# Open sets

1. Jan 17, 2009

### 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

2. Jan 17, 2009

### HallsofIvy

Staff Emeritus
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.

3. Jan 17, 2009

### poutsos.A

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

4. Jan 17, 2009

### NoMoreExams

distance formula?

5. Jan 18, 2009

### HallsofIvy

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

6. Jan 27, 2009

### 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> $$\l^2$$