# Distance from Subset of Real Numbers to Element

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## 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 $S$.

Then:

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

## Proof 1

From the definition of distance:

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

Thus:

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

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

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

So:

- $\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\size {x - y} } \ge 0$

If $x \in S$ then:

- $\size {x - x} = 0 \in T$

and so:

- $\displaystyle 0 \le \map {\inf_{y \mathop \in S} } {\map d {x, y} }$

Thus:

- $\displaystyle \map d {x, S} = \map {\inf_{y \mathop \in S} } {\map d {x, y} } = 0$

$\blacksquare$

## Proof 2

Recall from Real Number Line is Metric Space that the set of real numbers $\R$ with the distance function $d$ is a metric space.

The result is then seen to be an example of Distance from Subset to Element.

$\blacksquare$

## Examples

### From $3$ to $\openint 2 3$

Consider the open real interval $I = \openint 2 3$.

The real numbers $2$ and $3$ are both zero distance from $I$ but are not elements of $I$.