Currently I'm reading "A Course in Functional Analysis" by John B. Conway.In the section "Orthonormal Sets of Vectors and Bases" he writes "For an infinite-dimensional vector space , a basis is never a Hamel basis" without any further explanation. Can anyone throw some light on this matter? Thanks in advance.
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This may help. – David Mitra Aug 26 '17 at 07:14
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An infinite dimensional Banach space cannot have a countable Hamel basis by the Baire category theorem. However, other bases (orthonormal, Schauder, etc) can be countable. – Prahlad Vaidyanathan Aug 26 '17 at 12:43
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The explanation depends on whether he's already made it clear earlier in the text that when he uses the word "basis", he's specifically referring to an orthonormal basis rather than a Hamel basis. If so, then the answer below gives the mathematical proof of his statement. If not, then he's just being sloppy with language and his claim is meaningless. – tparker Aug 14 '18 at 15:57
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If $H$ is your Hilbert space and if $(e_j)_{j\in J}$ is an orthonormal basis of $H$, let $j(1),j(2),\ldots$ be a countable subset of $J$. Then$$\sum_{n=1}^\infty\frac1ne_{j(n)}\in H,$$but it cannot be expressed as a finite sum of elements of $(e_j)_{j\in J}$. Therefore, $(e_j)_{j\in J}$ is not a Hamel basis.
José Carlos Santos
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Thank you very much for the answer. Is the reason $\displaystyle\sum_{n=1}^\infty\frac{1}{n}e_{j(n)}$ converges because $\displaystyle\sum_{n=1}^k\frac{1}{n}e_{j(n)},,k\in\mathbb{N}$ is Cauchy? – Mar 05 '21 at 22:53
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1@JunpyoChoi That's indeed a way if justifying the convergence. – José Carlos Santos Mar 05 '21 at 22:56
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If H be finite dimensional space then each orthonormal bases will be Hamel bases, by Kronecker delta its easy to see. But for infinite dimensional spaces there is orthonormal bases that is not Hamel in general.
Martin Sleziak
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