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.
A normed space embedding question
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functional-analysis
normed-spaces
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0What does $\dim Y=\dim X+1$ mean if $X$ (and $Y$) is infinitely dimensional? – 2012-10-09
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2Perhaps it is better write codimension(Y)=1 ? – 2012-10-09
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0Yes, codim(Y)=1! – 2012-10-12
1 Answers
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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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0That $\,\Bbb R^n\,$ 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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0Ahh.. the word 'any' misled me.. you are right – 2012-10-09
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0I 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