Is the set {(x,y): 0≤xy≤1} bounded and closed?

[Somefantastik]

Homework Statement

$$A = \left\{(x,y): 0\leq xy \leq 1\right\}, A \in R^{2}$$

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

Homework Equations

if X = (x,y)

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

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 X_n = (x,y) $$\in$$ A such that X_n -> X' and X_n $$\neq$$ X'.

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

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

 

[Martin Rattigan]

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

 

[Somefantastik]

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

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

 

[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.

 

[Mark44]

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.

 

[Somefantastik]

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

 

[VeeEight]

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

 

[Martin Rattigan]

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