bio  website  tuwien.ac.at/goldstern 

location  
age  
visits  member for  4 years, 2 months 
seen  4 hours ago  
stats  profile views  2,694 
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 
LöwenheimSkolem for manysorted theories
Consider the twosorted 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 joinincomplete 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 settheoretic principle $\diamondsuit$ called $\diamondsuit$?
@JoelDavidHamkins Just a stupid joke. The two halves of the diamond look like twodimensional corner reflectors (en.wikipedia.org/wiki/Corner_reflector), so one reflects downwards, the other upwards. 
Jul 14 
accepted  Why is the settheoretic principle $\diamondsuit$ called $\diamondsuit$? 
Jul 14 
comment 
Why is the settheoretic 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 settheoretic 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 11 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 11 map from the Cantor space into $X$?". It turns out that a ZFCanswer 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 nonisolated 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.) 