Set Closure is Smallest Closed Set/Normed Vector Space

Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.

Let $S$ be a subset of $X$:


 * $S \subseteq X$

Let $S^-$ be the closure of $S$.

Then $S^-$ is the smallest closed set which contains $S$.

Proof
Let $F$ be a closed set in $X$.

Suppose $S \subseteq F$.

$S^-$ is contained in $F$
Let $L$ be a limit point of $S$.

Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S \setminus \set L$ which converges to $L$.

In other words:


 * $\forall n \in \N : x_n \in S \setminus \set L$.

Furthermore:


 * $S \setminus \set L \subseteq S \subseteq F$

Since $F$ is closed, $L \in F$.

So all limit points of $S$ belong to $F$.

Hence, $S^- \subseteq F$.

$S^-$ is closed
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $S^-$.

Let $\sequence {x_n}_{n \mathop \in \N}$ converge to $L$:


 * $\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall n \in \N: n > N \implies \norm {x_n - L} < \epsilon$

We can have either $L \in S$ or $L \notin S$.

Suppose $L \in S$.


 * By definition, $L \in S^-$.

Suppose $L \notin S$.

Define a new sequence $x_n'$ using $x_n$ as follows:


 * $(1): \quad$ if $x_n \in S$, then $x_n' := x_n$;


 * $(2): \quad$ if $x_n \notin S$, then $x_n$ is a limit point of $S$.


 * Then there is open ball $\ds \map {B_{\frac 1 n}} {x_n}$ which has an element of $S$.


 * Define $x_n'$ such that $x_n' \in S$ and $x_n' \in \map {B_{\frac 1 n}} {x_n}$

Suppose $x_n \in S$.

Then:


 * $\norm {x_n' - L} = \norm {x_n - L}$

Suppose $x_n \notin S$.

Then:

Thus, $\sequence {x_n'}_{n \mathop \in \N}$ is a sequence in $S \setminus \set L$ which converges to $L$.

So $L$ is a limit point of $S$:


 * $L \in S^-$.

By definition, $S^-$ is closed.

$S^-$ is the smallest closed set containing $S$
there exists a closed set $Q$ smaller than $S^-$ which contains $S$.

$S^-$ differs from $S$ only by limit points of $S$.

If $Q$ is smaller than $S^-$, it has to contain fewer limit points of $S$ than $S^-$.

Hence, $Q$ would not contain all its limit points.

By definition, $Q$ would not be closed.

This is a contradiction.