# Show subsets of the plane are open

1. Jan 22, 2004

### babbagee

Show that the subsets of the plane are open:

1.) A= {(x,y)|-1<x<1,-1<y<1}
2.) C= {(x,y)|2<x2 + y 2<4}

I have another question. What does this notation imply. f(x,y)=some function.

Last edited: Jan 22, 2004
2. Jan 23, 2004

### HallsofIvy

Staff Emeritus
Homework? Shouldn't this be in the homework section?

Okay, what is your definition of "open set"? In proofs like these you need to use the precise words of the definition and I know several.

The most fundamental definition of "open set" is that it is a member of the given "topology". I suspect your teacher would not accept that. In fact, I suspect that you are not even "given" the topology.

One definition that can be used in a metric space is that a set is open if every member of the set is an "interior point" of the set. Of course, to use that you need to know what an "interior point" is. A point, p, is called an interior point of set A if and only if for some &delta;>0, the set of all points closer to p than &delta; (the "&delta; ball of p") is contained in A. In R2 with the "standard topology", a "&delta; ball of p" is the interior of a circle of radius &delta; with center p.
For this problem you might say "Suppose p= (x,y) is in A={(x,y)|-1<x<1,-1<y<1}. The &delta;1= x+1, &delta;2= 1-x, &delta;3= y+1, and &delta;4= 1-y are all positive numbers. The "ball around (x,y) with radius= smallest of (&delta;1,&delta;2,&delta;3,&delta;4)" is contained in A. (Of course, you would need to show that.)

Yet another definition is that none of the sets "boundary points" are in the set. Here p is a "boundary point" of A is every ball (see above) centered on p has some point in A and some points not in A. What are the "boundary points" of the two sets you are given? Are any of them in the sets?

3. Jan 24, 2004

### babbagee

Our teacher taught us about the disk being inside the set or outside the set. He also talked about bounries. But i really didnt understand what he was talking about. That is why i am asking, because i dont understand.

4. Jan 26, 2004

### matt grime

let's think pictorially for the first one. the area is that bounded by four lines and forms a square. none of the points on the boundary is in the set, so if we pick some point in the set, it is some non-zero distance away from any of the sides. it's now posssible to draw a circle centred on that point and not going outside the set. draw the picture, and show it to yourself. now try and prove it rigorously - how much rigour is needed will depend on the level of the course: is it an epsilon proof you're used to?

5. Jul 23, 2004

### mathwonk

basically any set defined by an inequality like f< a is open, because of you choose a point in there then it will have f = some value k which is less than a. But then by continuity any point very near to it will also have value closer to that value than a-k, so the triangle inequality the value of the new point will still be less than a.

thus any point near the given point of the set will also be in the set. that is the definition, of open. moral memorize the definition of open then try to use it.

6. May 16, 2011

### kjartan

For part two, all we need to do is believe what we are given and apply the triangle inequality.

Suppose (a,b) are in C, then 2<a2+b2<4

notice that:
|x-a|= sqrt[(x-a)2] <= sqrt[(x-a)2 + (y-b)2]
Similarly for y.

Observing that C is an annulus with an inner radius of sqrt[2] and an outer radius of 2, we consider an open disc Dr about (a,b) in C and look at the cases where it is either closer to the inner radius or the outer radius. Then for an arbitrary (x,y) in our r-disc about (a,b) we have:
sqrt[(x-a)2 + (y-b)2] < r,
where r = min(sqrt[a2+b2]-sqrt[2],2-sqrt[a2+b2])

Now, since |x-a|< sqrt[a2+b2]-sqrt[2]
And in general since |s|<t <=> -t<s<t, we can simply add a to both sides and square to find a relation for x2. We proceed similarly for y. If we then sum we will find a relation for x2+y2.

Now all we need to do is consider where these upper and lower bounds have maxima or minima.

For example, using the relation |x-a| < sqrt[a2+b2]-sqrt[2], we have:
4+3(a2+b2)+2sqrt[2](a+b)-2sqrt[a2+b2](a+b+2sqrt[2]) < x2+y2
We find the upper bound of the left hand side to be (|a|,|b|)=(1,1). We evaluate this expression and find that 2 < x2+y2.

Using the same strategy, we proceed using the relation:
|x-a| = sqrt[(x-a)2] <= sqrt[(x-a)2 + (y-b)2] < r,
where r = 2-sqrt[a2+b2]
And find the lower bound of the right hand side.

Then we see that 2<x2+y2<4.
Therefore, for an arbitrary (x,y) in our r-disc, (x,y) is in C. Therefore Dr(a,b) is a subset of C, which in turn implies that C is open.

Last edited: May 16, 2011
7. May 16, 2011

### HallsofIvy

Staff Emeritus
It "implies" exactly what it says- that f(x,y) is ames some function of the two variables, x and y.

8. May 16, 2011

### dimitri151

To show that a set is open (at this level) have to show that if you pick any point in the set, there is a neighborhood around that point so that every point in the neighborhood is in the set. You can look at the neighborhood as a little circle or square around the point at this point. Just to start you off, if your set is {x:-1<x<1}, to show that {x:-1<x<1} is open you have to show that if x is in (-1,1) then there is a neighborhood d (which you will represent with the greek delta but which i will here denote with a d because of certain limitations with Laytex on my part) such that (x-d,x+d) is a subset of (-1,1). In this case d is the minimum of {x-(-1), 1-x} which is written d=min{x-(-1), 1-x}. So for instance if your point is -.9 then d=min{-.9-(-1),1-(-.9)}=min{.1,1.9}=.1. (If you're just starting on this matter it's easy for the notation to obscure the idea. Basically if your point is in the interval (-1,1) then d is the lesser of the distances to the two endpoints). Then you have to show that (x-d,x+d) is a subset of {x:-1<x<1}. To do this you have to show that if x is in (x-d,x+d) then x is in (-1,1). That's the general strategy for showing a set A is a subset of a set B. Hope this helps.