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    $\begingroup$ Axioms of separability: It seems hard to tell the difference between your $ T _ 0 $ and your $ T _ 1 $. So to be precise: a space is $ T _ 0 $ iff, whenever $ x \ne y $, there is a verifiable property that can be proved of $ x $ but not of $ y $ or there is a verifiable property that can be proved of $ y $ but not of $ x $. The space is $ T _ 1 $ iff, whenever $ x \ne y $, there is a verifiable property that can be proved of $ x $ but not of $ y $ and there is a verifiable property that can be proved of $ y $ but not of $ x $. $\endgroup$ Commented Jan 14, 2012 at 4:05
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    $\begingroup$ In Sierpiński space, there are two points, $ \top $ and $ \bot $, and one nontrivial verifiable property, which can be proved of $ \top $ but not of $ \bot $. So it is $ T _ 0 $. But there is no verifiable property that can be proved of $ \bot $ but not of $ \top $, so it is not $ T _ 1 $. We may take this as set of Turing machines modulo halting status, so $ \top $ is the (equivalence class of) machines that halt and $ \bot $ is the (equivalence class of) machines that run forever; the nontrivial verifiable property is that the machine halts. $\endgroup$ Commented Jan 14, 2012 at 4:09
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    $\begingroup$ Another separation axiom: (not ‘separability’, I just noticed): a space is $ T _ 2 $ iff, whenever $ x \ne y $, there are two mutually contradictory verifiable properties, one of which holds of $ x $ and one of which holds of $ y $. Equivalently, in the product space, that $ x \ne y $ is itself verifiable. (In Sierpiński space, it's not: you'll never verify that $ \top \ne \bot $, even after you see $ \top $ halt.) The other separation axioms can't be stated in such elementary terms, I think. $\endgroup$ Commented Jan 14, 2012 at 4:39
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    $\begingroup$ Compactness (as requested): If a verifiable property holds of at least one value, then this property of the property can be verified (by checking that value). If a property holds of all values, then this property of the property might be hard to verify (since a priori one would need to check all values). But for a compact space, this can be verified. Of course, a finite space is compact, since we can simply check each value in turn. $\endgroup$ Commented Jan 14, 2012 at 4:48
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    $\begingroup$ Compactness: As far as affirmative properties are concerned, any disjunction of cases boils down to a finite one. $\endgroup$ Commented Oct 26, 2017 at 21:07