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, $S^- \subseteq F$.
$\Box$
$S^-$ 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 of closure, $L \in S^-$.
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:
\(\ds \norm {x_n' - L}\) | \(=\) | \(\ds \norm {x_n' - x_n + x_n - L}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \norm {x_n' - x_n} + \norm {x_n - L}\) | Norm Axiom $\text N 3$: Triangle Inequality | |||||||||||
\(\ds \) | \(<\) | \(\ds \frac 1 n + \norm {x_n - L}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \frac 1 n + \epsilon\) |
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.
$\Box$
$S^-$ is the smallest closed set containing $S$
Aiming for a contradiction, suppose 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.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces