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I would like to understand correctly what does mean separable space, from Wikipedia I am reading that.

In mathematics a topological space is called separable if it contains a countable, dense subset; that is, there exists a sequence $\{ x_n \}_{n=1}^{\infty}$ of elements of the space such that every nonempty open subset of the space contains at least one element of the sequence.

Also related to Hilbert space:

A Hilbert space is separable if and only if it has a countable orthonormal basis, it follows that any separable, infinite-dimensional Hilbert space is isometric to $ℓ^2.$

Can we apply last condition for all space? I mean space is separable if it contains finite orthonormal basis? because this is much simple definition then previous, even I don't know exact definition of dense subset, that why if it is possible to use orthonormal basis related to separable space? Thanks in advance.

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    $\begingroup$ Well, no: we can't since for "orthonormal basis" we need to have a vector space, which isn't available in the general case of topological spaces... $\endgroup$ – DonAntonio Feb 20 '14 at 5:04
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    $\begingroup$ @datodatuashvili: You are welcome. $\endgroup$ – Mhenni Benghorbal Feb 20 '14 at 5:20
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    $\begingroup$ I think the simplest explanation is the one you already gave in your question, @datodatuashvili . $\endgroup$ – DonAntonio Feb 20 '14 at 5:23
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    $\begingroup$ In addition to DonAntonio's comment, "orthonormal" doesn't even make sense in most vector spaces. You need to have an inner product structure - not even all normed spaces have an inner product. $\endgroup$ – user61527 Feb 20 '14 at 5:34
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    $\begingroup$ Can you explain more what the question is? The definition you gave is good... Maybe think about what separable means for the open sets. $\endgroup$ – Ryan Feb 20 '14 at 8:56

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