Closure of Symmetric Subset of Normed Vector Space is Symmetric

Theorem

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

Let $A \subseteq X$ be symmetric.

Then the topological closure of $A$ is symmetric.

Proof

Let $A^-$ be the topological closure of $A$.

Let $a \in A^-$.

Then from Point in Closure of Subset of Metric Space iff Limit of Sequence, we have:

there exists a sequence $\sequence {a_n}_{n \mathop \in \N}$ in $A$ such that $a_n \to a$.

Since $A$ is symmetric, we have:

$-a_n \in A$ for each $n \in \N$.

We also have, from Multiple Rule for Sequences in Normed Vector Space:

$-a_n \to -a$

So we have:

$-a \in A^-$

$\blacksquare$