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

Let $(V,\parallel \cdot \parallel )$ 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
Canonical name	ClosureOfAVectorSubspaceInANormedSpaceIsAVectorSubspace
Date of creation	2013-03-22 15:00:16
Last modified on	2013-03-22 15:00:16
Owner	gumau (3545)
Last modified by	gumau (3545)
Numerical id	7
Author	gumau (3545)
Entry type	Result
Classification	msc 15A03
Classification	msc 46B99
Classification	msc 54A05
Related topic	ClosureOfAVectorSubspaceIsAVectorSubspace2
Related topic	ClosureOfSetsClosedUnderAFinitaryOperation