Is the Set C+Fx Closed in a Metric Space?

[iknowone]

Let X be a normed vector space. If C is a closed subspace x is a point in X not in C, show that the set C+Fx is closed. (F is the underlying field of the vector space).

 

[Marin]

Hi there!

Here's my suggestion, but it needs an overview from someone familiar with functional analysis, to be sure it's correct or incorrect:

As X is a normed space, C and Fx are also normed as subspaces of a normed space. So the convergence in these subspaces is set with respect to the original norm of X.

Let c_n, n\in N;c_n\longwrightarrow\c,c\in C be a convergent sequence in C+Fx with limit c. Then, following the structure of the space (because x does not lie in C, C+Fx is a direct sum), we could write c_n=a_n+\lambda_n x; a_n\in C,\lambda_n\in F both convergent sequences a_n\rightarrow a ;\lambda_n\rightarrow\lambda (otherwise c_n could not be convergent).

Since C is closed then a\in C and (assuming F is closed or complete)*** \lambda\in F. So the limit takes the form c_n\rightarrow a+\lambda x\in C+Fx and therefore lies in the space. *** I am not sure, but I think F must also be given as closed. (as a counter example consider the space R^2 over the rationals Q: Set C to be the disc around the origin (closed) and x be some point not in C)

I hope this proof is true at least to some extend.

[edit: spelling :)]

 

[g_edgar]

Marin is right, you need F to be a complete field, for example the reals or complexes. Otherwise, from \lambda_n x convergent we cannot conclude the limit is a scalar multiple of x.

 

[Marin]

Hi again!

As I look at the proof once again (assuming the proof is correct), the condition X to be a normed space could actually be reduced. The same proof will also be true for any metric space (X,d) with a metric d , as the metric induces the same metric on the subspaces.