Will cardinality of orthonormal basis will always be strictly less than cardinality of Hamel Basis. It is true in case of seperable spaces. (Because Hilbert space is always uncountable but orthonormal basis in seperable spaces is countable.)

  • $\begingroup$ Related: math.stackexchange.com/questions/1055062/…. $\endgroup$ – Dietrich Burde Jun 19 '15 at 18:55
  • $\begingroup$ This was actually _answered in the post Dietrich suggests as "related". Very clever: If $card(R)=c$ then $card(\ell^2(R))=c$ as well, and hence the cardinality of a Hamel basis for $\ell^2(R)$ is no larger than $c$. $\endgroup$ – David C. Ullrich Jun 19 '15 at 19:09
  • $\begingroup$ @DavidC.Ullrich Yes I got your argument. But why statement in post mentioned true: if you have a Hilbert space of Hilbert dimension at least c, then the Hilbert dimension and Hamel dimension do coincide. $\endgroup$ – Sushil Jun 20 '15 at 4:14
  • $\begingroup$ @Sushil Won't fit in a comment - posted a reply $\endgroup$ – David C. Ullrich Jun 20 '15 at 16:31

I didn't say "at least c", btw. Say $card(S)=c$. Then $S$ has exactly $c$ countable subsets: $c$ choices for the first element, $c$ choices for the second, etc, for a total of $c^{\aleph_0}=c$. Given a countable subset, how many ways are there to assign numbers to make an element of $\ell^2(S)$? There are $c$ choices for the first coordinate, etc, again it comes out to $c$. So $card(\ell^2(S)=c$. Now a Hamel basis is a subset of $\ell^2(S)$, so the cardinality of a Hamel basis is no more than $c$. But the cardinality of a Hamel basis is at least as large as the cardinality of an orthonormal basis. So the cardinality of a Hamel basis is $c$.


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