Timeline for An ordinal invariant for spaces based on a hierarchy of closed sets from z-sets
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20 events
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| Jan 30, 2011 at 3:28 | comment | added | Not Mike | The way I see it is, not about the 'zero-set' dependence, its about how the new set appear, you will only add a subset at a new level when there is a zero-set which is not the zero-set for anything else considered to that point. so as soon as you run out of objects which pass this test, you end the iteration. thus avoid cycles, because you never double back. | |
| Jan 30, 2011 at 3:00 | comment | added | David Feldman | I don't think its hard to do a little better. With some coarsening, instead of getting the discrete topology, you could get a space homeomorphic to the upper-half plane with the tangent disk topology. And you can even do so that the "x-axis" of the subspace form a dense uncountable (etc.) set of x-axis. So $\nu=3$, and then iterating, $\nu=n<\infty$. BTW, I don't see a tree, because I don't see why a set couldn't be a zero-set of zero-set in at least two distinct ways, giving rise to (ignoring direction) cycles... | |
| Jan 30, 2011 at 2:15 | comment | added | Not Mike | Indeed, I think so, because of how the tangent discs intermingle. So that would do it, that would be an example of a space with $\nu(X)=2$ | |
| Jan 30, 2011 at 1:32 | comment | added | David Feldman | Michael, in $X$, the tangent disk topology on the upper half-plane, the $x$-axis is a zero set and discrete in the subspace topology. So every subset of the $x$-axis is a zero set of a zero set. But the irrationals are not a zero set of $X$ itself. Right? | |
| Jan 30, 2011 at 0:30 | comment | added | Not Mike | The reason I say this is because what you are really asking for, is whether or not its possible to have three closed sets $U \subset V \subset X$ such that $U$ is a zero set of $V$, $V$ is a zero set of $X$ and $U$ is not a zero set of $X$. | |
| Jan 30, 2011 at 0:25 | comment | added | Not Mike | also one last thing, a hierarchy is not the best way to view this thing. You should really consider it a tree, order by inclusion, and have the level be the index. | |
| Jan 30, 2011 at 0:21 | comment | added | Not Mike | Or something, idk... | |
| Jan 30, 2011 at 0:20 | comment | added | Not Mike | spent about an hour trying to patch my proof, but I guess I should have thrown in the towl. Anyway, after playing around with it, I'd conjecture that $\nu(X)=1$ obtains, because the relative topology is always finer than the full spaces. | |
| Jan 30, 2011 at 0:16 | comment | added | Not Mike | Real-made-discrete? not sure I follow what you mean by that. | |
| Jan 30, 2011 at 0:14 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 30, 2011 at 0:10 | comment | added | David Feldman | You write: Then, for every $f \in C(X_1, [0,1])$, we have that $f \in C(X_0, [0,1])$ (as $f\circ i$ is continuous) and so $Z$ is a zero set for $X_0$ implies that $Z$ is a zero set for $X_1$, and every zero set of $X_1$ is contained in a zero set for $X_0$. Real-made-discrete maps ($X_0$) continuously to Real-standard-topology ($X_1$). The rationals are a zero set in $X_0$, but not in $X_1$. Of course it's true that zero set in $X_1$ is a zero set in $X_0$. | |
| Jan 30, 2011 at 0:08 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 23:49 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 21:11 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 18:55 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 18:46 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 16:31 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 10:28 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 10:04 | history | edited | Not Mike | CC BY-SA 2.5 |
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| Jan 29, 2011 at 9:47 | history | answered | Not Mike | CC BY-SA 2.5 |