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What is the simplest/most general commonly used (type of) topological space in analysis?

For instance, every example I can think of in analysis is first-countable.

I don't think it can be metric spaces, since there are non-metrizable spaces (for example spaces of weak-* functionals) of use in analysis.

I was thinking perhaps separable topological vector spaces which are either T0, T1, or T2. (I'm not sure which separation axiom is most appropriate though.)

Finally, which connectedness/compactness properties would we want?

Would local path-connectedness (or even local connectedness) be too strong in general? What about local compactness or paracompactness? Local finiteness?

EDIT: Considering the points brought up by Calvin Khor and Qiyu Wen below, I changed "separable" to "first-countable".

Related, but different question: Measuring the set-theoretical complexity of sets/spaces encountered in general analysis
In contrast, I am asking about topological properties of spaces in analysis.

EDIT: See also: https://math.stackexchange.com/a/1843091/327486.

These questions are also (indirectly) related: Is Every (Non-Trivial) Path Connected Space Uncountable?, Why study non-T1 topological spaces?, What is the motivation behind the arbitrary union topological axiom?

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    $\begingroup$ At the minimum you typically want something Hausdorff from a topological space not to be boring. But many of the spaces in analysis which are interesting have way more structure (e.g. normed spaces in functional analysis, or at least the minimum a topological vector space (which are typically taken to be at least T3.5 (complete regular) ) ). $\endgroup$ – Batman Jun 14 '16 at 23:22
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    $\begingroup$ The Hausdorff property is probably necessary. Separability would be nice, although we don't always have it. $\endgroup$ – Qiyu Wen Jun 14 '16 at 23:31
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    $\begingroup$ $L^∞$ is not separable. $\endgroup$ – Calvin Khor Jun 14 '16 at 23:31
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    $\begingroup$ Non-Hausdorff boring? That's where the fun starts. $\endgroup$ – Lee Mosher Jun 15 '16 at 1:12
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    $\begingroup$ For a topological group $T_0$ implies $T_2$(Hausdorff) and $T_{3½}$ (completely regular). Basically because the group structure makes the space homogeneous. In particular this holds for all topological vector spaces. $\endgroup$ – Jyrki Lahtonen Jun 15 '16 at 5:04
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For future reference of anyone who might have found this question accidentally and wants to know the answer: https://math.stackexchange.com/a/1843091/327486.

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