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Let $X$ be a metric space and $A$ and $B$ two subsets of $X$. If $A$ is closed, $B$ is a compact, and $A\cap B=\emptyset$, is it true that there is $d>0$ such that $\operatorname d (x,y)\ge d$ for all $y\in A$ and $x\in B$?

If so, is it the same as saying that the set $A\cup B$ is not connected?

Is it still true if $B$ is only closed? If not, can you give a counterexample in $\Bbb R^2$?

User
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    The first one is true --- there should be a proof in any book or notes on metric spaces. The last question is not true: think of the parabola xy=1, and either one of the coordinate axes. –  Feb 13 '14 at 11:11
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    See this for your first question. – David Mitra Feb 13 '14 at 11:13
  • Thanks! What about the not-connectedness? – User Feb 13 '14 at 11:17
  • For every point $p$ in $B$ we can take an open ball $O_p$ of radius less than $d$. Now the union of such $O_p$s is open, contains $B$ and does not intersect $A$. Take then $X\setminus B$: it is an open set containing $A$ that does not intersect $B$. So $A\cup B$ is not connected because $A$ and $B$ are open in $A\cup B$. Is it right? – User Feb 13 '14 at 11:45

1 Answers1

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  1. If $A$ is closed and $B$ is compact, then it is true. For if it not there would be $\{x_n\}\subset A$ and $\{y_n\}\subset B$ with $d(x_n,y_n)<1/n$. But as $B$ is compact, then $\{y_n\}$ has a convergent subsequence $y_{n_k}\to y$. But $$ d(x_{n_k},y)\le d(x_{n_k},y_{n_k})+d(y_{n_k},y)\to 0, $$ and thus $x_{n_k}\to y$. But this implies that $y\in A\cap B$, which contradicts the fact that $A$ and $B$ are disjoint.

  2. Take $A=\{(x,y)\in \mathbb R^2: y\le 0 \}$ and $B=\{(x,y)\in \mathbb R^2: y(1+x^2)\ge 1 \}$. Then they are both closed and their distance in zero, as for $u_n=(n,0)\in A$ and $v_n=\big(n,(1+n^2)^{-1}\big)$, their distance is $(1+n^2)^{-1}$ which can become arbitrarily small.

  3. In both cases $A\cup B$ is not connected, but that connected components of a disconnected set my have zero distance between each other, as in 2.

Yiorgos S. Smyrlis
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