Interval Divided into Subsets

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Theorem

Let $\mathbb I$ be a real interval.

Let $S$ and $T$ be non-empty subsets of $\mathbb I$ such that $\mathbb I = S \cup T$.

Then one of $S$ or $T$ contains an element at zero distance from the other.

$\mathbb I = S \cup T \implies \forall x \in \mathbb I: x \in S \lor x \in T$ from the definition of union.

That is, every element of $\mathbb I$ belongs either to $S$ or to $T$.

The distance of a point $c \in \R$ from a subset $S$ of $\R$ is given as:

$\displaystyle d \left({c, S}\right) = \inf_{x \in S} \left({\left|{c - x}\right|}\right)$

Assume that $S$ and $T$ have no elements in common, otherwise the result is trivial.

Suppose that $\exists s \in S, t \in T$ such that $s < t$.

(If not, then $\exists s \in S, t \in T$ such that $s > t$, and the following argument may be amended appropriately.)

Let $T_0 = \left\{{x: x \in T: x > s}\right\}$.

As $t \in T_0$ it follows that $T_0 \ne \varnothing$. Also, $T_0$ is bounded below by $s$.

Let $b = \inf \left({T_0}\right)$.

If $b \notin T$ then $b \in S$.

But from Distance from Subset of Real Numbers‎, it follows that $d \left({b, T_0}\right) = 0$.

Thus we have found a point in $S$ which is zero distance from $T$.

Otherwise, $b \in T$.

Then $b > s$, and the open interval $\left({s . . b}\right)$ is a non-empty subset of $S$.

Hence $b$ is a point in $T$ which is zero distance from $S$.

$\blacksquare$