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$M$ closed $\Rightarrow$ If $u_n\in$ M and $u_n\to u$ then $u \in M$

$M$ dense in $H$ $\Rightarrow$ $\forall u \in H \ \exists u_n\in M: u_n\to u$

$M$ subspace of $H$ $\Rightarrow$ $M \subset$ H

How I prove that $M$ is in fact the Hilbert space $H$?

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On any topological space $X$, the only subset $S$ which is both closed and dense is $X$ itself. That's so because $S=\overline S=X$.

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Let $u\in H$. Then, since $M$ is dense, there exists $(u_n)\subset M$ such that $u_n\to u$. Since $M$ is closed, $u\in M$. So, $H\subset M\subset H$, i.e., $M = H$.

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