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seen Feb 28 at 6:06

Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Mar
24
awarded  Notable Question
Jan
25
awarded  Tumbleweed
Jan
19
comment Construct an algebra from its finitely generated algebras
@JimBelk I thought that would come up, which is why I added the last7 words. I meant take "smartly chosen" direct products and get something isomorphic. This comes from a question, in which I need to prove whether a given algebra is in a variety (of its same type) iff all of its finitely generated subalgebras are in the variety to begin with. I am exploring the idea of accomplishing this only with direct producs, I feel it'll be enough.
Jan
19
asked Construct an algebra from its finitely generated algebras
Jan
18
asked What is the $K$-free algebra for the class of implication algebras, over a finite set
Nov
22
awarded  Nice Question
Nov
13
accepted Question on Galois Connections
Nov
12
comment Question on Galois Connections
@rschwied Great, thanks. Last question, then. Is Cl$(P)$ (the closed sets in $P$), closed under meets and joins? You used that fact in a few steps.
Nov
12
comment Question on Galois Connections
And feel free to use $F$ and $G$, if you want.
Nov
12
comment Question on Galois Connections
Oh, and another question... the problem has 4 sentences that have to be proved, for each combination of the set and the operation ($\{P,Q\}\times\{\vee,\wedge\}$). Do the other 3 statements follow directly from symmetry/duality from the one I have started working on (see previous comment)?
Nov
12
comment Question on Galois Connections
So... $p\wedge q\leq(p\wedge q)^{*'}\leq p^{*'}$, right? "Applying" $*$ to each side, we get that $(p\wedge q)^*\geq p^{*'*}=p^*$. Correct? Similarly, $(p\wedge q)^*\geq q^*$. Can we conclude from this that $(p\wedge q)^*\geq p^*\vee q^*$? That's half the battle :). On the right track? Any hints on getting the other inequality?
Nov
11
asked Question on Galois Connections
Oct
23
awarded  Yearling
May
15
comment Number of automorphisms of saturated models
@tomasz Makes sense, I'll try thinking about this approach. Two questions, perhaps related: why is saturation needed? How could you get such a long indiscernible sequence? (I'm not too familiar with indiscernibles, but I remember something called the Standard Lema?)
May
13
asked Number of automorphisms of saturated models
May
9
awarded  Caucus
May
6
accepted Classification of models
May
5
comment Classification of models
Oh, and I get why arbitrary types turn out to not be so complicated. But my previous comment stands. Even after showing $T$ is model complete, what does that have to do with saturation?