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 $\mathbb I \subseteq S \cup T$.


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


Proof

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

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


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

$\displaystyle \map d {c, S} = \map {\inf_{x \mathop \in S} } {\size {c - x} }$


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

Without loss of generality, 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 $T_0 = \set {x: x \in T: x > s}$.

As $t \in T_0$ it follows that $T_0 \ne \O$.

Also, $T_0$ is bounded below by $s$.

Let $b = \map \inf {T_0}$.


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

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

Thus we have found an element in $S$ which is zero distance from $T$.


Otherwise, $b \in T$.

Then $b > s$, and the open interval $\openint s b$ is a non-empty subset of $S$.

Hence $b$ is an element in $T$ which is zero distance from $S$.

$\blacksquare$


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