Closed and bounded

1. May 1, 2010

Somefantastik

1. The problem statement, all variables and given/known data
$$A = \left\{(x,y): 0\leq xy \leq 1\right\}, A \in R^{2}$$

I'm trying to determine if this set is bounded and/or closed.

2. Relevant equations

if X = (x,y)

euclidean metric: $$||X|| = \sqrt{x^{2}+y^{2}}$$

3. The attempt at a solution

I know a bounded set => ||X|| $$\leq$$ k

so I need to show somehow

$$||X|| = \sqrt{x^{2}+y^{2}} \leq k$$ (somehow)

and closed => every limit point belongs to the set. So take an arbitrary X'= (x',y') $$\in$$ A'. Then there exists Xn = (x,y) $$\in$$ A such that Xn -> X' and Xn $$\neq$$ X'.

Xn $$\in$$ A => $$0 \leq xy \leq 1$$

Need to show X' is such that $$0 \leq x'y' \leq 1$$ (somehow)

2. May 1, 2010

Martin Rattigan

If x=0 for what values of y would you have 0<=xy<=1?

3. May 1, 2010

Somefantastik

I see, all values of y, so this set is not finite and therefore not bounded?

4. May 1, 2010

Martin Rattigan

Correct.

Have you sketched it? The sketch wouldn't prove anything, but it can be helpful to suggest a proof for the next bit.

5. May 1, 2010

Staff: Mentor

Martin's advice of sketching the set seems to me a good one. For each nonzero value of k, with 0 < k <= 1, you have xy = k, or y = k/x, a hyperbola.

6. May 1, 2010

Somefantastik

In general though, "not finite" does not imply "not bounded", correct?

7. May 1, 2010

VeeEight

Correct. Just take [0,1] in the reals

8. May 2, 2010

Martin Rattigan

Yes, sorry, I shouldn't have written, "Correct". But I knew what you meant.