# Real Number at Distance Zero from Closed Real Interval is In Interval

## Theorem

Let $S$ be a subset of the set of real numbers $\R$.

Let $x \in \R$ be a real number.

Let $\map d {x, S}$ be the distance between $x$ and $S$.

Let $I \subseteq \R$ be a closed real interval.

Then:

$\map d {x, I} = 0 \implies x \in I$

## Proof

From the definition of distance:

$\forall x, y \in \R: \map d {x, y} = \size {x - y}$

Thus:

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

Because $I$ is an interval, if $x \notin I$ then $x$ is either an upper bound or a lower bound for $I$.

Suppose $x$ is an upper bound for $I$.

Let $B$ be the supremum of $I$.

Then because $I$ is closed, $B \in I$.

So:

 $\displaystyle \forall y \in I: \ \$ $\displaystyle \size {x - y}$ $=$ $\displaystyle x - y$ $\displaystyle$ $=$ $\displaystyle x - B + B - y$ $\displaystyle$ $=$ $\displaystyle x - B + \size {B - y}$

Now from Infimum Plus Constant:

$\inf_{y \mathop \in S} \size {x - y} = x - B + \inf_{y \mathop \in S} \size {B - y}$

But we also have:

$x - B \ge 0$
$\map d {B, S} \ge 0$
$\map d {x, S} = 0$

So it follows that $x = B$ and so $x \in I$.

A similar argument applies if $x$ is a lower bound for $I$.

$\blacksquare$