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:


 * $d \left({x, y}\right) = d \left({S, T}\right) > 0$

Here, $d$ denotes the Euclidean metric, and $d \left({S, T}\right)$ 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:


 * $d \left({S, T}\right) \le d \left({x_n, y_n}\right) < d \left({S, T}\right) + \dfrac 1 n$

so $\displaystyle \lim_{n \to \infty} d \left({x_n, y_n}\right) = d \left({S, T}\right)$.

By definition of bounded space, there exists $a \in S$ and $K \in \R$ such that for all $x \in S$, we have $d \left({x, a}\right) \le K$.

It follows that $\left\langle {x_n}\right\rangle$ is a bounded sequence.

Then $\left\langle {y_n}\right\rangle$ is also a bounded sequence, as:

The sequence $\left\langle{\left({x_n, y_n}\right) }\right\rangle$ in $\R^{2n}$ is also bounded, as:

From Bounded Sequence in Euclidean Space has Convergent Subsequence, it follows that $\left\langle{\left({x_n, y_n}\right) }\right\rangle$ has a subsequence $\left\langle{\left({x_{n_r}, y_{n_r} }\right) }\right\rangle_{r \in N}$ that converges to a limit $\left({x, y }\right) \in \R^{2n}$.

Then $\displaystyle \lim_{r \to \infty} x_{n_r} = x$, and $\displaystyle \lim_{r \to \infty} y_{n_r} = y$.

From Closed Set iff Contains all its Limit Points, it follows that $x \in S$, and $y \in T$.

Then $\displaystyle \lim_{r \to \infty} d \left({x_{n_r}, y_{n_r} }\right) = d \left({x, y}\right)$, as a Metric is Continuous.

As a Convergent Sequence in Metric Space has Unique Limit, we have


 * $\displaystyle d \left({x, y}\right) = \lim_{r \to \infty} d \left({x_{n_r}, y_{n_r} }\right) = \lim_{n \to \infty} d \left({x_n, y_n}\right) = d \left({S, T}\right)$

As $S$ and $T$ are disjoint, it follows that $x \ne y$.

Hence, $0 < d \left({x, y}\right) = d \left({S, T}\right)$.