every finite dimensional subspace of a normed space is closed


Theorem 1

Proof. Let (V,) be such a normed vector space, and SV a finite dimensional vector subspace.

Let xV, and let (sn)n be a sequence in S which converges to x. We want to prove that xS. Because S has finite dimension, we have a basis {x1,,xk} of S. Also, xspan(x1,,xk,x). But, as proved in the case when V is finite dimensional (see this parent (http://planetmath.org/EverySubspaceOfANormedSpaceOfFiniteDimensionIsClosed)), we have that S is closed in span(x1,,xk,x) (taken with the norm induced by (V,)) with snx, and then xS. QED.

0.0.1 Notes

The definition of a normed vector space requires the ground field to be the real or complex numbers. Indeed, consider the following counterexample if that condition doesn’t hold:

V= is a - vector space, and S= is a vector subspace of V. It is easy to see that dim(S)=1 (while dim(V) is infinite), but S is not closed on V.

Title every finite dimensional subspace of a normed space is closed
Canonical name EveryFiniteDimensionalSubspaceOfANormedSpaceIsClosed
Date of creation 2013-03-22 14:58:56
Last modified on 2013-03-22 14:58:56
Owner Mathprof (13753)
Last modified by Mathprof (13753)
Numerical id 14
Author Mathprof (13753)
Entry type Corollary
Classification msc 46B99
Classification msc 15A03
Classification msc 54E52