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Is it true that in every Hilbert space (not separable) there exists an orthonormal basis (complete and countable orthogonal system)? So equivalently asked can we have a Hilbert space in which it is not sufficient a countable set for which the closure of it's span to be the entirely Hilbert space?

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    $\begingroup$ It is true that every Hilbert space has an orthonormal basis (assuming the axiom of choice), but it is not true that every Hilbert space has a countable orthonormal basis. $\endgroup$ – Lorenzo Quarisa Aug 29 '18 at 10:26
  • $\begingroup$ A Hilbert space has a countable orthonormal basis if and only if it is separable. $\endgroup$ – gerw Aug 29 '18 at 10:59
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If $H$ is the space of all functions from $[0,1] \to \mathbb R$ such that $\sum_{0\leq t \leq 1} |f(t)|^{2} <\infty$ with the inner product $\langle f, g \rangle =\sum_{0\leq t \leq 1} f(t)g(t)$ the there is no countable orthonormal basis. A Hilbert space has a countable orthonormal basis iff it is separable. [In my example $\sum_{0\leq t \leq 1} |f(t)|^{2}$ is defined as the supremum of all finite sums. When this supremum is finite all but countable number of values $\{f(t),0\leq t \leq 1\}$ are zero which makes the definition of inner product meaningful. What I have done is to give explicit construction of a non -separable Hilbert space.

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Let $H$ be a Hilbert space (and $H \ne \{0\}$).

Let $u \in H$ with $||u||=1$ and put $S_0 = \{u\}$. Then $S_0$ is an orthonormal systen in $H$.

Let $M$ be the set of all orthonormal systems $S$ with $S_0 \subseteq S$. On $M$ we can define an order relation by inclusion. By Zorn's Lemma, $M$ contains a maximal element, which is an orthonormal basis of $H$.

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  • $\begingroup$ Why the downvote ??????????????????????????? $\endgroup$ – Fred Aug 29 '18 at 10:33
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    $\begingroup$ I think you relied on the title instead of reading the question fully. OP is asking for a countable ONB. $\endgroup$ – Kavi Rama Murthy Aug 29 '18 at 10:34
  • $\begingroup$ @Kavi: Hi, Sheriff. The OP asked: "Is it true that in every Hilbert space (not separable) there exists an orthonormal basis". In my answer I said: yes it is true (and I gave a short proof). So why you are beefing about my answer ???? $\endgroup$ – Fred Aug 29 '18 at 10:42
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    $\begingroup$ To clarify matters I did not downvote you. I am just trying to guess why you were downvoted. I still think you have not interpreted the question correctly. TH eOP does not know that countability of an ONB is tied to separability and he is asking if non-separable spaces also have countable ONB's. I am sorry if my interpretation is wrong. If I find a mistake in an answer I make a comment. I don't downvote. $\endgroup$ – Kavi Rama Murthy Aug 29 '18 at 11:48
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Let $\mathcal{H}$ consist of all functions $x : S\rightarrow\mathbb{C}$ such that $x(s)=0$ except for at most countably many values of $x$, and such that $\|x\|^2=\sum_{s\in S}|x(s)|^2 < \infty$. Then $\mathcal{H}$ is a Hilbert space under $\langle x,y\rangle = \sum_{s\in S}x(s)\overline{y(s)}$. Let $e_s$ be the function that is $0$ except at $s$, where it is $1$. Then $\{ e_s \}_{s\in S}$ is an orthonormal basis of $\mathcal{H}$ with the same cardinality as $S$. $\mathcal{H}$ is separable iff $S$ is countable or finite.

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