Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does any normed space X can be embedded into another normed space Y, such that X is density in the Y and dim(Y)=dim(X)+1.

share|cite|improve this question
What does $\dim Y=\dim X+1$ mean if $X$ (and $Y$) is infinitely dimensional? – Davide Giraudo Oct 9 '12 at 11:39
Perhaps it is better write codimension(Y)=1 ? – Tomás Oct 9 '12 at 12:00
Yes, codim(Y)=1! – Strongart Oct 12 '12 at 5:47

You mean 1 codimensional?

Well, the answer is no, the usual $\mathbb R^n$ is not going to be dense in $\mathbb R^{n+1}$ (every norm on finite dimension determines the same topology).

Ahh.. you asked whether exists such a situation? So, in finite dimension it cannot exist by the above argument, but in infinite dimension, of course:

Take any proper dense subspace $Y$ of an infinite dimension normed space (I bet, such always exists, but for example $X:=L_1[0,1]$ and $Y:=C[0,1]$ with the $L_1$-norm), and extend algebraically its basis -using axiom of choice- to a basis of $X$ and leave one basis vector.

share|cite|improve this answer
That $\,\Bbb R^n\,$ is not going to be dense in $\,\Bbb R^{n+1}\,$ does not prove yet that what the OP asked cannot be attained. – DonAntonio Oct 9 '12 at 12:03
Ahh.. the word 'any' misled me.. you are right – Berci Oct 9 '12 at 21:59
I find a claim: a finite codimension space must be a summand, so the space Y must be closed in the space X, it is also no, but your example means yes, what is the matter? – Strongart Oct 12 '12 at 5:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.