Timeline for If any open set is a countable union of balls, does it imply separability?
Current License: CC BY-SA 3.0
21 events
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| Sep 28, 2014 at 21:51 | comment | added | Todd Trimble | Ah, very good, Joel -- thanks for looking into this. I had begun recording these arguments in the nLab, and now I'm convinced. :-) | |
| Sep 28, 2014 at 18:58 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Fixed issue noticed by Todd. (Thanks, Todd!) Should use $\frac r4$ not just $\frac r2$.
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| Sep 28, 2014 at 18:51 | comment | added | Joel David Hamkins | @ToddTrimble I think you are right! I should argue more carefully. Pick $p_\alpha$ within $\frac r4$ of $x_\alpha$, and add the ball $B_{d(x_\alpha,p_\alpha)}(x_\alpha)$ to $U$. If this is a union of balls, and some ball $B$ centered within $\frac r4$ of some $x_\gamma$ contains some $x_\alpha$ and $x_\beta$, then the radius of $B$ must exceed $\frac r2$, in which case it will contain $p_\gamma$ by an instance of the triangle inequality. So if $U$ is a union of open balls, then each ball contains at most one $x_\alpha$, and so $U$ is not a countable union of open balls. | |
| Sep 28, 2014 at 14:14 | comment | added | Todd Trimble | Joel, I'm probably being thick, but why must a ball $B_s(x)$ that contains two $x_\alpha$, $x_\beta$ also exclude $p_\alpha$? | |
| Sep 19, 2014 at 19:24 | comment | added | Avshalom | So does the slightly different question about Sierpinski's proof now arise: if every nonseparable metric space contains a sequence of subsets with no convergent subsequence, does CH hold? | |
| Sep 19, 2014 at 12:33 | comment | added | Joel David Hamkins | How funny! I totally forgot that! | |
| Sep 19, 2014 at 12:30 | comment | added | Ramiro de la Vega | Joel, it seems you had answered this same question a few years ago math.stackexchange.com/a/94301. Perhaps you should remove the CH hypothesis over there too. | |
| Sep 19, 2014 at 12:18 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 324 characters in body
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| Sep 19, 2014 at 0:34 | comment | added | Joel David Hamkins | For the purpose of forcing a consistent counterexample, one can imagine undertaking a forcing iteration with countable support (or finite support) of length $\omega_2$, so that every countable set of balls is added by some stage. And then...? I'm not sure. | |
| Sep 18, 2014 at 22:46 | comment | added | Joel David Hamkins | Perhaps one can try to build a counterexample under the assumption that $2^\omega=2^{\omega_1}$, which is to say, Luzin's hypothesis. | |
| Sep 18, 2014 at 22:34 | comment | added | Joel David Hamkins | Does Sierpinski's argument go through with the weaker assumption that $2^\omega<2^{\omega_1}$? | |
| Sep 18, 2014 at 22:07 | comment | added | Timothy Chow | I did a quick MathSciNet search to see if this result was known. The closest hit I got was a paper by Wacław Sierpiński, "Sur l'inversion du théorème de Bolzano-Weierstrass généralisé," Fund. Math. 34 (1947), 155–156, which shows that if the continuum hypothesis is true, then every nonseparable metric space contains a sequence of subsets with no convergent subsequence. While this result doesn't seem directly relevant to the question at hand, it is sort of interesting that Sierpiński found it necessary to assume CH. So maybe CH is needed here as well, after all. | |
| Sep 18, 2014 at 21:58 | comment | added | Włodzimierz Holsztyński | @Joel, I'll read and think more to see all this clearly. Thank you. | |
| Sep 18, 2014 at 21:56 | comment | added | Włodzimierz Holsztyński | I meant that in the case one would the extra step (not in the above proof) of considering one open set per each subset of the subspace then one would be exposed to something like @Fedor's example mentioned earlier. | |
| Sep 18, 2014 at 21:55 | comment | added | Joel David Hamkins | @WłodzimierzHolsztyński In my final case, the points are isolated in the whole space, not just in the set of $x_\alpha$'s, so I don't think your comment is right. And for an open set to be a union of balls, the centers of those balls should be in the open set. | |
| Sep 18, 2014 at 21:52 | comment | added | Włodzimierz Holsztyński | I feel that the above Joel's proof is incomplete. The mentioned 1-point sets are open in the respective subspace but not in the whole space. And the centers of the ball should belong to the whole space, not just to the subspace. The given argument is open to an objection in the style of @Fedor's example from his comment under his question (under my comment under that question), | |
| Sep 18, 2014 at 20:51 | comment | added | Joel David Hamkins | You are right! My argument is more algorithmic, like the proof of Zorn, rather than using Zorn itself. | |
| Sep 18, 2014 at 20:49 | comment | added | Fedor Petrov | It is not very much important, but in order to construct uncountable set with pairwise distances at least $r$ you may just take the maximal (by inclusion) set with this property for small enough $r$: if for any $r=1/n$ such a maximal set is countable, the space is clearly separable. | |
| Sep 18, 2014 at 19:56 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 180 characters in body
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| Sep 18, 2014 at 19:49 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 180 characters in body
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| Sep 18, 2014 at 19:41 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |