Neighbourhood of Point Contains Point of Subset iff Distance is Zero

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Theorem

Let $M = \struct {X, d}$ be a metric space.

Let $A \subseteq X$ be a non-empty subset of $X$.

Let $x \in X$.


Then every neighborhood of $x$ contains a point of $A$ if and only if:

$\map d {x, A} = 0$

where $\map d {x, A}$ denotes the distance from $x$ to $A$.


Proof

Sufficient Condition

Let $x \in X$.

Let every neighborhood of $x$ contain a point in $A$.

Every open ball $\map {B_\epsilon} x$ centered at $x$ is seen to be a neighborhood of $x$ in $M$.


But then this implies that for every $\epsilon \in \R_{\gt 0}$ there must exists a $y \in A$ such that:

$\map d {x, y} < \epsilon$


From this it is seen that $\map d {x, A} = 0$.

$\Box$


Necessary Condition

Let $\map d {x, A} = 0$ for $x \in X$.

Let $S$ be a neighborhood of $x$ in $M$.


From the definition of a neighborhood, there exists an open ball $\map {B_\epsilon} x$ centered at $x$ such that:

$\map {B_\epsilon} x \subseteq S$


We have that:

$\epsilon < 0$

and $\map d {x, A} = 0$

Hence there must exists a $y \in A$ such that:

$\map d {x, y} < \epsilon \implies \map {B_\epsilon} x \cap A \ne \O$

If otherwise then $\epsilon$ would be a lower bound for $\set {\map d {x, y} : y \in A}$ that is greater than $0$, which is a contradiction.


Then from Set Intersection Preserves Subsets we have:

$\map {B_\epsilon} x \cap A \subseteq S \cap A$

and so:

$\map {B_\epsilon} x \cap A \ne \O \implies S \cap A \ne \O$


Hence the result.

$\blacksquare$


Sources

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