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1d
answered A group topology which commutes with closed subgroups
2d
comment Does a continuous map $f$ from the $n$-ball $B$ into $R^n$ such that $B\subset f(B)$ have a fixed point?
The new answer is even nicer than mine.
Jul
28
revised Online introduction to Lattice Theory?
not to be confused: number theory, universal algebra
Jul
23
comment Löwenheim-Skolem for many-sorted theories
Consider the two-sorted theory of "sets" and "elements". If your theory contains the extensionality axiom, then any model with $\kappa$ many "elements" has at most $2^\kappa$ many "sets". -- This can be iterated finitely many times.
Jul
22
comment Contracting join-incomplete lattice endomorphisms
You use $f$ both for the given homomorphism and for the one you are looking for. Please call one of them $g$.
Jul
15
comment Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
@JoelDavidHamkins Just a stupid joke. The two halves of the diamond look like two-dimensional corner reflectors (en.wikipedia.org/wiki/Corner_reflector), so one reflects downwards, the other upwards.
Jul
14
accepted Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
Jul
14
comment Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
@JoelDavidHamkins So $\curlyvee$ and $\curlywedge$ stand for upward and downward reflection?
Jul
13
answered Does a continuous map $f$ from the $n$-ball $B$ into $R^n$ such that $B\subset f(B)$ have a fixed point?
Jul
13
asked Why is the set-theoretic principle $\diamondsuit$ called $\diamondsuit$?
Jul
8
answered Incomplete lattice homomorphisms between complete lattices
Jun
27
revised Preserving $\omega_1$ is Inaccessible to the reals
typos corrected: Schindler
Jun
25
comment Characterising subsets of the reals as ordered spaces
As Joel has pointed out, it is not literally true - only if you modify the notion of "dense" appropriately, or if you restrict to dense linear orderings (i.e., without successive points, at least without too many). The fact that every dense order isomorphic to a subset of the reals has a countable dense subset is exercise 2.29 in Rosenstein's book. I do not have a reference for the other direction, but Souslin must already have known this in 1920 when he posed his famous problem in Fund.Math.: matwbn.icm.edu.pl/ksiazki/fm/fm1/fm1125.pdf
Jun
20
comment $\mathbb{P}_{\kappa}$ forces non$(\mathcal{M})$=cov$(\mathcal{M})=\kappa$
Have a look at 2.4 in the book by Bartoszyński and Judah.
Jun
20
comment $\mathbb{P}_{\kappa}$ forces non$(\mathcal{M})$=cov$(\mathcal{M})=\kappa$
A bit of background would be useful here. What do you know, and why do you want to prove that?
Jun
18
revised Finitely generated group with $\aleph_0<X_G<2^{\aleph_0}$ normal subgroups?
spacing; very minor change
Jun
18
comment Countable group with uncountable number of subgroups $< 2^{\aleph_0}$
I think that questions of the form "Does $|X|\ge 2^{\aleph_0}$, i.e., is there a 1-1 map from the Cantor space into $X$?" should always be expanded to "Does $X$ contain a perfect set in some natural T2 topology", or equivalently "Is there a CONTINUOUS 1-1 map from the Cantor space into $X$?". It turns out that a ZFC-answer of "yes" to the first question almost always (in particular: here) is proved by showing that even the second question has a positive answer. The perfect set answer is more interesting because it shows a structural result about $X$, more than a mere cardinality estimate.
Jun
15
comment A compact T1 topological space has a proper dense subset to which it is homeomorphic. What can be said about the space?
An example: Take any compact space with a non-isolated point $x_0$, and replace $x_0$ by infinitely many copies. - This example suggests the following (still fuzzy) question: If $X$ is a space as required, can you always find a continuous image $Y$ which is compact Hausdorff, where the map $f:X\to Y$ has small (in some sense) fibers?
Jun
15
awarded  Citizen Patrol
Jun
11
comment Mathias forcing with Ramsey ultrafilters, and Cohen reals
A possibly simpler way of getting the Cohen real: Let $g\in \omega^\omega$ be the generic real, a strictly increasing sequence. Let $c(i)=0$ if $g(i)$ and $g(i+1)$ are in the same class of the partition $(A_k:k\in \omega)$, and $c(i)=1$ otherwise. Then $c\in 2^\omega$ is a Cohen real. (The proof uses the fact that the intersections of $ran(g)$ with the $A_k$ are not bounded.)