Let $X$ be a normed space and let $Y\subset X$ be a subspace of codimension 1. Let $\operatorname{cl}_{\widehat{X}}(Y)$ be the closure of $Y$ in the completion $\widehat{X}$ of $X$. Is still the dimension of $\widehat{X}/\operatorname{cl}_{\widehat{X}}(Y)$ equal to one?
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2No. Your subspace $Y$ can be dense in $X$, and in this case you have $\overline Y=\widehat X$. – Etienne Jul 21 '13 at 21:21
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Take any non-zero discontinuous linear functional on $X$, say $f$. Then $\operatorname{Ker}(f)$ is dense in $X$. It is remains to put $Y=\operatorname{Ker}(f)$, so we have $\operatorname{cl}_{X}(Y)=X$. Since $\operatorname{cl}_{\widehat{X}}(X)=\widehat{X}$ by construction of completion, then $$ \operatorname{cl}_{\widehat{X}}(Y)=\operatorname{cl}_{\widehat{X}}(\operatorname{cl}_{\widehat{X}}(Y))=\operatorname{cl}_{\widehat{X}}(X)=\widehat{X} $$ and we conclude $\operatorname{dim}(\widehat{X}/\operatorname{cl}_{\widehat{X}}(Y))=0$.
