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 Feb6 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. Feb5 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? Aug8 awarded Curious Aug7 accepted Existence of families of sets whose elements are incomparable in terms of $\in$ Aug7 revised Existence of families of sets whose elements are incomparable in terms of $\in$ edited body Aug7 comment Existence of families of sets whose elements are incomparable in terms of $\in$ Right, corrected. Aug7 asked Existence of families of sets whose elements are incomparable in terms of $\in$ Apr11 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 :) Apr11 accepted A characterization of recursive functions via arithmetical formulas Apr7 awarded Commentator Apr7 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? Apr7 asked A characterization of recursive functions via arithmetical formulas Jan26 accepted Can we find a formula defining a recursively enumerable set? Jan23 asked Can we find a formula defining a recursively enumerable set? Oct29 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. Oct25 awarded Yearling Oct25 comment Is every Boolean algebra a separative partial order? Thanks a lot for a quick response! Oct25 accepted Is every Boolean algebra a separative partial order? Oct25 asked Is every Boolean algebra a separative partial order? Oct17 awarded Scholar