Distance from Subset of Real Numbers

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

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

Let $$d \left({x, S}\right)$$ be the distance between $$x$$ and $$S$$.

Then:
 * $$x \in S \implies d \left({x, S}\right) = 0$$;
 * If $$S$$ is bounded above and $$x = \sup S$$, then $$d \left({x, S}\right) = 0$$;
 * Similarly, if $$S$$ is bounded below and $$x = \inf S$$, then $$d \left({x, S}\right) = 0$$;
 * If $$I$$ is a closed real interval, then $$d \left({x, I}\right) = 0 \implies x \in I$$;
 * If $$I$$ is an open real interval apart from $$\varnothing$$ or $$\R$$, then $$\exists x \notin I: d \left({x, I}\right) = 0$$.

Proof
From the definition of distance, $$\forall x, y \in \R: d \left({x, y}\right) = \left|{x - y}\right|$$.

Thus $$d \left({x, S}\right) = \inf_{y \in S} \left({\left|{x - y}\right|}\right)$$.


 * $$x \in S \implies d \left({x, S}\right) = 0$$:

Consider the set $$T = \left\{{\left|{x - y}\right|: y \in S}\right\}$$.

This has $$0$$ as a lower bound as Absolute Value Bounded Below by Zero.

So $$d \left({x, S}\right) = \inf_{y \in S} \left({\left|{x - y}\right|}\right) \ge 0$$

If $$x \in S$$ then $$\left|{x - x}\right| = 0 \in T$$ and so $$0 \le \inf_{y \in S} \left({d \left({x, y}\right)}\right)$$.

Thus $$d \left({x, S}\right) = \inf_{y \in S} \left({d \left({x, y}\right)}\right) = 0$$.


 * If $$S$$ is bounded above and $$x = \sup S$$, then $$d \left({x, S}\right) = 0$$:

Let $$x = \sup S$$.

Then $$\forall y \in S: \left|{x - y}\right| = x - y$$

So we need to show that no $$h > 0$$ can be a lower bound for $$T = \left\{{\left|{x - y}\right|: y \in S}\right\}$$.

Suppose this is false, and $$\exists h > 0: \forall y \in S: x - y \ge h$$.

But then $$\forall y \in S: y \le x - h$$ and hence $$x - h$$ is a lower bound for $$T$$ smaller than $$x = \sup S$$ which is supposed to be the supremum, i.e. the smallest upper bound.

So there is no such $$h > 0$$ and so $$d \left({x, S}\right) = 0$$.


 * If $$S$$ is bounded below and $$x = \inf S$$, then $$d \left({x, S}\right) = 0$$:

Consider $$d \left({-x, S'}\right)$$ where $$S' = \left\{{-x: x \in S}\right\}$$.

By Negative of Infimum, $$x = \inf S \implies-x = \sup S'$$.

Thus from the above, $$d \left({-x, S'}\right) = 0$$ and hence the result.


 * If $$I$$ is a closed real interval, then $$d \left({x, I}\right) = 0 \implies x \in I$$:

Since $$I$$ 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:

$$ $$ $$

Now from Infimum Plus Constant, $$\inf_{y \in S} \left|{x - y}\right| = x - B + \inf_{y \in S} \left|{B - y}\right|$$.

But we also have:
 * $$x - B \ge 0$$;
 * $$d \left({B, S}\right) \ge 0$$;
 * $$d \left({x, S}\right) = 0$$.

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

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


 * If $$I$$ is an open real interval apart from $$\varnothing$$ or $$\R$$, then $$\exists x \notin I: d \left({x, I}\right) = 0$$:

As $$I \ne \varnothing$$ and $$I \ne \R$$ it follows that one of the following applies:


 * $$\exists a, b \in \R: I = \left({a \, . \, . \, b}\right)$$;
 * $$\exists a \in \R: I = \left({a \, . \, . \, \infty}\right)$$;
 * $$\exists b \in \R: I = \left({-\infty \, . \, . \, b}\right)$$.

It follows by the definition of open real interval that $$I$$ has either an infimum $$a$$, or a supremum $$b$$, or both. Thus the required value of $$x$$, from what has been proved above, is either $$a$$ or $$b$$.