# closure of a vector subspace in a normed space is a vector subspace

Let $(V,\|\cdot\|)$ be a normed space, and $S\subset V$ a vector subspace. Then $\overline{S}$ is a vector subspace in $V$.

Proof

First of all, $0\in\overline{S}$ because $0\in S$. Now, let $x,y\in\overline{S}$, and $\lambda\in K$ (where $K$ is the ground field of the vector space  $V$). Then there are two sequences in $S$, say $(x_{n})_{n\in\mathbb{N}}$ and $(y_{n})_{n\in\mathbb{N}}$ which converge to $x$ and $y$ respectively.

Then, the sequence $(x_{n}+\lambda\cdot y_{n})_{n\in\mathbb{N}}$ is a sequence in $S$ (because $S$ is a vector subspace), and it’s trivial (use properties of the norm) that this sequence converges to $x+\lambda\cdot y$, and so this sum is a vector which lies in $\overline{S}$.

We have proved that $\overline{S}$ is a vector subspace. QED.

Title closure  of a vector subspace in a normed space is a vector subspace ClosureOfAVectorSubspaceInANormedSpaceIsAVectorSubspace 2013-03-22 15:00:16 2013-03-22 15:00:16 gumau (3545) gumau (3545) 7 gumau (3545) Result msc 15A03 msc 46B99 msc 54A05 ClosureOfAVectorSubspaceIsAVectorSubspace2 ClosureOfSetsClosedUnderAFinitaryOperation