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I am having troubles with figuring if this space (with the Euclidean metric) is complete or not. $$ \left\{{(x,y)\in \mathbb{R^2}} : x > 0, y \geq \frac{1}{x}\right\}$$

I tried Cauchy sequences, but I was not able to find one which converges and is contained within the set. Also, the set is not closed, so I cannot use that either. Can someone help me out?

surfer1311
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  • What do you know about the convergence of Cauchy sequences? – science Mar 21 '15 at 19:48
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    Assuming that the metric of this space is the euclidean one, that space is indeed CLOSED in $\mathbb{R^2}$ so it's complete. You can prove that it's closed (for example) proving directly that its complement is open, it's a computational task. – Daniel Mar 21 '15 at 19:50
  • See this. – science Mar 21 '15 at 19:51

3 Answers3

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Hints:

A subset of a complete metric space is closed if and only if it's complete.

There's no notion of completely closed. Is your set closed or is it not? Try sketching it to get a better idea (then prove your guess).

If you still have trouble, look at $\{(x,y)\in \mathbb{R}^2 : xy \geq 1\}$. Is this closed? How does it relate to your set?

One last hint, if you're still having trouble:

$$\{(x,y) \in \mathbb{R}^2 : xy \geq 1, x > 0\} = \\ \{(x,y)\in \mathbb{R}^2 : xy \geq 1\} \cap \{(x,y) \in \mathbb{R}^2: x > 0\} = \\ \{(x,y)\in \mathbb{R}^2 : xy \geq 1\} \cap \{(x,y) \in \mathbb{R}^2: x \geq 0\}$$

aes
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  • With completely closed I meant that one part of the product is closed and the other is not, but it was vague indeed. – surfer1311 Mar 21 '15 at 19:50
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    @surfer1311 Actually it turns out that it is closed. This is probably easier to see by showing that its complement is open. – Ian Mar 21 '15 at 19:51
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Define $f(x,y)=xy$. This is a continuous function $f:\mathbb R^{\geq 0}\times\mathbb R^{\geq 0}\rightarrow\mathbb R^{\geq 0}$. Your set is the set $f^{-1}\left([1,+\infty)\right)$. Since $[1,+\infty)$ is closed, your set is closed in $\mathbb R^{\geq 0}\times\mathbb R^{\geq 0}$, which is a complete metric space.

If you'd rather only deal with $\mathbb R$, you can define two functions:

$$f(x,y)=xy\\g(x,y)=x+y$$Then your set is equal to $$f^{-1}\left([1,+\infty)\right)\cap g^{-1}\left([0,+\infty)\right)$$

which is the intersection of two sets which are closed in $\mathbb R\times \mathbb R$, since $f$ and $g$ are continuous.

Thomas Andrews
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Note that

A closed subset of a complete metric space (your case it is $\mathbb{R^2}$) is complete.

science
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