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In Methods of modern mathematical physics, they construct the Tensor product of two $\mathbb{K}$-Hilbert spaces $H_1\otimes H_2$ as the inner product space given by the completion of $E$, where $E$ is the space of all finite linear combinations of simple tensors. That is $$ E=\left\{\sum_{i=1}^n a_i \phi_{i}\otimes \psi_{i}: n\in\mathbb{N},\, a\in \mathbb{K}^n,\, \phi\in H_1^n,\, \psi\in H_2^n\right\}, $$ In trying to understand the following proposition

Proposition: If $\{\phi_k\}$ and $\{\psi_j\}$ are orthonormal bases for $H_1$ and $H_2$ respectively, then $\{\phi_k\otimes \psi_j\}$ is an orthonormal basis for $H_1\otimes H_2$

Proof: Assume wlog that $H_1$ and $H_2$ in infinite dimensional and separable. since the set $\{\phi_k\otimes \psi_j\}$ is clearly orthonormal we need only to show that E is contained in the closed space spanned by $\{\phi_k\otimes \psi_j\}$. [...].

So if the let $B=\{\phi_k\otimes \psi_k\}$ and $\iota:E\to H_1\otimes H_2$ be the isometric embedding into the completion. I am fully aware that proving that $\iota(E)\subset \mathrm{span}(B)$ results in $H_1\otimes H_2=\overline{\mathrm{span}(B)}$, but why does this prove this $B$ is a (orthonormal) basis?

TL;DR: Assume that a set $B$ is orthornormal in a Hilbert space $H$. Why does $\overline{\mathrm{span}(B)}=H$ prove that $B$ is a basis for a Hilbert space $H$?

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  • $\begingroup$ Snip of proof: imgur.com/a/L03On $\endgroup$ – Martin Dec 10 '16 at 9:53
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    $\begingroup$ When we say "basis" for a Hilbert space $H$, we aren't talking about a Hamel basis. We just mean an orthonormal family whose span is dense in $H$, since this tends to be more useful than having an actual "basis" in the purely linear-algebraic sense. So checking that the closed span of $B$ is $H$ is precisely what you need to conclude that $B$ gives a basis (in the Hilbert sense). $\endgroup$ – Josh Keneda Dec 10 '16 at 9:57

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