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• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeJun 10th 2010

Tim van Beek has written about unbounded posets at partial order.

Where is this used?

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeJun 10th 2010

Note that [being unbounded] is independent of being bounded

That seems like particularly unfortunate terminology then!

• CommentRowNumber3.
• CommentAuthorTim_van_Beek
• CommentTimeJun 10th 2010
• (edited Jun 10th 2010)

Sorry, I did not know of a better name, it is used in the definition of causal index sets, which themselves are needed to define Haag duality, see the thread “conformal net”. Edit: To be more precise, I don’t know of any name of this property, it is just one of the axioms of index sets for example in the book Wollenschläger/Baumgärtel: “causel nets of operator algebras”. So any better alternative would do.

• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeJun 11th 2010
• (edited Jun 11th 2010)

Could you say that the partial order $P$ under consideration is ’long’? Then it may have a top and bottom, but there is an infinite sequence of distinct elements in $P$ that form a linear order. Or more cleanly since such a thing is an injective map $\omega \to P$ of partial orders ($\omega = 1 \lt 2 \lt \ldots$), that $P$ has a long linear suborder (an LLS, as it were?). Clearly being infinite is a necessary conditions. For example, $\mathbb{Q}\cap [0,1]$ has an LLS and is bounded, but any finite-height tree with an infinite number of leaves with the obvious order, with a top element adjoined, is bounded but with no LLS.

• CommentRowNumber5.
• CommentAuthorDavidRoberts
• CommentTimeJun 11th 2010

Then there is the question of whether a partial order has ’enough’ long linear suborders. For example, I would say $\mathbb{Q}\cap [0,1]$ does, but the top-closure of the finite-height tree from my previous comment, with $\omega+1$ joined at the top and bottom, has an obvious family of LLSs (given by injections $\omega \to \omega \subset \omega +1$; I would not say it had ’enough’ LLSs, as all the other branches of the tree don’t get a look in.

Any thoughts, TIm? Does this sort of reasoning appear in the literature, or is it superfluous? I’m afraid I don’t have a good characterisation of how many long linear suborders is ’enough’.

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeJun 11th 2010

The statement at poset calls for a single infinite sequence such that every element is bounded above by some element of the sequence.

Actually, the statement does not require the sequence to be increasing, or even injective. So my claim that this condition is unrelated to being bounded is wrong; in fact, every bounded poset is ’unbounded’! (Take every element of the sequence to be the top element.) If you require the sequence to be increasing (injective is insufficient), however, then no bounded poset is ’unbounded’ (although still many posets are neither).

It actually seems strange to me that you expect a single sequence to bound everything. Following David, take a tree with infinitely many leaves (but of finite height, so it’s a causet if that matters), which then has $\omega$ (not $\omega + 1$) stuck onto each leaf. That fails to satisfy the condition, but only because it spreads too far. (Even if it has finitely many leaves, it fails if the sequence must be increasing.) Really, this condition is a kind of weak boundedness condition!

Tim, you gave a reference in the other thread, but I can’t get a copy right now. Can you quote its definition and motivation of this condition?

Incidentally, I note that the examples at causal index set are in fact all bounded posets.

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeJun 11th 2010

@Toby,

I believe the intent of the infinite sequence is that it models time in some way: it doesn’t just stop after finitely many steps, so increasing (as you say, not injective) is probably implied. I obviously didn’t read the section at partial order in any detail! this is why I got confused and started talking about ’enough long linear sequences’. You are right, Toby, about requiring a single sequence to bound everything. I would lean towards requiring the existence of a set of increasing sequences such that each element of the partial order is bounded by some element of some sequence. But perhaps the condition is meant to rule out wild branching as as in your example.

• CommentRowNumber8.
• CommentAuthorTobyBartels
• CommentTimeJun 11th 2010

Yes, I would also think that the idea of indefinite time means that, given any $x$, there exists an infinite increasing sequence $y_1,y_2,\ldots$ such that $x \leq y_i$ for some $i$. Assuming dependent choice, it is equivalent to say that, for any $x$, there exists some $y$ such that $x \lt y$. (You make that $y_0$, then apply the condition repeatedly to get $y_1,y_2,\ldots$. Since the next element in the sequence is not given uniquely, we need dependent choice to choose them all at once.)

Now that I would be inclined to call ’unbounded’! (Even better: ’unbounded above’.)

• CommentRowNumber9.
• CommentAuthorDavidRoberts
• CommentTimeJun 11th 2010

(Even better: ’unbounded above’.)

I would agree - there would be analogous statements for ’below’ and ’above and below’. Although it would be interesting to see what sort of spacetimes would be allowed given properties of the corresponding ’causal index sets’ (still not happy with that name - any ideas on a better one?) Really the idea that this is an index set is a little artificial, because we can just take the set $B$ of bounded open regions in spacetime: no need to index them. How about calling it a causal site? (I think there is a coverage hiding there, but I haven’t checked)

• CommentRowNumber10.
• CommentAuthorTim_van_Beek
• CommentTimeJun 11th 2010
• (edited Jun 11th 2010)

Toby wrote:

Tim, you gave a reference in the other thread, but I can’t get a copy right now. Can you quote its definition and motivation of this condition?

The definition of “index set” is an abstraction of an index set consisting of bounded open subsets of Minkowski spacetime. The existence of an infinite unbounded (Edit: unbounded is the word that is used in my reference, but I see that that is the wrong term in the given context, because the sequence does not need to be bounded in the sense of a poset) sequence is the abstraction of the fact that there is a countable exhaustion of Minkowski spacetime by bounded open sets. But there is no upper bound, because Minkoswki spacetime itself is not bounded :-) So this poset is not bounded, correct? And unless I am mistaken the example with the finite subspaces of an infinite Hilbert space on causal index set is not bounded, too, because the upper bound would be the Hilbert space itself, which is not finite dimensional.

David wrote:

I believe the intent of the infinite sequence is that it models time in some way…

Yes, in some way, it models that every bounded region of Minkowski spacetime is embedded in a larger region: spacetime is not compact.

’causal index sets’ (still not happy with that name - any ideas on a better one?)

You do not like “causal index set”? Why not?

Really the idea that this is an index set is a little artificial, because we can just take the set B of bounded open regions in spacetime: no need to index them.

The bounded open regions are the elements of the index set, they are not indexed themselves :-)

• CommentRowNumber11.
• CommentAuthorTim_van_Beek
• CommentTimeJun 11th 2010
• (edited Jun 11th 2010)

What about: “a poset has a majorizing sequence” instead of “bounded poset”?

BTW 1: The book “Causal Nets of Operator Algebras” is incredibly useful, but very hard to get hold of, because it has been out of print for a long time. If I find something more accessible I’ll cite that.

BTW 2: I’m glad that no one asked me where the axiom with the “unbounded infinite sequence” is actually used in the book, because I don’t know :-) I’ll keep my eyes open, though.

• CommentRowNumber12.
• CommentAuthorTobyBartels
• CommentTimeJun 11th 2010
• (edited Jun 11th 2010)

The existence of an infinite unbounded sequence is the abstraction of the fact that there is a countable exhaustion of Minkowski spacetime by bounded open sets.

Ah, now I think that I understand the idea!

I will write a longer post in a moment, but here is the short version:

• The definition at poset really is the right one for this notion (except maybe for the empty poset).
• The correct term for this notion is ‘$\sigma$-bounded’ (rather than ‘unbounded’), or more precisely ‘$\sigma$-bounded above’.
• CommentRowNumber13.
• CommentAuthorTim_van_Beek
• CommentTimeJun 11th 2010

The correct term for this notion is ‘σ-bounded’…

Doh! should have remembered that. I made another mistake, the authors assume that all index sets they consider are infinite, that is they implicitly assume that infinite elements of the sequence are pairwise different.

• CommentRowNumber14.
• CommentAuthorTobyBartels
• CommentTimeJun 11th 2010

The existence of an infinite unbounded sequence is the abstraction of the fact that there is a countable exhaustion of Minkowski spacetime by bounded open sets.

Ah, now I think that I understand the idea!

And yes, it is (as I remarked in #6) really a weak boundedness condition, not an unboundedness condition at all.

That is, while Minkowski spacetime is not bounded, still it is a countable union of bounded subspaces. By the usual logic of terminology in topology, any space[*] which is a countable union of bounded subspaces is a $\sigma$-bounded space.

[*] Precisely, by ‘space’ here I mean bornological space, since those are the spaces with a notion of bounded subspace.

And so, while the poset $P$ of bounded subspaces of Minkowski spacetime is not a bounded poset, nevertheless there is a countable subset $Y$ of $P$ such that, in the lattice completion of $P$, the join of (the elements of) $Y$ is the top element in the lattice completion. (This lattice completion, of course, is the poset of all subspaces of Minkowski spacetime.) Stated in terms of $P$ itself rather than its lattice completion, every element of $P$ is bounded above by (the top element of the lattice completion, hence by the join of $Y$ in the lattice completion, hence by) some element of $Y$. Just as Tim said.

Actually, I would want to make one very small tweak to the definition; rather than speak of an infinite sequence, I would rather speak (as I did above) of a countable set. Since the sequence is not required to be injective (much less increasing), the only difference that this makes is for the empty poset. (I would call it $\sigma$-bounded; although it has no infinite sequence of elements, the empty subset is the required countable subset.)

And unless I am mistaken the example with the finite subspaces of an infinite Hilbert space on causal index set is not bounded, too

Oops, I missed the bit about finite subspaces.

We could make the other example similar by requiring $X$ to be countable (analogous to using a separable Hilbert space) and then using only finite subsets of $X$. (Using finite subsets while allowing $X$ to be uncountable would fail the definition.)

In any case, a bounded poset is $\sigma$-bounded, as it should be.

• CommentRowNumber15.
• CommentAuthorTobyBartels
• CommentTimeJun 11th 2010

Since Tim has approved ‘$\sigma$-bounded’, I edited poset, causal index set, and causal complement to match. (Any others?)

I also refined the example at causal index set as I suggested in my previous post (requiring $X$ to be countable and using only finite subsets of $X$). That really is analogous to the Hilbert space example, so I like it. The original example (the fulll power set of any set whatsoever) is $\sigma$-bounded for the much more trivial reason that it is bounded.

• CommentRowNumber16.
• CommentAuthorTim_van_Beek
• CommentTimeJun 11th 2010
• (edited Jun 11th 2010)

Thanks!

Any others?

None that I know of, these were the ones where I used the concept.