Closed Linear Span is Closed Vector Subspace

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \tau}$ be a topological vector space over $\GF$.

Let $S \subseteq X$.

Let $\map \span S$ be the linear span of $S$.

Let $\cl$ be the closure taken in $\struct {X, \tau}$.

Let $\map \cl {\map \span S}$ be the closed linear span of $S$ in $\struct {X, \tau}$.


Then $\map \cl {\map \span S}$ is a closed vector subspace in $\struct {X, \tau}$.


Proof

From Linear Span is Linear Subspace, $\span S$ is a vector subspace of $\struct {X, \tau}$.

From Closure of Linear Subspace of Topological Vector Space is Linear Subspace, $\map \cl {\map \span S}$ is a vector subspace in $\struct {X, \tau}$.

From Topological Closure is Closed, $\map \cl {\map \span S}$ is closed.

Hence $\map \cl {\map \span S}$ is a closed vector subspace in $\struct {X, \tau}$.

$\blacksquare$

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