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i'm sorry if my question is really trivial but this one is really bugging me out..

So let's have a partially ordered set $I$ with the topology in which the open sets are the increasing ones: $i\in U$ and $j\ge i$ implies $j\in U$.

In the article i am reading the author is making this two assumptions:

1) There is a global bound in the lengths of all chains in $I$

2) for each $i$ there are only finitely many $j$ comparable to it

I can see why these two assumptions are not equivalent to one another:

$1 \nRightarrow 2$ because i could have an infinite number of chains of finite length in which $i$ appears

$2 \nRightarrow 1$ because i could have bounds which are not $global$

What i don't get is the following:

If i assume (1) then $I$ is covered by the open sets $U_{\ge i_0}=\{j|j\ge i_0\}$ for $i_0$ minimal

What i don't get is where i need the hypothesis of the $global$ bound..

condition (2) then is needed to say that $\{U_{\ge i_0}|i_0$ is minimal$\}$ is locally finite which i think is trivial

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  • $\begingroup$ You don't need a global bound. If all chains are finite, the poset is well founded, hence every element is above a minimal element. $\endgroup$ – Emil Jeřábek Oct 24 '14 at 10:03
  • $\begingroup$ @EmilJeřábek: Perhaps the article is in a context where AC is not assumed. $\endgroup$ – Ramiro de la Vega Oct 24 '14 at 14:21
  • $\begingroup$ @Ramiro: OK, but still, (2) (specifically the fact that every element has only finitely many elements below it) implies the same thing without AC. $\endgroup$ – Emil Jeřábek Oct 24 '14 at 14:27
  • $\begingroup$ forgive my ignorance, but what do you mean by "AC"? $\endgroup$ – User28341 Oct 24 '14 at 17:19
  • $\begingroup$ The axiom of choice. $\endgroup$ – Emil Jeřábek Oct 24 '14 at 18:32
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I'm not sure I understand your point, I might be missing something basic here. Why should you need globality for this? (It might be that globality is demanded for other reasons in the paper.) I don't even see why you need the requirement 2.

Take the poset $I = \{\ast\} \cup \{\ast, 0\} \cup \{\ast, 1, 10\} \cup \{\ast, 2, 21, 210\} \cup\cdots$, where the elements are seen as strings ("$\ast$" for the empty one), and the order is defined by the prefix relation (so $\ast \leq m$, for all $m \in \mathbb{N}$ and $1 \leq 10$, $2 \leq 21 \leq 210$, and so on). Then you don't have a global bound on the chains (1 does not hold) and the minimal element (the empty string $\ast$) has infinitely many comparable nodes (2 does not hold). Still, the collection of all $U_\ast$'s covers $I$.

EDIT: Okay, I make at least one undue assumption above, namely, that you don't have infinitely descending chains. This might be the case for my petty example, but certainly not in any poset, and the requirement 2 ensures that it holds. I'm still with you though, on that I don't see why you would need 1.

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If we assume (at least a weak form of) the axiom of choice then certainly you don´t need a global bound. There are no infinite descending chains so every element must be above a minimal element as Emil mentioned in his comment.

On the other hand, without the axiom of choice one might have a partial order in which all chains are finite and there are no minimal elements at all.

Finally, if there is a global bound for the size of the chains, then we are back to the situation where every element is above a minimal element, and now you don´t need the axiom of choice at all.

It would help if you give the reference of the article you are reading to see what exactly is going on.

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