Singleton in Normed Vector Space is Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $X$ be a vector space over $\R$ or $\C$.

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

Let $\set x \subseteq X$ be a singleton.


Then $\set x$ is closed.


Proof

Note that either $\set x = X$ or $\set x \subset X$.

Suppose, $X = \set x$.

Then:

$X \setminus \set x = \empty$

We have that Empty Set is Open in Normed Vector Space.

$\set x$ is a complement of $\empty$ in $X$.

By definition, $x$ is closed.

Suppose, $X \ne \set x$, i.e. $\set x \subset X$.

Define $U := X \setminus \set x$.

Let $y \in U$.

Let $\epsilon := \norm {x - y}$.

Suppose, $\epsilon > 0$.

By Normed Vector Space is Open in Itself:

$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq X$

Suppose, $z \in \map {B_\epsilon} y$.

Then:

$\norm {y - z} < \epsilon$.

We have that:

\(\ds \norm {z - x}\) \(\ge\) \(\ds \norm {x - y} - \norm {y - z}\) Triangle inequality
\(\ds \) \(\ge\) \(\ds \epsilon - \norm {y - z}\)
\(\ds \) \(>\) \(\ds \epsilon - \epsilon\)
\(\ds \) \(=\) \(\ds 0\)

Hence, $z \ne x$.

Then $z$ is in the complement of $\set x$ in $X$, i.e. $z \in U$.

In other words:

$\forall y \in U : \exists \epsilon = \norm {x - y} > 0 : \map {B_\epsilon} x \subseteq U$

So $U$ is open, and $\set x = X \setminus U$ is closed.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Singleton_in_Normed_Vector_Space_is_Closed&oldid=593811"