7/30/2019 An application of Tychonoff's theorem, to prove the compactness theorem for propositional logic.

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Proof. Suppose that X is compact. Then, as each projection pi is a continuous surjection,we get from Proposition 2 that each Xi is compact.

Now suppose that each Xi is compact. We would like to show that every ultrafilter ofX converges; this is equivalent to compactness by Proposition 1. So let U be an ultrafilteron X. Since each projection pi is surjective, it follows that pi(U) = {pi(b) | b U} is an

ultrafilter on Xi. As each Xi is compact, we may introduce xi Xi such that pi(U) convergesto xi. The we define X by (i) = xi. By Lemma 3, we obtain that U converges to .

Compactness of Propositional Logic Tychonoffs theorem about the compactness ofproduct spaces may be used to prove the compactness theorem for propositional logic. Firstlets recall what propositional logic is. We have a set = {pi | i I} of proposition letters.We have L = {, , } an algebraic signature. And the set of formulas F is the free L-algebra with as generators. A valuation is a map v : 2, and as F is the free L-algebrawith as generators, valuations extend uniquely to L-homomorphisms v from F to 2.

Let X be the set which consists of all the valuations, i.e. maps v : 2. For each

F we let B := {v | v() = 1}. Since B12 = B1 B2 and B = and B = X, wesee that B = {B | F} forms a base for a topology T on X. Further B is a clopen base,since X B = B.

T is compact in the topological sense iff F is compact in the logical sense. In detail, T iscompact iff every collection of basic closed sets that is consistent has nonempty intersection.Since B is a closed base, this is equivalent to saying that for every F, the existence of avaluation v for each finite subset 0 of that satisfies 0 implies the existence of a valuationv that satisfies all of . This is the logical version of compactness for F.

Let the L-algebra 2 be given the discrete topology (all sets are open). Of course, beingfinite, 2 is compact. And so 2 =

v 2 is compact by Tychonoffs theorem. However, we

claim that 2 is homeomorphic to X, from which it follows that X is compact as well. First

note that the underlying sets of 2

and X are the same, and so it suffices to show that theidentity map id: 2 X is continuous, and its inverse is continuous.

To see that id1 is continuous, we give ourselves a basic open set Oi1 Oin of 2 and

show that it is open in X. We may assume that each Oij is either {0} or {1} (2 = {0, 1}),because if Oij = 2, we can remove it without changing the basic open set (or its the onlything to remove and the basic open set is 2 = X and so open in X), and if its , then thebasic open set is and so open in X by default. Now define J {1, . . . , n} such that j Jiff Oij = {1}. Then define an element of F by:

:=

jJ

pij

j{1,...,n}J

pij

It follows that B = Oi1 Oin .Conversely, to see that id is continuous, its enough to show that the inverse image of

each basic open set in X is open in 2. I.e., we want to show that B is open in 2 for each

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7/30/2019 An application of Tychonoff's theorem, to prove the compactness theorem for propositional logic.

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F. We may write in disjunctive normal form

i

j

pij

and then for each pij with a + there is a corresponding open set Oij = p

1

pij({1}) andsimilarly for each pij with a there is Oij = p1pij({0}). Then

B =

i

j

Oij

is open in 2.

5 Separation Axioms

A space is hausdorff if every pair of distinct points x1

and x2

can be separated by disjointopen sets. I.e., there exists disjoint open sets O1 and O2 such that x1 O1 and x2 O2. Aspace is regular if points can be separated from closed sets by open sets. I.e., if C is a closedset and x C, then there exists disjoint open sets O1 and O2 such that x O1 and C O2.Finally, a space is said to be normal if closed sets can be separated from closed sets by opensets. I.e., if C1 and C2 are disjoint closed sets, then there are disjoint open sets O1 and O2such that C1 O1 and C2 O2.

Every hausdorff normal space is regular, because in a hausdorff space singletons are closedsets.

Proposition 5. Every compact hausdorff space Y is normal.

Proof. First we show that Y is regular. Let C be a closed set and y C. For each x Clet (Ux, Vx) be disjoint open sets such that x Ux and y Vx. As C is a closed subset of acompact space, it is compact, so we may introduce x1, . . . , xn such that Ux1 Uxn C.Then the open sets

ni=1 Uxi,

ni=1 Vxi disjointly separate C and y.

Now let C and D be two disjoint closed sets. For each y C, by regularity we mayseparate it from D via disjoint open sets (Uy, Vy). Then using compactness again we get Uy1 Uyn C, and the disjoint open sets

ni=1 Uyi,

ni=1 Vyi give the desired separation.

Theorem 6 (Urysohn). Let X be any topological space. X is normal iff there exists, forevery pair of disjoint closed sets (A, B), a continuous function f: X [0, 1] such thatf(A) = {0} and f(B) = {1}

Proof. First assume that there is such a function for every pair of disjoint closed sets (A, B).Then f1([0, 1/2)) and f1((1/2, 1]) are disjoint open sets that separate A and B.

Now assume that X is normal. Let two disjoint closed sets A and B be given. In order todefine f, we shall find it helpful to first define for every rational number an open set Up ofX.For p < 0, we set Up := , and for p > 1 we set Up := X. Next, we define U1 := B. Since

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