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Example to show the distance between two closed sets can be 0 even if the two sets are disjoint

Let $(X,d)$ be a metric space and $A,B$ be two distinct closed set in $X$ such that $dist(A,B)=0$. Does it imply $A\cap B=\emptyset$ ?

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By choosing $A=B$ we see that it does not imply. Did you mean instead that does it imply $A\cap B\neq \emptyset$? –  Thomas E. May 16 '12 at 12:36
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@GEdgar: At present, no, it's just a triviality. –  Chris Eagle May 16 '12 at 13:21
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marked as duplicate by GEdgar, Jennifer Dylan, sdcvvc, Old John, Michael Greinecker Aug 29 '12 at 16:09

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2 Answers

Can you find two closed, unbounded subsets of, say, ${\bf R}^2$ that come arbitrarily close to each other as they go off to infinity?

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Yeah as Gerry says, this isn't always true: consider the $x$-axis and $\{(x, e^{-x}) \; x \in \mathbb{R}\}$ in $\mathbb{R}^2$: the second asymptotically approaches $y = 0$ as $x \rightarrow \infty$, but never quite hits it.

However, this is true once one of either $A$ or $B$ is compact, say $A$.

To see this, as the distance between them is 0, we can take sequences of points $a_i \in A, b_i \in B$ such that $d(a_i, b_i) < \frac{1}{i}$, now extract a convergent subsequence $a'_i$ from the $a_i$, we have $a'_i \rightarrow a$, $b'_i \rightarrow a$, hence $a \in A \cap B$, as desired.

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