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T RAZADOS LITERARIOS Y AUDIOVISUALES

In document Fronteras de la literatura y el cine (página 173-175)

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1. T RAZADOS LITERARIOS Y AUDIOVISUALES

In this section we evaluate the Bartha topology onBST92*models. The frame- work for our exploration is the results found in Lemma 1.20 – we would like to see how the Bartha topology interacts with its history-relative counterparts, whether the spaces are connected, and so on. Throughout this section, we fix (W,≤) as aBST92*model, and leth1andh2be arbitrary histories ofW. The

Bartha topology onW is defined as in Definition 1.16.1 Again we will denote

byτW

B the Bartha topology onW, and byτBh the Bartha topology relative to

the historyhofW.

1 Namely, a subset U ofW is open in the Bartha topology iff for each point inU

and each maximal chainC passing throughx, there are elementsc1 andc2 inC

such that c1 < x < c2 and the causal diamonddc1c2 is contained withinU (see

Our first result is an analogue to Lemma 1.18. We omit the proof, since it is identical to theBST92 case.

Proposition 4.1.Let (W,≤) and (W′,) be BST92* models. If f : W

Wis an order-isomorphism, then f is also a homeomorphism between the

topological spaces(W, τW

B ) and(W

, τW

B ).

Before proving any analogues to the results of Lemma 1.20, we first need a few facts regarding maximal≤-chains.

Lemma 4.2.Suppose that C is a maximal chain of W, and h1 and h2 are

histories of W such that Ch16=∅andC6⊆h1 andCh2. Then

suph2(C∩h1) =inf(C\h1).

Proof. Denotea:=suph2(C∩h1) andb:=inf(C\h1). We show that botha andb are members ofC.

Consider firsta. Let cC. If cCh1, then by definition ca. If

c /h1, then cC\h1, and thus Ch1 ≤c (otherwise, Ch16≤c implies

that there is some dCh1 such thatdc, hencecdand thusch1).

Henceac. It follows thatC∪{a}is a chain, and sinceCis maximal, we may conclude thataC. Consider nowb, and letcCbe arbitrary. IfcC\h1

thenbcby definition. Ifch1 thatcCh1, and thuscC\h1(if not,

dc for somedC\h1, hencedh1 follows from the downward closure of

h1). It follows thatC∪ {b} is a chain, and hencebC.

Suppose towards a contradiction that a 6= b. Observe that aC\h1,

since otherwise any element dC\h1 such that d < a would contradict

a = suph2(C∩h1). Hence a < b, and since W is dense (see BST1), there is some c in W such that a < c < b. We now show that this elementc lies in the chain C. Consider some dC. If dh1, then da < c and if

d /h1, thendC\h1andc < bd. It follows thatC∪ {c}is a chain, hence

cC. However this leads to a contradiction – ifch1, thencCh1thus

ca, and ifc /h1, thencC\h1 thus bc. In either case, we contradict

a < c < b. We may then conclude thata=b, as required. ⊓⊔

From the above result, we can identify hot pairs using maximal chains. This is done as follows.

Lemma 4.3.If C is a maximal chain ofW, andh1 andh2 are two histories

such that Ch1 andC6⊂h1 andCh2, then the pair {suph1(C∩h1), suph2(C∩h1)}

forms a hot pair for h1 andh2.

Proof. If C 6⊆ h1 then there is some cC such that c /h1. Then this

there exists (possibly two) history-relative suprema a:=suph1(C∩h1) and

b:=suph2(C∩h1). We show that{a, b} ∈H(h1, h2).

Observe first thatCh1⊂h1∩h2. So, to show thataandbform a hot pair

forh1 andh2, it suffices to show thata6=b. Suppose towards a contradiction

that a=b, and consider the subchainC\h1. This is a chain in the difference

h2\h1, so by PCP* there exists some choice pair {x, y} ∈ C(h1, h2) such

that xC\h1 (where here yh1 and xh2). By the previous lemma,

b=inf(C\h1), hencexb=a. Since we have supposed thata=b, andais

by definition a member ofh1, we may use the downwards-closure of histories

to conclude thatxh1. However, this means that both elements of the choice

pair{x, y}lie in the historyh1, which contradicts Lemma 2.5.4. Thus a6=b,

and we may conclude that{a, b} ∈H(h1, h2). ⊓⊔

Using the above lemma, we can prove the following result, which says that every open set in τh

B is also an open set inτBW.

Lemma 4.4.Let hbe a history ofW andτh

B its associated Bartha topology.

Thenτh BτBW.

Proof. Let U be some element of τh

B. Consider some element x in U and

some chain C that is a maximal chain of W. If Ch, then C is also a maximal chain of h, and the result follows immediately. So, suppose that C 6⊆h. Then there is some cC such that c /h. By Lemma 4.3, it follows that {suph(C∩h), suph′(C∩h)} forms a hot pair for the historieshandh′, where his some history such that C h. Since sup

h(C∩h) is compatible

with every element ofCh, we can extend (Ch)∪ {suph(C∩h)}to a chain

D that is maximal in h. By assumption U is open in the Bartha topology on h, and since xChD, and D is a maximal chain inh, it follows that there are c1, c2∈D such that c1 < x < c2 anddc1c2 ⊂U. Without loss of generality we can assume that c2 < suph(C∩h) (i.e. c2 ∈ Ch), since

otherwise we can take some c3∈Chsuch thatx < c3< suph(C∩h)c2

and use thatdc1c3 ⊂dc1c2 ⊂U. Since c1 andc2are members of C, it follows that UτW

B as required. ⊓⊔

An immediate corollary of the above is the following.

Corollary 4.5.Every historyhof W is open in the topologyτW B .

This is in stark contrast to theBST92case (cf. Lemma 1.20.2), and this is one of the key reasons that we have adoptedBST92*overBST92in this thesis. We also have the following useful characterisation ofτh

B, which is suspected

to be false in theBST92setting.

Lemma 4.6.The Bartha topologyτh

Bon a history coincides with the subspace

topologyτh

Proof. LetVτh

B. Then by Lemma 4.4VτBW, henceV =VhτSh, and

consequentlyτh

BτSh. For the converse, suppose thatVτSh, i.e.V =Uh

whereUτW

B . LetxV andCa maximal chain inh. Since every maximal

chain inhis also a maximal chain inW, it follows from our assumption that there are c1 and c2 in C such thatxdc1c2 ⊂U. SinceUτ

W

B and xU,

there are c1 and c2 in C such that xdc1c2 ⊂ U. Since c2 ∈ Ch, the downward closure ofH implies thatdc1c2 ⊂h, and thusdc1c2 ⊂Uh=V. It follows thatV is an open set ofτh

B, from which the result follows. ⊓⊔

Corollary 4.7.A subset U of W is open in the Bartha topology on W iff for each historyh of W, the set Uhis open in the history-relative Bartha topologyτh

B.

Proof. The direction from left-to-right follows from the previous lemma. So, suppose thatUhis open inτh

B for allhinH(W). By Lemma 4.4, the sets

Uh are also open in τW

B . Then U =

S

hH(W)(U ∩h) is a union of open

sets, and is thus open inτW

B as well. ⊓⊔

From the above result we can conclude that the analogue of Lemma 1.20.3 holds in theBST92*case.

We will now move on to discussing the connectedness of the Bartha topol- ogy. In the next two results, we will prove the BST92*-analogue to Lemma 1.20.4. We start with a proof that all histories of a BST92* model are con- nected. Before introducing the argument, it should be noted that the following proof is an adaptation of that found in [19, Fact 53].2

Lemma 4.8.The topology(h, τh

B)is connected.

Proof. Suppose towards a contradiction that there are two open subsets U, Vτh

B such that UV =∅ and UV = h. Since both U and V are

empty, there are two elementsx, yhsuch thatxU andyV. Sincehis directed, there is somezhthat upper-bounds bothxandy. By assumption U andV coverh, sozmust lie in one of these sets. Without loss of generality, suppose that zU. Sinceyz, we can extend {y, z} to a chainC that is maximal inh. We can restrict the chainCto the segment D:=C∩ ↓z∩ ↑y, which can be seen as amaximal chain with endpointsy and z.3

Consider now the subchain DU. Since this is a chain that is lower- bounded byy, by BST3there exists an infimum inf(D∩U). We now show 2 Unfortunately, a modification of [19, Fact 53] is quite necessary. This is because

Placek applies the axiom BST3 to a chain without first confirming that the chain in question is lower-bounded. As it turns out, the chain in question is

lower-bounded, but this complicates things sufficiently as to warrant a reshaping of the argument.

3 Here we do not use “maximal” in our normal sense, but instead we mean thatD

is a maximal element of the poset of bounded chains inhwhose endpoints arey

that the element inf(D∩U) is an element of the chain D. Let a be some element inD. We proceed via case distinction:

Case 1: aU. ThenaDU, so by definition we have thatinf(D∩U)a. Case 2: aV andalower-bounds the sub-chainDU. Sinceinf(D∩U) is

thegreatest lower-bound of DU, we have thatainf(D∩U). Case 3: aV andadoes not lower-boundDU. Then there is somebDU

such that b < a. Since inf(D∩U) lower-bounds DU, we have that inf(D∩U)≤b < a, whenceinf(D∩U)< a.

We can conclude from the above cases thatinf(D∩U) is comparable with every element ofD. Moreover, the first case above ensures thatinf(D∩U)< z, and the second case above ensures that yinf(D∩U). We then have that D∪{inf(D∩U)}is a chain whose endpoints areyandz, from which it follows from the maximality ofDthat inf(D∩U)∈D.

Now that we have proved thatinf(D∩U) lies inD, consider again our original chainC, which is maximalh. Sinceinf(D∩U)∈DC, we can now apply the Bartha condition in the relevant places to obtain our contradiction. Since U and V cover h, it must be the case that inf(D∩U) lies in one of these two covering sets.

Ifinf(D∩U)∈U, then we can apply the Bartha condition to C andU to conclude that there are c1, c2 ∈ C such that c1 < inf(DU)< c2 and

dc1c2 ⊂U. Observe that y < c1 (since otherwise,c1 < y < inf(D∩U)< c2 implies that yU). Then c1 is an element of the subchain DU that is

strictly belowinf(D∩U), which is a contradiction.

Similarly, ifinf(D∩U)V, then we can again apply the Bartha condition toCandV to conclude that there aree1, e2∈Csuch thate1< inf(D∩U)<

e2 and de1e2 ⊂V. Observe thate2 lower-bounds the sub-chainDU (since otherwise, there would be some elementaDUsuch thatinf(D∩U)≤a < e2, soaV). Theninf(D∩U)< e2< DU, which is again a contradiction.

In either case, we arrive at a contradiction. Thus we may conclude that no such U andV exist, and thereforehis indeed connected. ⊓⊔

Lemma 4.9.The topology(W, τW

B )is connected.

Proof. Suppose towards a contradiction that there are two non-empty, open subsets U, VW that coverW. Since both U and V are non-empty, there are elements xU andyV. By Lemma 1.12.3, there are histories h1 and

h2ofW such thatxh1 andyh2.

Consider first the historyh1. As witnessed by x, the set h1∩U is non-

empty. Moreover, sinceU is open inW, by Lemma 4.7 the seth1∩U is open

in the history-relative Bartha topology onh1. It then has to be the case that

h1∩V =∅, else we contradict the connectedness ofh1. Since h1 and V are

disjoint, and U, V coverW, we may conclude that h1⊆U.

We can perform a similar argument forh2 to conclude thath2 ⊆V. By

assumptionUandV are disjoint, which means thath1andh2are also disjoint.

We will now finish this section with a look at the Hausdorff property on W. The following result tells us that hot pairs are Hausdorff-violating. Lemma 4.10.Leth1 andh2 be histories ofW. Then every hot pair{x, y} ∈

H(h1, h2)violates the Hausdorff property.

Proof. Without loss of generality suppose that xh1 and yh2. LetU, V

be open sets ofW such thatxU and yV. We show that U and V are not disjoint. Sincexandy form a hot pair, there is some chain Ch1∩h2

such that suph1(C) =xandsuph2(C) =y. Let C1 andC2 be two maximal

extensions4 of C containing xand y respectively. Observe5 that C = {z

C1 |z < x}and similarlyC={zC2 |z < y}.

Since U is open, xU and xC1, by definition there are two points

x1, x2∈C1 such thatx1 < x < x2 anddx1x2 ⊂U. Sincex1< x, it must be the case that x1 ∈ C, hence x1 ∈CU. Similarly, sincex < x2 it must be

the case that x2 ∈ C1\C. We can repeat the same argument for y, C2 and

V to conclude that there is some y1, y2 ∈ C2 such that y1 < y < y2, with

y1∈CV anddy1y2⊂V.

We would like to show the existence of at least some point in the inter- section UV. In fact, any pointzC that lies above both x1 andy1 will

suffice6. Indeed, we havex

1 ≤z < x < x2 and y1 ≤z < y < y2, thus both

zdx1x2 ⊂V anddy1y2⊂V, and thuszUV as required. Since we chose U andV arbitrarily, it follows thatxandyviolate the Hausdorff property. ⊓⊔

The following theorem summarises the results of this section, and serves as a direct comparison to Lemma 1.20.

Theorem 4.11.Let (W,≤) be a model of BST92*, and h a history of W.

Then 1.τh

BτBW

2. All historieshof W are open.

3.U is open inW iffUhis open in (h, τh

B)for each hinH(W).

4. Both W andhare connected.

5. IfW is a multi-history model, thenW is not Hausdorff. 6. The history-relative Bartha topologyτh

B coincides with the subspace topol-

ogy induced fromτW B .

4 These always exist by Zorn’s lemma. 5 Ifz C thenz < x(z6=xsince x /h

2 andCh1∩h2). Ifz /C thenC < z

(otherwisezcand thenzC) hencexz.

6 An elementz such as this always exists: sincex

1andy1 are both inC, they are

comparable, so we can takez=y1in the case thatx1≤y1, orz=x1in the case

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