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In a topology, a base is defined to be a class of subsets such that every open set is the union of some members of it.

In a convexity structure, a base is defined to be a class of subsets such that every convex set is the union of an up-directed subcollection of it.

For a sigma algebra, an attempt to define a base is dismissed here. But I guess it may still be open for other possibilities to define what a base is.

A filter base generates a (proper) filter by including all sets which contain a set of the filter base.

So I was wondering whether a base can be defined for a general set system? When does a set system admit a base, and when doesn't?

Thanks and regards!

Some sources I have just found, although they haven't provided definitions:

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I feel somewhat uncomfortable with this question, since the notion of a "base" doesn't seem to be well-defined. For example, in algebraic structures, a "base" means something qualitatively different than what is meant in topology. Perhaps you could clarify your question by telling us what properties you would expect from a "base" (maybe you should also add tag for category theory) –  William Jan 4 '13 at 3:38
    
@William: (1) I am not sure what properties I will look for. Maybe closed under part of the operations under which the set systems are close? (2) I think a "base" for a set system may be different from a base for an algebraic structure. But who knows if they will merge into the same concept under category theory? –  Tim Jan 4 '13 at 3:42
    
Same question in the title –  leo Jan 4 '13 at 16:54

1 Answer 1

I second William's comment that the concept of "base" means different things in different areas of math. So let me focus on the concept of "base" thought as "minimal set of elements that generate a structure".

There is a nice and simple categorical approach to this concept. Categorically, to give an algebraic structure, is to give a monad on a category $T \colon \mathcal{C} \rightarrow \mathcal{C}$. An algebra for a monad is an object in the category $\mathit{Alg}(T)$ of the Eilenberg-Moore resolution for the monad. This resolution comes equipped with a pair of adjoint functors: the forgetful functor $U \colon \mathit{Alg}(T) \rightarrow \mathcal{C}$ and its left adjoint "the free functor" $F \colon \mathcal{C} \rightarrow \mathit{Alg}(T)$ such that $T = U \circ F$.

Now, the crucial thing is that we may compose these functors in the other direction obtaining a comonad $D = F \circ U$ on the category $\mathit{Alg}(T)$. As it is the case with every comonad, $D$ has its own coresolution in the category of the Eilenberg-Moore coalgebras $\mathit{coAlg}(D)$. Therefore, we may consider coalgebras over algebras. It turns out that in many cases such induced coalgebras over algebras correpond to the usual notion of base. For example: coalgebras over vector spaces decompose vectors into coefficients according to a base; coalgebras over algebras for the power-set monad give decomposition of elements of an atomic (complete) lattice into their descending atoms.


It may help to work out a simple example. Let us consider $\mathcal{C} = \mathbf{Set}$ together with a finite sequences monad: $$T(X) = X^*$$ The Eilenberg-Moore resolution for $T$ gives the category of monoids $\mathbf{Mon}$. The forgetful functor $U \colon \mathbf{Mon} \rightarrow \mathbf{Set}$ assigns to a monoid its carrier: $$U(\langle M, \bullet, \iota\rangle) = M$$ and the free functor $F \colon \mathbf{Set} \rightarrow \mathbf{Mon}$ gives a monoid action via concatenation: $$F(X) = \langle X^*, \circ, [\;]\rangle$$ The induced comonad $D = F \circ U$ associates with every monoid the free monoid build upon the same carrier: $$D(\langle M, \bullet, \iota\rangle) = \langle M^*, \circ, [\;]\rangle$$ There is the counit of the comonad $\epsilon \colon \langle M^*, \circ, [\;]\rangle \rightarrow \langle M, \bullet, \iota\rangle$ defined by folding with monoid's multiplication: $$\epsilon([a_1, a_2, \dotsc, a_n]) = a_1 \bullet a_2 \bullet \cdots \bullet a_n$$ and the comultiplication of the comonad $\delta \colon \langle M^*, \circ, e\rangle \rightarrow \langle M^{**}, \circ, e\rangle$ which decomposes words on singletons: $$\delta([a_1, a_2, \dotsc, a_n]) = [[a_1], [a_2], \dotsc, [a_n]]$$ By definition, a coalgebra over a monoid $\langle M, \bullet, \iota\rangle$ is a monoid homomorphism $h \colon \langle M, \bullet, \iota\rangle \rightarrow \langle M^*, \circ, [\;]\rangle$ such that:

  • $\epsilon \circ h = \mathit{id}$, that is: $h(r)_1 \bullet h(r)_2 \bullet \cdots \bullet h(r)_n = r$, where $h(r)_k$ is the $k$-th element of sequence $h(r)$
  • $D(h) \circ h = \delta \circ h$, that is: $h(h(r)_k) = [h(r)_k]$

These conditions say that coalgebras over monoids are tantamount to finite decompositions of elements on indecomposable elements.

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Just a note: perhaps it would be better to use $\mathcal{C}$ instead of $\mathbb{C}$ to avoid confusion with the set of complex numbers. –  William Jan 4 '13 at 15:27
    
William, I do not know how to type the font you suggested. Feel free to make any corrections :-) –  Michal R. Przybylek Jan 4 '13 at 16:50
    
Mockup, the latex command for that is \mathcal{C}. It seems that you have managed to edit your post already. –  William Jan 5 '13 at 0:56
    
Thank! I don't have enough knowledge to understand your reply yet. By "let me focus on the concept of base thought as minimal set of elements that generate a structure", did you only talk about basis in algebraic structures? I agree basis in algebraic structures and in set systems mean different, and I am not sure if they can be unified at higher level such as in category theory. But my question is more about base for a set system, such as for a topology, sigma algebra, not about basis for a vector space or another algebraic structure. –  Tim Jan 13 '13 at 17:32
    
Hi Tim! You are right, the above approach works only for (in some sense) algebraic categories. I should also have said that I learnt this approach from Bart Jacobs. If I recall, the relevant paper is: B. Jacobs "Coalgebras and approximation", though I have never read it --- maybe you will find more information there. –  Michal R. Przybylek Jan 13 '13 at 22:36

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