# Problem involving a hyperplane and affine subspace II

I am trying to solve this little problem.

Suppose you have a normed vector space $E$. Let $H$ be a hyperplane ( $H=\{x\in E: f(x)= \alpha\}$ for some linear functional $f$ and some real number $\alpha$) and let $V$ be an affine subspace (i.e. $V=U+a$ where $a \in E$ and $U$ is a vector subspace of $E$) that contains $H$.

In my previous post I proved the following: either $H=V$ or $V=E$.

Check it out here: A problem involving a hyperplane and an affine subspace

Now, I want to deduce from this that H is either closed or dense in $E$. Any thoughts?

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If $f$ is not identically $0$, then let $x_0$ such that $f(x_0)=\alpha$. Then for $x\in\ker f$: $f(x+x_0)=\alpha$ so $\ker f+x_0\subset H$ and if $y\in H$ then $f(y)=\alpha=f(x_0)$ so $y-x_0\in\ker f$ and $y\in \ker f+x_0$. So $H=\ker f+x_0$, and so $H\subset \overline{\ker f}+x_0$, and $\overline{\ker f}$ is a linear subspace of $E$. So either $\overline{\ker f}+x_0=H$ hence $H$ is closed, or $\overline{\ker f}+x_0=E$, hence $\overline{\ker f}=E$ and $H$ is dense.