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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.

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What does $\dim Y=\dim X+1$ mean if $X$ (and $Y$) is infinitely dimensional? –  Davide Giraudo Oct 9 '12 at 11:39
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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

1 Answer 1

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.

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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

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