Show that the set S is Closed but not Compact

In summary, the set S of all (x,y) ∈ ℝ2 such that x2+xy+y2=3 is closed but not compact. The attempt at a solution showed that it is closed, but the problem statement needed to show that it is compact.
  • #1
emergentecon
57
0

Homework Statement



Show that the set S of all (x,y) ∈ ℝ2such that 2x2+xy+y2
is closed but not compact.

Homework Equations


set S of all (x,y) ∈ ℝ2such that 2x2+xy+y2

The Attempt at a Solution


I set x = 0 and then y = 0
giving me
[0,±√3] and [±√3,0] which means it is closed

However, for it to be Compact, it needs to be Closed & Bounded.
For it to be bounded, it must have both an upper and lower bound, which to me it appears to have?
The bound for me are when x and y = ±√3

Clearly I am wrong, given how the question is structured. Any ideas?
 
Physics news on Phys.org
  • #2
You'll have to be more clear o. What exactly your set S is. ##{(x,y) \in R^2 : 2x^2+xy+y^2}## is what you've defined. But, I can plug any points x, y into that since you've not told us what the equation must satisfy.
 
Last edited by a moderator:
  • #3
Most correct, my omission.
The function equals 3.
 
  • #4
OK, so if the function must equal 3, then x=0 and y=0 is not even in the set. Your definition of compact is closed and bounded. So you need to show that the set is closed but not bounded.
 
  • #5
emergentecon said:
Most correct, my omission.
The function equals 3.

Are you saying that your ##(x,y)## must satisfy ##2x^{2+xy+y^2} = 3##? That is the function you wrote. Or, did you mean ##2 x^2 + xy + y^2 = 3##?
 
Last edited:
  • #6
Sorry, I responded on my phone so I'm not sure how the LaTex parts turned out.
 
  • #7
I need to apologise to everyone . . . immediately after I posted, I needed to run to lectures . . . I never noticed the error in my post, and wasn't able to correct it from my mobile phone. Really did not mean to waste your time. Stated correctly, it should read as:

Show that the set S of all (x,y) ∈ ℝ2 such that x2+xy+y2 = 3
is closed but not compact.

(Have been trying to correct the original post, but doesn't seem possible?)
 
  • #8
Ray Vickson said:
Are you saying that your ##(x,y)## must satisfy ##2x^{2+xy+y^2} = 3##? That is the function you wrote. Or, did you mean ##2 x^2 + xy + y^2 = 3##?
Please see my apology below.
You are correct, it should read:

Show that the set S of all (x,y) ∈ ℝ2 such that x2+xy+y2 = 3
is closed but not compact.
 
  • #9
MostlyHarmless said:
OK, so if the function must equal 3, then x=0 and y=0 is not even in the set. Your definition of compact is closed and bounded. So you need to show that the set is closed but not bounded.
Yes, which I think I did.
I tried to show it was closed, but don't know how to show that it is bounded or not, as the case may be?
 
  • #10
emergentecon said:
My original post contains errors and I cannot see how to correct it. So here is the corrected post:

1. Homework Statement


Show that the set S of all (x,y) ∈ ℝ2such that x2+xy+y2=3
is closed but not compact.

Homework Equations


set S of all (x,y) ∈ ℝ2such that x2+xy+y2=3

The Attempt at a Solution


I set x = 0 and then y = 0
giving me
[0,±√3] and [±√3,0] which means it is closed

However, for it to be Compact, it needs to be Closed & Bounded.
For it to be bounded, it must have both an upper and lower bound, which to me it appears to have?
The bound for me are when x and y = ±√3

Clearly I am wrong, given how the question is structured. Any ideas?
 
  • #11
##x=y=\sqrt(3)## Is not in the set, nor is 0. You'll need to show that they are upper and lower bounds.

Another definition of bounded is that you can find a ball in which the whole set is contained in.
 
  • #12
  • #13
Let u= x+ y and v= x- y. Then x= (u+ v)/2 and y= (u- v)/2. Replace x and y in the equation. That will eliminate the "xy" term and it is easy to see what kind of conic you have.
 
  • #14
emergentecon said:
I need to apologise to everyone . . . immediately after I posted, I needed to run to lectures . . . I never noticed the error in my post, and wasn't able to correct it from my mobile phone. Really did not mean to waste your time. Stated correctly, it should read as:

Show that the set S of all (x,y) ∈ ℝ2 such that x2+xy+y2 = 3
is closed but not compact.

(Have been trying to correct the original post, but doesn't seem possible?)

I think there's a further error in the problem statement: S as you have defined it is actually compact.
 

FAQ: Show that the set S is Closed but not Compact

What does it mean for a set to be "closed"?

A set is considered closed if it contains all of its boundary points. In other words, if a point is on the edge of the set, it must also be included in the set itself.

What does it mean for a set to be "compact"?

A set is considered compact if it is both closed and bounded. This means that the set contains all of its boundary points and is also limited in size.

How can a set be closed but not compact?

A set can be closed but not compact if it is missing one of the two criteria for being compact. For example, a set can be closed but unbounded, meaning it contains all of its boundary points but is not limited in size.

What are some examples of sets that are closed but not compact?

One example of a set that is closed but not compact is the set of all real numbers. This set is closed because it contains all of its boundary points, but it is not bounded as it extends infinitely in both the positive and negative directions. Another example is the set of integers, which is also closed but not compact.

Why is it important to distinguish between a set being closed and being compact?

Distinguishing between a set being closed and being compact is important because these properties have different implications in mathematical proofs and analysis. Closed sets have different properties and theorems associated with them than compact sets, and understanding the distinction between the two can lead to a deeper understanding of the structure of the set.

Back
Top