If $H$ is a Hilbert space and $M$ is a nonempty,closed, bounded and convex subset(not necessarily a subspace)of $H$, then is it true that $M$ is complete? If it is, then can we use it without proof? I mean is it a theorem?
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http://math.stackexchange.com/questions/244661/showing-that-if-a-subset-of-a-complete-metric-space-is-closed-it-is-also-comple – Ilham May 03 '15 at 00:54
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Doesn't it affect that $H$ is not a metric space? – Extremal May 03 '15 at 00:57
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4A Hilbert space is a complete metric space with the metric induced by the inner product. – Ilham May 03 '15 at 00:57
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Every closed subset (in your case $M$) of a complete space (in your case $H$) is itself complete as a metric space.
Take a Cauchy sequence $(x_n)_{\{n\in\mathbb{N}\}}$ in $M$. This is also a Cauchy sequence in $H$, so it has some limit $x\in H$. Since $M$ is closed, this limit is in $M$, so $M$ is itself complete.
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