1

Let $G$ be an open subset of $\mathbb C$. I would like to prove that this set $\{z\in G; d(z,\mathbb C-G)\ge 1/n\}$, where $n\in \mathbb R$, is well-defined. In another words, I would like to know if this distance is always well-defined for an open set $G\subset \mathbb C$.

user42912
  • 23,582
  • It is well-defined for all $G$. If $G=\mathbb{C}$, then $d(G,\mathbb{C})=\infty$ if you take infimum in $[0,\infty]$. To make $d(G,\mathbb{C})<\infty$, it should be a proper open subset $G$ in $\mathbb{C}$. – Rubertos May 30 '16 at 10:03
  • @Rubertos Why is this distance well-defined? thanks! – user42912 May 30 '16 at 10:04
  • @Rubertos is it some well-known fact? – user42912 May 30 '16 at 10:05
  • 1
    In a metric space $X$ and $A\subset X$ and $x\in X$, $d(x,A)$ is just defined as the infimum of ${d(x,a):a\in A}$. Depend on the number system you are taking infimum on this set, it is well-defined. Usually, people restrict $A$ to be a nonempty set. – Rubertos May 30 '16 at 10:07
  • @Rubertos in this case, we have had $d(G,\mathbb C)=0$, no? – user42912 May 30 '16 at 10:43

1 Answers1

2

Conventionally, in a metric space $(X,d)$ (in your case it's just the complex plane with Euclidean metric), the distance between any two nonempty sets $A,B$ can be defined as $$d(A,B):=\inf_{x\in A,y\in B} d(x,y).$$ (The infimum is of course defined because of the non-emptiness.)
[EDIT: Also note that the inf is not necessarily attained in general. A quick example is $A=\{1/n\},B=\{0\}$. Hence, even if two sets are zero-distanced, it doesn't imply that they intersect.
However, if one set is closed and the other one is compact, then zero distance is indeed equivalent to intersection. This follows from the facts that:
1). $d(z,B)$ when regarded as a function of $z$ is continuous.
2). Continuous functions on compact sets attain their minimum values.
3). Limit points of closed sets don't overflow. ]

If $A=\{x\}$, containing exactly one point, then we conventionally write $d(x,B)$ for $d(A,B)$.

If, furthermore, $B=\{y\}$ is also a single-point set, then we will write $d(x,y)$, which you should check is compatible with the original definition of the metric $d$.

Vim
  • 13,640
  • If every distance between sets is achieved why have we to care to prove this question http://math.stackexchange.com/questions/845799/showing-that-the-minimum-distance-between-a-closed-and-compact-set-is-attained ? – user42912 May 30 '16 at 10:49
  • 1
    @user42912 minimum $\ne$ infimum. Distance is defined to be the inf, not necessarily attained, but it always exists. – Vim May 30 '16 at 10:50
  • Of course, thanks! – user42912 May 30 '16 at 10:51
  • Can I say that the infimum always will exist because the distance function is lower bounded? – user42912 May 30 '16 at 10:53
  • @user42912 there are many cases in which the distance between two nonempty sets is zero but they don't. For instance, let $A={\frac1n},B={0}$. However, if one is closed and another is compact, then distance=0 implies intersection. – Vim May 30 '16 at 10:54
  • @user42912 yes. It's alway non-negative. (Of course, assuming we're only interested in nonempty sets). – Vim May 30 '16 at 10:55
  • I think this is the key point I've just realized. Maybe it should be interesting add this fact in your answer for future viewers. – user42912 May 30 '16 at 10:57
  • @user42912 no problem. I'll write an addendum concerning this distinction. – Vim May 30 '16 at 10:59