Specifically, I need to find if the metric space $(A,d)$ is complete or not.
$A = \{ (x,y) \in \mathbb{R}^2: x>0 , xy \ge 1\}$ and $d$ is the usual metric.
Find if a metric space is complete
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metric-spaces
complete-spaces
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0Do you have a specific question? What do you think the answer is? – 2017-02-01
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0Is $A$ closed.? – 2017-02-01
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0$A$ is neither closed nor open. – 2017-02-01
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0A subspace $M$ of a complete metric space $X$ is itself complete if and only if the set $M$ is closed in $X$. – 2017-02-01
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0Try writing $A$ as the inverse image of a closed set under some continuous mapping. Completeness follows immediately. – 2017-02-01
1 Answers
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Lemma. If the subspace $A$ of metric space $B$ is complete, then $A$ is closed in $B$.
Proof. If it is not closed then there exists such $x\in B\backslash A$ and a sequence $a_i\in A$ that $a_i\to x$. This sequence is Cauchy in $A$. By completeness of $A$ it has a limit $y\in A$, but then in $B$ it has two limits $x$ and $y$, which cannot be true. $\square$
The converse also holds when $B$ is complete itself. For if we have any Cauchy sequence in the closed set $A$, it is Cauchy in the complete space $B$. Thus it has a limit in $B$, but $A$ is closed and so this limit must lie in $A$.
You probably know that $\mathbb R^2$ is complete and so you just need to determine if the given $A$ is closed. If you draw it, you'll see that it is just the upper part of those 2 parts of the plane, which the upper right branch of hyperbola $y=1/x$ subdivides the plane on, including the border. It is clear that this set is closed, though you need some formal argument to show it. But indeed if you have a converging sequence of the points $(x_i,y_i)$ for which $x_i>0$ and $x_iy_i\ge 1$, then for the limit $(x,y)$ we have $x\ge 0$ and $xy\ge 1$. And $x\ne 0$, because otherwise $xy=0<1$. Therefore $(x,y)$ is also in our set, qed.