Closure of Open 1-Ball in Standard Discrete Metric Space

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Theorem

Let $M = \struct {A, d}$ be the standard discrete metric space on a set $A$.

Let $x \in A$.

Let $\map {B_1} x$ be the open $1$-ball of $x$ in $M$.


Then:

$\map \cl {\map {B_1} x} = \set x$

while:

$\set {y \in A: \map d {x, y} \le 1} = A$


Proof

By definition of the standard discrete metric:

$\map d {x, y} = \begin {cases}

0 & : x = y \\ 1 & : x \ne y \end {cases}$

That is:

$\forall \tuple {x, y} \in A: \map d {x, y} \le 1$

Thus:

$\set {y \in A: \map d {x, y} \le 1} = A$


From Open Ball in Standard Discrete Metric Space:

$\map {B_1} x = \set x$

Let $y \in A: y \ne x$.


By definition of closure:

$\map \cl {\map {B_1} x} = \paren {\map {B_1} x}^i \cup \paren {\map {B_1} x}'$

where:

$\paren {\map {B_1} x}^i$ denotes the set of isolated points of $\map {B_1} x$
$\paren {\map {B_1} x}'$ denotes the set of limit points of $\map {B_1} x$.


But:

from Point in Standard Discrete Metric Space is Isolated, all points in $\map {B_1} x$ are isolated:
$\paren {\map {B_1} x}^i = \map {B_1} x$
from Set in Standard Discrete Metric Space has no Limit Points, $\map {B_1} x$ has no limit points:
$\paren {\map {B_1} x}' = \O$


Hence:

\(\ds \map \cl {\map {B_1} x}\) \(=\) \(\ds \map {B_1} x \cup \O\)
\(\ds \) \(=\) \(\ds \map {B_1} x\) Union with Empty Set
\(\ds \) \(=\) \(\ds \set x\)

$\blacksquare$


Sources

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