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I have come across the following example of a non-separable Hilbert space:

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Why do I need the discrete topology on $I$? Or more generally: why do I need a topology? If we talk about $L^p$ spaces in general, we only want a measure space and we don't mention a topology because $f \in L^p$ doesn't have to be continuous. Thanks for your help.

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We do not need any topology here – Norbert Aug 2 '12 at 16:19
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You are welcome! – Norbert Aug 2 '12 at 16:21
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Maybe topology is useful in an other part of the example (if the whole example is displayed, I don't see). A remark: the Hilbert space we define in such a way can be separable (when $I$ is finite or countable), otherwise it isn't. – Davide Giraudo Aug 2 '12 at 16:22
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Maybe earlier examples were about $L^p$-spaces with respect to Radon measures on locally compact spaces? The discrete topology is locally compact and $\mathbb P(I)$ is the Borel $\sigma$-algebra, etc. hence compactly supported (=finitely supported) functions are dense. – t.b. Aug 2 '12 at 16:27
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I don't see a typo here. It is in the mindset of many people to equip a set with the discrete topology by reflex when it has none and to say that explicitly. That's why you find "Let $G$ be a discrete group" all over the place even when there's no topological group whatsoever in sight. – t.b. Aug 2 '12 at 17:28
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