every normed space with Schauder basis is separable


dense in

Here we show that every normed space that has a Schauder basis is separablePlanetmathPlanetmath (http://planetmath.org/Separable). Note that we are (implicitly) assuming that the normed spaces in question are spaces over the field K where K is either or . So let (X,) be a normed space with Schauder basis, say S={e1,e2,}. Notice that our notation implies that S is infinite. In finite dimensional case, the same proof with a slight modification will yield the result.

Now, set Q to be the set of all finite sums q1e1++qnen such that each qj=aj+bji where aj,bj. Clearly Q is countable. It remains to show that Q is dense (http://planetmath.org/Dense) in X.

Let ϵ>0. Let xX. By definition of Schauder basis, there is a sequence of scalars (αn) and there exists N such that for all nN we have,


But then in particular,


Furthermore, by density of in , we know that there exist constants a1,,aN,b1,,bN in such that,


By triangle inequalityPlanetmathPlanetmath we obtain:


Noting that


is an element of Q (by construction of Q) and that x and ϵ were arbitrary, we conclude that every neighborhoodMathworldPlanetmathPlanetmath of x contains an element of Q, for all x in X. This proves that Q is dense in X and completesPlanetmathPlanetmathPlanetmathPlanetmath the proof.

Title every normed space with Schauder basis is separable
Canonical name EveryNormedSpaceWithSchauderBasisIsSeparable
Date of creation 2013-03-22 17:36:07
Last modified on 2013-03-22 17:36:07
Owner asteroid (17536)
Last modified by asteroid (17536)
Numerical id 11
Author asteroid (17536)
Entry type Theorem
Classification msc 42-00
Classification msc 15A03
Classification msc 46B15