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 Curious
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Feb
6
comment Difference between basis and subbasis in a topology?
@Prayutsh I think that it would be better to give a bit deeper explanation. moose question arised because David goes from subbase to topology directly, not mentioning basis. If you first take basis $B=\{\{0\},\{0,1\},\{0,2\}\}$ generated by $S$ (as all finite intersections of elements of $S$), then you can see how to generate emptyset, since in case of $B$ you form unions over its subsets, not elements (like in case of intersections and S). So $\emptyset\subseteq B$ and $\bigcup\emptyset=\emptyset$. By the way, empty intersection doesn't work.
Feb
5
comment Difference between basis and subbasis in a topology?
@Pratyush What do you mean by "the empty union of empty intersection"? Could you clarify this please?
Aug
8
awarded  Curious
Aug
7
accepted Existence of families of sets whose elements are incomparable in terms of $\in$
Aug
7
revised Existence of families of sets whose elements are incomparable in terms of $\in$
edited body
Aug
7
comment Existence of families of sets whose elements are incomparable in terms of $\in$
Right, corrected.
Aug
7
asked Existence of families of sets whose elements are incomparable in terms of $\in$
Apr
11
comment A characterization of recursive functions via arithmetical formulas
OK, I misunderstood something but your comment helped me to grasp the idea. Thanks a lot :)
Apr
11
accepted A characterization of recursive functions via arithmetical formulas
Apr
7
awarded  Commentator
Apr
7
comment A characterization of recursive functions via arithmetical formulas
I do not understand one thing. Could you explain a case when you hit $a_1,\ldots,a_n,b$ such that $f(a_1,\ldots,a_n)\neq b$? Why the procedure will terminate?
Apr
7
asked A characterization of recursive functions via arithmetical formulas
Jan
26
accepted Can we find a formula defining a recursively enumerable set?
Jan
23
asked Can we find a formula defining a recursively enumerable set?
Oct
29
comment Topology question, very basic
Complexity of proof partially depends on what you start from. But you can show what you want without reference to points. Recall that A set $A$ is closed iff $\bar{A}=A$. Then use this and monotonicity of closure operation.
Oct
25
awarded  Yearling
Oct
25
comment Is every Boolean algebra a separative partial order?
Thanks a lot for a quick response!
Oct
25
accepted Is every Boolean algebra a separative partial order?
Oct
25
asked Is every Boolean algebra a separative partial order?
Oct
17
awarded  Scholar