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I cannot prove by my self the following remark

By the adjoint functor theorem for posets, having either all joins or all meets is sufficient for the other [1]

So far I got this:

Let $A$ be a poset and suppose it has all lubs, for every $X \subseteq A$ let $LX$ be the lowest upper bound. When $A$ is considered as a category, $L$ is a functor from the powerset of $A$ to $A$ (morphisms are mapped to the univeral map). Since every triangle commutes in a poset, (co)limits of (co)cones are just (lubs) glbs thus $A$ is co-complete. Also $L(\{\})$ is initial. The functor $L$ preserves colimits so it has a right adjoint $G$. Thus we have adjunction, for all $X \in \mathcal P A$, $a \in A$,

$$\frac{X \le G a}{LX \le a}$$

taking the dual of this (and using the fact the opposite in powerset is just compliment) gives

$$\frac{a^{op} \le LX^{op}}{(G a)^c \le X^c}.$$

Can I conclude from this that $A^{op}$ has colimits (lubs)? I don't know how.

Otherwise I thought of using this lemma $A$ has limits of shape $J$ iff the constant functor $\Delta : A \to [J,A]$ has a right adjoint, but I don't know how to do this argument either.

[1] http://ncatlab.org/nlab/show/complete+lattice

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1 Answer 1

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I suppose that could work. Let me write it out cleanly. Let $A$ be a poset and let $I(A)$ be the set of all downward-closed subsets of $A$. There is then a monotone embedding $\mathop{\downarrow} (-) : A \to I(A)$ and moreover this monotone embedding preserves all meets that exist in $\mathop{\downarrow} (-)$.

Now suppose $A$ has all meets. Then the adjoint functor theorem says $\mathop{\downarrow} (-)$ has a left adjoint, say $\sup : I(A) \to A$. It is not hard to see that $I(A)$ is closed under unions (hence $I(A)$ has all joins!), and left adjoints preserve meets, so we can show that, for any subset $S \subseteq A$, $$\sup \bigcup_{s \in S} \mathop{\downarrow} (s) = \bigvee_{s \in S} \sup \mathop{\downarrow} (s) = \bigvee_{s \in S} s$$ hence $A$ itself has all joins.

Conversely, if $A$ has all joins, then the dual of this argument shows that $A$ has all meets.

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wait a second you did it backwards (or forwards if you consider me backwards) starting with $\downarrow$ (what I called $G$)? it seems like knowing explicitly what $\downarrow$ is (downward closed sets) the stuff about adjoint functor theorem distracts from the core of the proof, is that right? Thanks for your answer. In fact all this category theory is irrelevant if we somehow knew the idea of sup union downarrow s = join s. –  ryu jin Jan 8 '13 at 17:29
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Yeah, using the adjoint functor theorem to prove this is somehow overkill. –  Zhen Lin Jan 8 '13 at 19:13

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