$\inf{d(x,y);x∈A, y∈B}>0$

Question

Let $A$ be a subset of sequence of points which converges to point a∈$R^n$. For a closed subset $B$ of $\mathbb R^n$ satisfying closure of $A$ and $B$ has no intersection,can we say $\inf{d(x,y);x∈A, y∈B}>0$? I guess true, but I cannot proof this.. Any help would be appreciated, thank you.

P.S Sorry,the first question was trivial..closure of A and B has no intersection, sorry..

Does this answer your question? $A$ and $B$ disjoint and closed implies there is positive distance between both sets — Martin R, Jul 07 '20 at 17:43
Consider $n=1$, and $B=[0,1]$ so closed and $A=(1,2)$. Then the infimum is $0$ near $1$ — Henry, Jul 07 '20 at 17:45
Take $A$ to be the graph of $y=\frac1x$ and $B$ to be the $x$-axis in $\Bbb R^2$. — Brian M. Scott, Jul 07 '20 at 17:45
What does close mean here? Compact and without boundary or any limit point belongs to the set? — C.F.G, Jul 07 '20 at 17:52

score 2 · Accepted Answer · answered Jul 07 '20 at 17:45

No. Consider the closed set $B=\{(x,0): x\in\mathbb{R}\}\subset\mathbb{R}^2$ and $A=\{(0,1/n): n\geq1\}\subset\mathbb{R}^2$. Then $\inf_{x\in A, y\in B}d(x,y)=0$.

If both $A,B$ are closed, this is still not true (Brian's comment provides a counter-example).

On the other hand: the result is true if one set is closed and the other is compact.

$\inf{d(x,y);x∈A, y∈B}>0$

1 Answers1