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##|2 \rangle=[0 \; 1 \; 0 ...]^{\mathsf{T}}##, ##|3 \rangle=[0 \; 0 \; 1 \; 0...]^{\mathsf{T}}##... is called topological basis?

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- Thread starter LagrangeEuler
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- #1

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##|2 \rangle=[0 \; 1 \; 0 ...]^{\mathsf{T}}##, ##|3 \rangle=[0 \; 0 \; 1 \; 0...]^{\mathsf{T}}##... is called topological basis?

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martinbn

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fresh_42

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But they are not open, aren't they? And they aren't suited as a basis for closed sets. Your argument shows, that they cannot be a vector space basis since we needed finite linear combinations. Any open set is a union of ##|n\rangle ##, however, they are not part of the topology except for the discrete.

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martinbn

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What I meant is that if you take the span of those vectors you don't get the whole space you need to take the closer of the span.But they are not open, aren't they? And they aren't suited as a basis for closed sets. Your argument shows, that they cannot be a vector space basis since we needed finite linear combinations. Any open set is a union of ##|n\rangle ##, however, they are not part of the topology except for the discrete.

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fresh_42

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They generate the direct product (not direct sum!) ##\ell_2=\mathbb{R}|1\rangle \times \mathbb{R}|2\rangle \times \mathbb{R}|3\rangle \times \ldots ##. To call it a topological basis is plain wrong in my opinion.

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martinbn

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Not sure what you mean! No one was talking about the direct sum nor the direct product. And, yes, it is just i name the author uses. I think it is quite standard.topological basisin contrast tovector space basis,in order to get rid of the finiteness constriction of vector space bases, but they are neither.

They generate the direct product (not direct sum!) ##\ell_2=\mathbb{R}|1\rangle \times \mathbb{R}|2\rangle \times \mathbb{R}|3\rangle \times \ldots ##. To call it a topological basis is plain wrong in my opinion.

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fresh_42

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Unfortunately. This is what it is about.No one was talking about the direct sum nor the direct product.

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martinbn

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No, it is not about bases of a topology, but about bases of topological vector spaces.The basis of a topology is defined as subset of open sets. To distinguish it from a topological basis is hopefully not standard! And if, I will no longer use english and topology at the same time.

I still don't understand the ##l_2## space is not the direct sum nor the direct product.Unfortunately. This is what it is about.

- #9

fresh_42

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In this case, the wording is more than confusing.No, it is not about bases of a topology, but about bases of topological vector spaces.

As ##\mathbb{R}##-module?I still don't understand the ##l_2## space is not the direct sum nor the direct product.

- #10

martinbn

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It can be, that's why the context is important.In this case, the wording is more than confusing.

The direct sum is the vector space of all finite sequences of real numbers. The direct product of all sequences. The ##l_2## of all square summable sequences.As ##\mathbb{R}##-module?

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fresh_42

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Calling it a topological basis when a basis of a topology is something else, is a very bad choice.

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Math_QED

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I also agree with @martinbn post #2. This is what the author had in mind.

Since ##l^2## is a Hilbert space, a more appropriate name would be "orthonormal basis".

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Office_Shredder

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Infrared

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https://ncatlab.org/nlab/show/basis+in+functional+analysis

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Infrared

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Office_Shredder

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Infrared

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Thanks, I should have read a bit further before posting...

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