# 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 |
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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 |