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If $E$ is a subspace of a normed space $X$, then $E^{\perp}$ is closed

Let $X$ be a normed space and let $E$ be a subspace of $X$. Show that $E^{\perp}$ is closed.

So to show that $E^{\perp}$ is closed it suffices to show that every converging net in $E^{\perp}$ converges. I’m not sure how to show this though. I thought about using the closed graph theorem, but this is only applicable if $E^{\perp}$ is open. Any hints on how to proceed?

A:

If $E$ is a subspace of $X$, then $E^\perp$ is the orthogonal complement of $E$ in the normed space $X$, whose topology is generated by the seminorms $p_x$ defined by $p_x(y):=\|x+y\|-\|x\|-\|y\|$. If $\phi_k \subset E^\perp$ converges to $\phi$, then the restriction $\phi_k|_E$ converges to $\phi|_E$. Since the topology of $E$ is the coarsest topology making all restriction maps continuous, it follows that $\phi_k|_E$ converges to $\phi|_E$.

Quantification of contrast-enhanced high-resolution t1-weighted imaging with gadolinium chelate-based blood perfusion at 3T in the presence of severe motion artifacts: a phantom study.
Objective To assess the feasibility of contrast-enhanced high-resolution t1-weighted imaging with gadolinium chelate-based blood perfusion (CE-HRT1-BW) at 3T in the presence of
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