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Obscure Names in Mathematics
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Obscure Names in Mathematics
@GerryMyerson Dagnabbit! I searched the page for "Byzantine" but I forgot to click on "show more comments". I will delete the generals from my answer. 
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Obscure Names in Mathematics
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Obscure Names in Mathematics
The Beer Glass Theorem says that, if three congruent circles in the plane intersect at one point, then the circle determined by the other three intersections is congruent to the original circle. Is there another name for this theorem? 
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Obscure Names in Mathematics
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Obscure Names in Mathematics
The Stable Marriage problem (or theorem) is about the stability of marriage, isn't it? 
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Firstorder axiomatization of free groups
@Avshalom I don't think two cases have to be handled separately. The needed property is that for each finite $n$ there are $n$ distinct elements which all commute with one another; then, using compactness we can get an uncountable set of commuting elements. 
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Obscure Names in Mathematics
After posting this, I tried to StartPage "beer glass theorem" and couldn't find anything, so maybe it's not so well known by that name. Too bad, it would be a nice accompaniment to the Ham Sandwich Theorem. 
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answered  Obscure Names in Mathematics 
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Separating points in the plane II
Your "open set" condition just says that any subset of $A$ containing $m+1$ points has diameter $\ge2$, right? 
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Lemma on Polish Spaces
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Lemma on Polish Spaces
You asked the question on StackExchange 9 hours ago: math.stackexchange.com/questions/982828/… 
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suggested  suggested edit on Lemma on Polish Spaces 
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Firstorder axiomatization of free groups
@Avshalom That's the sort of construction I had in mind, but I'm not a logician so I wasn't sure. Anyway, I guess the general compactness theorem is just as nonconstructive as an ultrafilter? Either way, I guess you don't quite need an infinite Abelian subgroup in the base group, unbounded finite Abelian subgroups would do as well. 
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Firstorder axiomatization of free groups
So any group with an infinite Abelian subgroup has an elementary extension with an uncountable Abelian subgroup. You used an ultrapower, but I guess you can also do this with the compactness theorem or the upward Löwenheimâ€“Skolem theorem? 
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The continuum hypothesis for packing shapes without overlapping
In fact, that's true even for homeomorphs of $Y$sets; only countably many can be packed in the plane. R. L. Moore's triod theorem is a generalization of that fact. 
Oct 15 
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Most 'unintuitive' application of the Axiom of Choice?
In the hatguessing game, $Y_i$ need only depend on the $X_j,j\gt i$, which is even less intuitive (a little bit less, anyway) than letting it depend on $X_j,j\ne i$. As stated in a longago Monthly problem, there is a function $f:s^\omega\to S$ such that, for each sequence $x_0,x_1,\dots\in S^\omega$, one has $x_n=f(x_{n+1},x_{n+2},\dots)$ for all but finitely many $n$. 
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Examples of seemingly elementary problems that are hard to solve?
I don't understand the "hard version" of Frankl's conjecture, would you mind restating it? Thanks. 
Oct 12 
answered  Choosing subsets to cover larger sets 
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Choosing subsets to cover larger sets
E.g., by Turán's theorem, $$f(2,3,n)=\binom n2\lfloor\frac{n^2}4\rfloor=\binom{\lfloor\frac n2\rfloor}2+\binom{\lceil\frac n2\rceil}2$$ 