# Closed respect a norm $\|\cdot\|$

What mean that expression "closed respect a norm". For example:

$$H^{m,p}(\Omega)= \overline{\left\{{u\in C^m(\Omega); \|u\|_{W^{m,p}}<\infty}\right\}}^{\|\cdot\|}$$

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The grammatical version of "closed respect a norm $\|\cdot\|$" is closed with respect to the norm $\|\cdot\|$" – rschwieb Nov 7 '12 at 14:00

It means that in the topology generated by this norm, the set is closed. Since norms are metrics, this means that all the $\|\cdot\|$-convergent sequences from this set have limits inside this set as well.
I will be able to write $$H^{m,p}(\Omega)= \overline{\left\{{u\in C^m(\Omega); \|u\|_{NORM1}<\infty}\right\}}^{NORM2}$$ You saw that differents norms? – user46060 Nov 7 '12 at 14:08
@user46060: I just saw now that those are different norms. That doesn't matter, though. This is just a definition of a particular set, and you claim that it is closed under $NORM_2$, meaning all $NORM_2$-convergent sequences find their limit within this set. – Asaf Karagila Nov 7 '12 at 14:10
It means that you take the closure of $\left\{{u\in C^m(\Omega); \|u\|_{W^{m,p}}<\infty}\right\}$ with respect to $\| \cdot \|$. The closure is the set itself together with all its limit points. And a limit point is a point such that every $\varepsilon$-ball around it has non-empty intersection with the set.