff for all xεS ,there exists r>0 such that, yεDisc(x,r) =====> yεS.


Let now x=(k,l) ,y=(m,n) then yεDisc(x,r) <=====>\sqrt{(k-m)^2+(l-n)^2}< r

AND the above becomes:



S is open iff [if (k,l)εS then there exists r>0 such that ,if \sqrt{(k-m)^2+(l-n)^2}< r then (m,n)εS].


But (k,l)εS <====> k< \l^2, (m,n)εS <====> m< \ n^2. And if w=(o,p) is a point on the curve THEN the distance between y and w is :

....\sqrt{(m-o)^2 + (n-p)^2}..........

AND if we choose r< \sqrt{(m-o)^2 + (n-p)^2}...the above becomes:



S is open iff .

......if... k< \l^2 and o=p^2 and \sqrt{(k-m)^2+(l-n)^2}< \sqrt{(m-o)^2 + (n-p)^2}... then m< \ n^2

Can anybody curry on from here?

Also note k< \l^2 is one case another is k> \l^2