Timeline for Why is a topology made up of 'open' sets?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 21, 2015 at 4:01 | comment | added | Toby Bartels | @JohannesHahn: Thanks, that's good to know. | |
| Apr 3, 2015 at 22:03 | comment | added | Johannes Hahn | @TobyBartels: The sequences of preinteriors (or dually preclosures) can become arbitrarly long. Consider any ordinal number $\alpha$ and define a closure operator by $A \mapsto [0,\sup(A)+1]\cap\alpha$. The sequence of closures of $\{0\}$ becomes stationary only at the $\alpha$-th step. And of course this is as bad as it gets: If $X$ has cardinality $\kappa$, then the chain will become stationary before the $\kappa^+$-th term. | |
| Jan 14, 2012 at 6:13 | comment | added | Toby Bartels | To fix this, one may add the axiom of that a preinterior must be open, or in your terms, every interior point of $ S $ must be an interior point of the set of interior points of $ S $. But this is significantly more complicated than the other axioms, and suggests that topological spaces are just a special case of pretopological spaces, much as Hausdorff spaces are just a special case of topological spaces. | |
| Jan 14, 2012 at 6:08 | comment | added | Toby Bartels | Define the preinterior $ S ^ \circ $ of $ S $ to be the set of all points of which $ S $ is a neighbourhood. Call $ S $ open if its preinterior is all of $ S $ (which is how you defined this). Then let the interior $ I S $ of $ S $ be the union of all of the open subsets of $ S $. Nothing in your axioms proves that $ I S $ is all of $ S ^ \circ $. Instead we have $ S \supseteq S ^ \circ \supseteq S ^ { \circ \circ } \supseteq \cdots I S = I I S $. (In principle one could transfinitely iterate the preinterior, but I don't know how long this sequence can get or if it must converge.) | |
| Jan 14, 2012 at 5:57 | comment | added | Toby Bartels | I do like this. But notice that specifying an ‘interior point’ condition gives more information than specifying the open sets! You have (using also Axiom 0, that an interior point of $ S $ must be a member of $ S $) written down the definition of a pretopological space. It is more standard (or at least that's what I've seen) to say that $ S $ is a neighbourhood of $ x $ instead of that $ x $ is an interior point of $ S $. One reason is that there are two notions of interior, and that is the crux of the matter. | |
| Jan 14, 2012 at 5:49 | history | edited | Toby Bartels | CC BY-SA 3.0 |
Changed ‘closed’ to ‘open’ in #3.
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| Dec 3, 2011 at 16:08 | comment | added | BSteinhurst | I don't think you have to worry about your example. It would simply be the situation where you can't wiggle 0 about for some specific example. In general you have to allow for this to happen in your view otherwise you have no idea which points are interior or not. I do like this way of motivating the axioms. | |
| Dec 3, 2011 at 8:57 | history | answered | Michael | CC BY-SA 3.0 |