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I want to prove the following: Let $X$ be an infinite dimensional normed space. For all integer $n\geq1$: $X$ has a dense subspace of codimension $n$, i.e. a subspace $L$ such that $\dim(X/L)=n$.

How can i do this? My first thought was: Take a basis $e_1,e_2,\ldots$ such that $X=Span(e_1,e_2,\ldots)$ and let $L=Span(e_2,e_3,\ldots)$ than $\dim(X/L)=1$. On the same way you can go on. Is this the right method to conclude the result?

Thank you.

user95344
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1 Answers1

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You can prove this by induction.

Basis of induction: Since $X$ is infinite dimensional there exists a discontinuous linear functional $f$ on $X$. Since $f$ is discontinuous, then $\ker f$ is dense in $X$. As $f$ is a functional we have $\dim (X/\ker f)=1$.

Step of induction: Assume we have constructed a dense subspace $X_n$ of codimension $n$, then apply the argument given above to get a dense in $X_n$ subspace $E$ of codimension $1$ in $X_n$. Obviously, $E$ is of codimension $n+1$ in $X$ and still dense in $X$. Now set $X_{n+1}:=E$.

Norbert
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  • Why is Ker(f) dense in X? Should i use $f^2$ to continue? – user95344 Sep 17 '13 at 14:32
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    These two links explain the assertions made in the answer: http://math.stackexchange.com/questions/99206/discontinuous-linear-functional, http://math.stackexchange.com/questions/123282/linear-functional-on-a-banach-space-is-discontinuous-then-its-nullspace-is-dense –  Sep 17 '13 at 14:32