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

Example 2.84. Let $I$ be a set, equipped with the discrete topology and the counting measure $\lambda_{\text{ count}}$ defined on the $\sigma$-algebra $\Bbb P(I)$ of all subsets of $I$. Then $\ell^2(I)=L^2\big(I,\Bbb P(I),\lambda_{\text{ count}}\big)$ is a Hilbert space, and it comprises all functions $a:I\to\Bbb R$ (or $\Bbb C$) for which the support $F=\{i\in I : a(i)\ne0\},$ is finite or countable, and for which $\sum_{i\in I}|a_i|^2=\sum_{i\in F}|a_i|^2\lt\infty$.

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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    I have downloaded it from here: http://www.math.ethz.ch/~einsiedl/FA-lecture.pdf But it seems to be a newer version - this one has it as Example 2.88. (It writes "Draft July 2, 2012" in the title.)$I$think that if your question is based on some book or lecture notes, it would be nice (for the benefit of other users) to mention it in the post. (And add a link, if it is available.)2012-08-04

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To summarize the comments in an answer:

You are correct, the topology on $I$ is irrelevant and was probably mentioned by mistake.

(If someone will upvote this answer, the question will stop being bumped.)

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In fact, "discrete topology on $I$ = $\sigma$-algebra of $\mathbb{P}(I)$ of all subsets of $I$".

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The reason why a discrete topology is required for $I$, is to make sure that the $\sigma$-algebra of the Borel sets is the whole power set of $I$, i.e. ${\mathscr P}(I)$.