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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?2012-10-09
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    Perhaps it is better write codimension(Y)=1 ?2012-10-09
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    Yes, codim(Y)=1!2012-10-12

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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.2012-10-09
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    Ahh.. the word 'any' misled me.. you are right2012-10-09
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    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?2012-10-12