Distance between Closed Sets in Euclidean Space
Theorem
Let $S, T \subseteq \R^n$ be closed, non-empty subsets of the real Euclidean space $R^n$.
Suppose that $S$ is bounded, and $S$ and $T$ are disjoint.
Then there exists $x \in S$ and $y \in T$ such that:
- $\map d {x, y} = \map d {S, T} > 0$
where:
- $d$ denotes the Euclidean metric
- $\map d {S, T}$ is the distance between $S$ and $T$.
Proof
By definition of distance from subset, we can for all $n \in \N$ find $x_n \in S, y_n \in T$ such that:
- $\map d {S, T} \le \map d {x_n, y_n} < \map d {S, T} + \dfrac 1 n$
so:
- $\ds \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
By definition of bounded space, there exists $a \in S$ and $K \in \R$ such that for all $x \in S$, we have $\map d {x, a} \le K$.
It follows that $\sequence {x_n}$ is a bounded sequence.
Then $\sequence {y_n}$ is also a bounded sequence, as:
\(\ds \map d {y_n, a}\) | \(\le\) | \(\ds \map d {y_n, x_n} + \map d {x_n, a}\) | Triangle Inequality for Vectors in Euclidean Space | |||||||||||
\(\ds \) | \(<\) | \(\ds \map d {S, T} + \dfrac 1 n + K\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \map d {S, T} + 1 + K\) |
The sequence $\sequence {\tuple {x_n, y_n} }$ in $\R^{2 n}$ is also bounded, as:
\(\ds \map d {\tuple {x_n, y_n}, \tuple {a, a} }\) | \(=\) | \(\ds \paren {\sum_{i \mathop = 1}^n \paren {x_i - a}^2 + \sum_{i \mathop = 1}^n \paren {y_i - a}^2}^{1 / 2}\) | Definition of Euclidean Metric on Real Vector Space | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{i \mathop = 1}^n \paren {x_i - a}^2 + \sum_{i \mathop = 1}^n \paren {y_i - a}^2\) | by Minkowski's Inequality for Sums | |||||||||||
\(\ds \) | \(\le\) | \(\ds 2 K + \map d {S, T} + 1\) |
From Bounded Sequence in Euclidean Space has Convergent Subsequence, it follows that $\sequence {\tuple {x_n, y_n} }$ has a subsequence $\sequence {\tuple {x_{n_r}, y_{n_r} } }_{r \mathop \in N}$ that converges to a limit $\tuple {x, y} \in \R^{2n}$.
Then $\ds \lim_{r \mathop \to \infty} x_{n_r} = x$, and $\ds \lim_{r \mathop \to \infty} y_{n_r} = y$.
From Closed Set iff Contains all its Limit Points, it follows that $x \in S$, and $y \in T$.
Then from Distance Function of Metric Space is Continuous:
- $\ds \lim_{r \mathop \to \infty} \map d {x_{n_r}, y_{n_r} } = \map d {x, y}$
As a Convergent Sequence in Metric Space has Unique Limit, we have:
- $\ds \map d {x, y} = \lim_{r \mathop \to \infty} \map d {x_{n_r}, y_{n_r} } = \lim_{n \mathop \to \infty} \map d {x_n, y_n} = \map d {S, T}$
As $S$ and $T$ are disjoint, it follows that $x \ne y$.
Hence:
- $0 < \map d {x, y} = \map d {S, T}$
$\blacksquare$
Sources
- 2001: Christian Berg: Kompleks funktionsteori: $\S A$