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Tychonoff theorem

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topology (point-set topology)

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Statement

What is known as the Tychonoff theorem or as Tychonoff’s theorem (Tychonoff 35) is a basic theorem in the field of topology. Assuming the axiom of choice, then it states that the product topological space (with its Tychonoff topology) of an arbitrary set of compact topological spaces is itself compact:

(iI(X icompact))(iIX icompact). \left( \underset{i \in I}{\forall} \left( X_i \, \text{compact} \right) \right) \;\Rightarrow\; \left( \underset{i \in I}{\prod} X_i \, \text{compact} \right) \,.

For a finite number of factors the proof of this statement is elementary (see below), but this already essentially implies for instance the Heine-Borel theorem. For an arbitrary set of factors the proof is non-trivial, but is conveniently given using nets (see below) ultrafilters (further below).

In its classical form, Tychonoff’s theorem is equivalent to the axiom of choice. (Compare the statement of the axiom of choice: a product of sets is inhabited if each set is inhabited.)

Notice that no choice whatsoever is needed to prove the analogous theorem for locales: even in constructive mathematics, a Cartesian product of locales is compact if each of the locales is compact. From that perspective, the statement equivalent to the axiom of choice is this: a product of locales is spatial if each of the locales is spatial. Either of these may be considered a Tychonoff theorem for locales.

To prove Tychonoff’s theorem for Hausdorff topological spaces only, the full axiom of choice is not needed; the ultrafilter theorem (and possibly excluded middle) are enough.

Proofs

Elementary proof of finitary case

The finitary case of Tychonoff’s theorem, where products with only a finite number of factors are considered, may be proven with elementary means from basic topology. Such a proof is spelled out here at Introduction to Topology -- 1:

Proof

of the finitary Tychonoff theorem

Let X,YX,Y be two twopological spaces, with product topological space X×YX \times Y. Let {U iX×Y} iI\{ U_i \subset X \times Y \}_{i \in I} be an open cover of the product space. We need to show that this has a finite subcover.

By definition of the product space topology, each U iU_i is the union, indexed by some set K iK_i, of Cartesian products of open subsets of XX and YY:

U i=k iK i(V k i×W k i)AAAAV k iτ XAandAW k iτ Y. U_i \;=\; \underset{k_i \in K_i}{\cup} \left( V_{k_i} \times W_{k_i} \right) \phantom{AAAA} V_{k_i} \in \tau_{X} \,\phantom{A}\text{and} \phantom{A}\, W_{k_i} \in \tau_{Y} \,.

Consider then the disjoint union of all these index sets

KiIK i. K \,\coloneqq\, \underset{i \in I}{\sqcup} K_i \,.

This is such that

()AA{V k i×W k iX×Y} k iK (\star) \phantom{AA} \left\{ V_{k_i} \times W_{k_i} \subset X \times Y \right\}_{k_i \in K}

is again an open cover of X×YX \times Y.

But by construction, each element V k i×W k iV_{k_i} \times W_{k_i} of this new cover is contained in at least one U j(k i)U_{j(k_i)} of the original cover. Therefore it is now sufficient to show that there is a finite subcover of ()(\star), consisting of elements indexed by k iK finKk_i \in K_{fin} \subset K for some finite set K finK_{fin}. Because then the corresponding U j(k i)U_{j(k_i)} for k iK fink_i \in K_{fin} form a finite subcover of the original cover.

In order to see that ()(\star) has a finite subcover, first fix a point xXx \in X and write {x}X\{x\} \subset X for the corresponding singleton topological subspace. This is homeomorphic to the abstract point space *\ast. and there is thus a homeomorphism of the form

{x}×YY. \{x\} \times Y \simeq Y \,.

Therefore, since (Y,τ Y)(Y,\tau_Y) is assumed to be compact, the open cover

{((V k 1×W k 1)({x}×Y)){x}×Y} k iK \left\{ \left( (V_{k_1} \times W_{k_1}) \cap (\{x\} \times Y) \right) \;\subset\; \{x\}\times Y \right\}_{k_i \in K}

has a finite subcover, indexed by a finite subset J xKJ_x \subset K.

Here we may assume without restriction of generality that xV k ix \in V_{k_i} for all k iJ xKk_i \in J_x \subset K, because if not then we may simply remove that index and still have a (finite) subcover.

By finiteness of J xJ_x it now follows that the intersection

V xk iJ xV k i V_x \;\coloneqq\; \underset{k_i \in J_x}{\cap} V_{k_i}

is still an open subset, and by the previous remark we may assume without restriction that

xV x. x \in V_x \,.

Now observe that by the nature of the above cover of {x}×Y\{x\} \times Y we have

{x}×Yk iJ xKV k i×W k i \{x\} \times Y \subset \underset{k_i \in J_x \subset K}{\cup} V_{k_i} \times W_{k_i}

and hence

{x}×Y{x}×k iJ xKW k i. \{x\} \times Y \subset \{x\} \times \underset{k_i \in J_x \subset K}{\cup} W_{k_i} \,.

Since by construction V xV k iV_x \subset V_{k_i} for all k iJ xKk_i \in J_x \subset K, it follows that we have found a finite cover not just of {x}×Y\{x\} \times Y but of V x×YV_x \times Y

V x×Yk iJ xK(V k i×W k i). V_x \times Y \subset \underset{k_i \in J_x \subset K}{\cup} \left( V_{k_i} \times W_{k_i} \right) \,.

To conclude, observe that {V xX} xX\{V_x \subset X\}_{x \in X} is clearly an open cover of XX, so that by the assumption that also XX is compact there is a finite set of points SXS \subset X so that {V xX} xSX\{V_x \subset X\}_{x \in S \subset X} is still a cover. In summary then

{V k i×W k iX×Y} xSXk iJ xK \left\{ V_{k_i} \times W_{k_i} \subset X \times Y \right\}_{{x \in S \subset X} \atop { k_i \in J_x \subset K }}

is a finite subcover as required. t

Elementary proof of general case via closed-projection characterization

Another equivalent characterization of compactness of a space XX is that for all spaces YY then the projection π Y:Y×XY\pi_Y \;\colon\; Y \times X \to Y out of the product topological space is a closed map, see at closed-projection characterization of compactness.

This characterization allows an elementary proof of the general Tychonoff theorem, see there.

Proof via net convergence

We give the proof of the Tychonoff theorem using the characterization of compactness via convergence of nets (ths prop). This proof is due to (Chernoff 92).

Proof

(of the Tychonoff theorem via convergence of nets)

By this prop. a topological space is compact precisely if every net in that space has a convergent subnet. This in turn is equivalently the case if every net has a cluster point. We will show that this is the case for the product of compact spaces.

So let

ν:AiIX i \nu \;\colon\; A \to \underset{i \in I}{\prod} X_i

be a net in the product space. We need to show that this has a cluster point.

For every finite subset JIJ \subset I, write

ν| I:AνiIX ipr JiJIX i \nu\vert_I \;\colon\; A \overset{\nu}{\longrightarrow} \underset{i \in I}{\prod} X_i \overset{pr_J}{\longrightarrow} \underset{i \in J \subset I}{\prod} X_i

for the image of the net under projection to the product space of factors with indices in the subset JJ.

Say that a partial cluster point of ν\nu is

  1. a finite subset JIJ \subset I;

  2. a cluster point c JiJIX ic_J \in \underset{i \in J \subset I}{\prod} X_i of ν| J\nu\vert_J.

Write PartialClusterPt(ν)PartialClusterPt(\nu) for the set of all partial cluster points of ν\nu. Equip this with the partial order \leq given by the evident extension of domains.

Observe that, by definition, a partial cluster point c Jc_J for which J=IJ = I is an actual cluster point of ν\nu

We claim now that if the partially ordered set (PartialClusterPt )(PartialClusterPt_{\leq}) has a maximal element then this is an actual cluster point of ν\nu.

To see this, assume it were not, hence that the maximal element were a partial cluster point c Jc_J with I\JI \backslash J \neq \emptyset. Then there were an element kI\Jk \in I\backslash J. Since X kX_k is a compact space by assumption, it would follow that the net ν| {k}\nu\vert_{\{k\}} had a cluster point c kc_k. But then adjoining that to c Jc_J would yield a partial cluster point with larger domain J{k}J \cup \{k\}, in contradiction with the assumption that c Jc_J is already maximal. Hence we have a proof by contradiction that every maximal element in PartialClusterPt(ν) PartialClusterPt(\nu)_{\leq} is an actual cluster point.

Therefore we may now conclude by showing that the partially ordered set PartialClusterPt(ν) PartialClusterPt(\nu)_{\leq} does have a maximal element. To this end we invoke Zorn's lemma. This says that we need to check that PartialClusterPt(ν)PartialClusterPt(\nu) is not empty, and that every subset L PartialClusterPt(ν) L_{\leq} \subset PartialClusterPt(\nu)_{\leq} on which the partial order restricts to a total order has an upper bound partial cluster point.

Now non-emptyness follows immediately because there is always the partial cluster point with empty domain c c_{\emptyset}.

Hence consider a totally ordered subset S={(c k) J k} kKS = \{ (c_k)_{J_k} \}_{k \in K} of partial cluster points.

For

J totkKJ k J_{tot} \coloneqq \underset{k \in K}{\cup} J_k

the union of the domains of the partial cluster points in the subset, then this induces a point

c totiJ totX i. c_{tot} \in \underset{ i \in J_{tot} }{\prod} X_i \,.

If this is a partial cluster point, then it is clearly an upper bound of SS. Hence we may conclude by showing that cc is indeed a partial cluster point.

This means that we need to show that for every Tychonoff-basic open neighbourhood UiJ totU \subset \underset{i \in J_{tot}}{\prod} of c totc_{tot} and every element aAa \in A (in the domain of the net) there exists bab \geq a such that ν bU\nu_b \in U.

But by definition of the Tychonoff topology, UU is a product iJ totU i\underset{i \in J_{tot}}{\prod} U_i of open subsets U iX iU_i \subset X_i such that only over a finite subset of J totJ_{tot} they differ from the total spaces X iX_i. By construction of c totc_{tot}, that finite subset of J totJ_{tot} must be contained in J kJ_k for some kKk \in K, and since c J kc_{J_k} is a partial cluster point, the claim follows.

Proof via ultrafilter convergence

One method of proof uses ultrafilter convergence. This is sometimes called “Bourbaki’s proof”, following Cartan 37.

First version

Proof

Let X α αA\langle X_\alpha \rangle_{\alpha \in A} be a family of compact spaces.

  1. Every ultrafilter U αU_\alpha on the underlying set of X αX_\alpha converges to some point x αx_\alpha. (Consider the collection of all closed sets belonging to U αU_\alpha. Finite subcollections have nonempty intersection (finite intersection property), so by compactness the intersection of the collection is also nonempty (this prop.). Let x αx_\alpha be a point belonging to that intersection. Every closed set disjoint from x αx_\alpha belongs to the complement of U αU_\alpha, hence every open set containing x αx_\alpha belongs to U αU_\alpha, i.e., U αU_\alpha converges to x αx_\alpha.)

  2. Let UU be an ultrafilter on αX α\prod_\alpha X_\alpha. Since the set-of-ultrafilters construction SβSS \mapsto \beta S is functorial, we have a push-forward ultrafilter U α=β(π α)(U)U_\alpha = \beta(\pi_\alpha)(U) on each X αX_\alpha, where π α\pi_\alpha is a projection map. Choose a point x αx_\alpha to which U αU_\alpha converges. Then it is easily checked that UU converges to the point x α\langle x_\alpha \rangle in the product space.

  3. Since every ultrafilter UU on αX α\prod_\alpha X_\alpha converges to a point, the product is compact. (If it were not, then we could find a collection of nonempty closed sets whose intersection is empty, but closed under finite intersections. This collection generates a filter which is contained in some ultrafilter UU, by the ultrafilter theorem. Since every ultrafilter UU converges to some xx, it cannot contain any closed set in the complement of xx, hence cannot contain some closed set in the original collection, contradiction.)

Remark

The axiom of choice is used in step 2 of the proof abov, to combine the x αx_\alpha into a single family x α\langle x_\alpha \rangle; if the X αX_\alpha were Hausdorff, then the x αx_\alpha would be unique, and we would not need the axiom of choice at this step. The ultrafilter theorem is used in step 3, and this is the only other place where a choice principle is needed. In other words, working over a choice-free set theory like ZF or even BZ (bounded Zermelo set theory), the ultrafilter principle (UF) implies Tychonoff’s theorem for Hausdorff spaces.

Second version

Bourbaki gives an alternative proof using ultrafilters, which may be formulated in terms of categorical lifting/extension properties. For any set XX and ultrafilter UU on XX, form a space X UX_U whose underlying set is X{}X \sqcup \{\infty\} (adjoin an extra point \infty) and where a subset of X{}X \sqcup\{\infty\} is open iff it either does not contain \infty or is of form Z{}Z \sqcup\{\infty\} where ZUZ\in U, i.e., is “big” according to the ultrafilter UU. The inclusion of the set XX with its discrete topology into X UX_U is a subspace inclusion.

Theorem (Bourbaki)

A map g:YZg: Y \to Z is proper if and only if it has the right lifting property with respect to the inclusion i:XX Ui: X \hookrightarrow X_U for any such pair (X,Uβ(X))(X, U \in \beta(X)) (in symbols: igi \rightthreetimes g). In other words: iff for each commutative diagram

X Y i g X U Z\array{ X & \to & Y \\ \mathllap{i} \downarrow & & \downarrow \mathrlap{g} \\ X_U & \to & Z }

there is a lift X UYX_U \to Y making the evident triangles commute.

Remark

Actually the topology on XX does not have to be discrete: the same result extends to any space XX and its inclusion into X{}X\sqcup\{\infty\} where we declare a subset in X{}X \sqcup \{\infty\} to be open iff it either does not contain \infty and is open in XX, or is of form Z{}Z\sqcup\{\infty\} where ZUZ\in U is and ZZ is open in XX.

A space KK is compact iff the map K{*}K\longrightarrow \{*\} is proper. Thus to see that a product iIY i\prod_{i \in I} Y_i of compact spaces is compact, it suffices to show that for any (X,U)(X, U) and any f:X iIY if: X \to \prod_{i \in I} Y_i, there is a continuous extension f˜:X U iIY i\tilde{f}: X_U \to \prod_{i \in I} Y_i. But (using the axiom of choice) this is clear from the universality property of the product: choose a continuous extension π if˜:X UY i\widetilde{\pi_i f}: X_U \to Y_i for each π if:XY i\pi_i f: X \to Y_i, and then assemble these into an extension

f˜π if˜ iI:X U iIY i.\tilde{f} \coloneqq \langle \widetilde{\pi_i f} \rangle_{i \in I}: X_U \to \prod_{i \in I} Y_i.

Proof via Taimanov theorem and lifting properties

Remark

One can avoid mentioning ultrafilters altogether, at least for Hausdorff spaces, using the Taimanov theorem.

Call a map ff ultrafilter-like iff ff has the left lifting property wrt each proper, equivalently closed, map of finite topological spaces. The Taimanov theorem implies that a map gg of normal (T4) spaces is proper iff it has the lifting property fgf \rightthreetimes g against each ultrafilter-like ff.

Using notation explained just below, this can expressed as follows: The Taimanov theorem says that a Hausdorff space KK is compact iff K{*}({{a}{ab}} 3 r) lrK\longrightarrow \{*\} \,\in\, (\{\{a\}\longrightarrow \{a{\searrow}b\}\}^r_{\le 3})^{lr}, or in fact

{{ab}{a=b},{ab}{a=b},{b}{ab},{aob}{a=o=b}} lr \{\, \{a\leftrightarrow b\}\longrightarrow \{a=b\},\, \{a{\searrow}b\}\longrightarrow \{a=b\},\, \{b\}\longrightarrow \{a{\searrow}b\},\,\{a{\swarrow}o{\searrow}b\}\longrightarrow \{a=o=b\}\,\,\}^{lr}

For a property C{C} of arrows (morphisms) in a category, define

C l{f:gCfg} C^l \coloneqq \left\{ {f} \colon \underset{g \in C}{\forall}\,{f} \,\rightthreetimes\, {g} \right\}
C r{g:fCfg} C^r \coloneqq \left\{ {g} : \underset{f \in C}{\forall}\,{f} \,\rightthreetimes\, {g} \right\}
C lr(C l) r,... C^{lr} \coloneqq (C^l)^r, ...

here fgf \,\rightthreetimes\, g reads “ ff has the left lifting property wrt gg ”, “ ff is (left) orthogonal to gg ”, i.e. for f:ABf:A\longrightarrow B, g:XYg:X\longrightarrow Y, fgf \,\rightthreetimes\, g iff for each i:AXi:A\longrightarrow X, j:BYj:B\longrightarrow Y such that ig=fjig=fj (“the square commutes”), there is j:BXj':B\longrightarrow X such that fj=ifj'=i and jg=jj'g=j (“there is a diagonal making the diagram commute”).

Finally,

C n:={f:fC, both the domain and range of f are finite of size less or equal than n}. C_{\le n} := \{ {f} : {f} \in {C},\text{ both the domain and range of }f \text{ are finite of size less or equal than }\,n \}.

and {a}{ab}\{a\}\longrightarrow \{a{\searrow}b\} denotes the inclusion of the open point aa into the Sierpinski space {ab} \{a{\searrow}b\}.

These observations give rise to the following question.

Question

Question. Is (({a}{ab}) 4 r) lr((\{a\}\longrightarrow \{a{\searrow}b\})^r_{\le 4})^{lr} the class of proper maps?

Considerations above show that it is contained in the class of proper maps and that it contains proper maps between normal (T4) spaces (this is what gives the standard proof of the Taimanov theorem).

Proof of converses

Now we will prove that Tychonoff’s theorem implies the axiom of choice, while Tychonoff’s theorem for Hausdorff spaces implies the ultrafilter theorem. This is done by judicious choice of examples.

Tychonoff’s theorem implies axiom of choice: let {X α} αA\{X_\alpha\}_{\alpha \in A} be a family of nonempty sets. Let Y αY_\alpha be obtained by adjoining a point pp to X αX_\alpha. Topologize Y αY_\alpha by taking the nontrivial open sets to be X αX_\alpha and {p}\{p\}. Then Y αY_\alpha is compact; assuming Tychonoff, Y= αY αY = \prod_\alpha Y_\alpha is compact. For each α\alpha, put

K α={yY:y αX α}K_\alpha = \{y \in Y: y_\alpha \in X_\alpha\}

Then K αK_\alpha is closed, and any finite intersection of the K αK_\alpha is nonempty (use pp in all but finitely many components). Hence

αX α= αK α\prod_\alpha X_\alpha = \bigcap_\alpha K_\alpha

is nonempty as well, by compactness. Thus the axiom of choice follows.

Tychonoff’s theorem for Hausdorff spaces implies ultrafilter theorem: let FF be a filter on a set XX, i.e., a filter in the Boolean algebra PXP X. An ultrafilter UU containing FF is tantamount to a Boolean algebra map 2 X22^X \to 2 which sends all of FF to 11, or equivalently to a Boolean algebra map ϕ:B=2 X/F2\phi: B = 2^X/F \to 2. Thus it suffices to prove the following result:

Proposition

For every (non-terminal) Boolean algebra BB, there exists a homomorphism B2B \to 2.

Proof

Let |B||B| denote the underlying set of BB. The set of all functions f:|B|2f: |B| \to 2 is a |B||B|-indexed product of copies of 22; considering 22 as a compact Hausdorff space, this set is compact Hausdorff, assuming Tychonoff’s theorem for Hausdorff spaces. For each finite subset S|B|S \subseteq |B|, let K SK_S be the subspace of functions f:|B|2f: |B| \to 2 for which the equations

f(xy)=f(x)f(y)f(¬x)=¬f(x)f(x \wedge y) = f(x) \wedge f(y) \qquad f(\neg x) = \neg f(x)

hold for all x,ySx, y \in S. As the space 2 |B|2^{|B|} is Hausdorff, K SK_S being defined by an equalizer is a closed subspace, and it is easy (and classical) that it is nonempty: the subalgebra B(S)BB(S) \subseteq B generated by SS is finite and in particular atomic and thus admits a homomorphism f:B(S)2f: B(S) \to 2 defined by f(x)=1f(x) = 1 iff axa \leq x for some given atom aa, and then we can take f(b)=1f(b) = 1 for any bb outside B(S)B(S). By compactness of 2 |B|2^{|B|}, the intersection of all the K SK_S is nonempty, and this is precisely the set of Boolean algebra maps ϕ:B2\phi: B \to 2.

  • This proof may be seen as a classical illustration of the compactness theorem for propositional logic. We may think of a Boolean algebra BB as a Lindenbaum algebra for a propositional theory, and a Boolean algebra map ϕ:B2\phi: B \to 2 as a model for that theory. A model exists iff a model exists for every theory generated by a finite subset of propositions, by the compactness theorem. In fact, the proof above suggests the topological provenance of the term “compactness theorem”.

Tychonoff’s theorem for locales

Given the close connection between Tychonoff’s theorems and choice principles, it may come as a surprise that Tychonoff’s theorem for locales – the product of a small collection of compact locales is compact, equivalently that the coproduct of a small collection of compact frames is compact – may be proved without the axiom of choice and even constructively.

More details to appear at Tychonoff theorem for locales.

References

The statement of Tychonoff’s theorem is made in

  • A. N. Tychonoff, p. 272 of Ein Fixpunktsatz, Mathematische Annalen, December 1935, Volume 111, Issue 1, pp 767–776

where he says that the proof is the same as the one he gave for a product of closed intervals in

  • A. N. Tychonoff, Über die topologische Erweiterung von Räumen, Mathematische Annalen, December 1930, Volume 102, Issue 1, pp 544–561

An explicit proof was then given in

  • Eduard Cech, p. 830 of On bicompact spaces, Annals of Mathematics 38 (1937)

The proof using ultrafilters is due to

  • Henri Cartan, Theorie des filtres, Comptes Rendus de l’Acad. Sci. (Paris) 205 (1937), 595-598

and reproduced in

  • Bourbaki, General Topology, I\S10.2, Thm.1(d), p.101

whence often known as “Bourbaki’s proof”.

  • Henri Cartan, Filtres et ultrafiltres, Comptes Rendus de l’Acad. Sci. (Paris) 205 (1937), 777-779.

and its modern version is due to

  • John Kelley, Convergence in topology, Duke Math. J. 17 (1950), 277-283.

The proof using convergence of nets is due to

  • Paul R. Chernoff, A Simple Proof of Tychonoff’s Theorem Via Nets, The American Mathematical Monthly Vol. 99, No. 10 (Dec., 1992), pp. 932-934 (jstor)

See also

Revised on May 10, 2017 11:11:04 by Urs Schreiber (131.220.184.222)