Tychonoff's theorem

Theorem

Tychonoff's theorem

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Statement

Let \(I\) be an arbitrary index set and let \(\{X_i\}_{i\in I}\) be a family of nonempty topological spaces, each of which is compact. Equip the product set \(X=\prod_{i\in I}X_i\) with the product topology (the coarsest topology making every projection \(\pi_i:X\to X_i\) continuous, i.e. the topology with subbase \(\mathcal{S}=\{\pi_i^{-1}(U):i\in I,\ U\text{ open in }X_i\}\)). Then \(X\) is compact. (Equivalently: every open cover of \(X\) has a finite subcover.) The result requires the Axiom of Choice when \(I\) is infinite, and in fact is equivalent to it.

Why it matters

Compactness is the topological substitute for "finiteness" that makes existence arguments (extrema are attained, limits of nets exist, systems of constraints are simultaneously satisfiable) work. Tychonoff's theorem says this good behaviour survives passage to arbitrary — even uncountable — products, which is very far from obvious: a product of infinitely many "small" spaces is a genuinely large, high-dimensional object, and the naive coordinatewise diagonal arguments that prove sequential compactness of \([0,1]^{\mathbb{N}}\) simply fail for uncountable index sets.

The theorem is the single load-bearing fact behind an enormous amount of analysis and logic: the Banach–Alaoglu theorem (weak\(^*\) compactness of the dual unit ball), the Stone–Čech compactification, the existence of Haar measure via compactness of spaces of means, and the compactness theorem for propositional and first-order logic can all be derived by embedding a space of interest into a product of compact spaces and invoking Tychonoff.

Hypotheses

Each factor \(X_i\) is compact. Drop it: \(\mathbb{R}^{\mathbb{N}}\) with the product topology is a countable product of copies of \(\mathbb{R}\) (not compact); it is not compact — the standard basic open cover by boxes \(\prod_n(-n,n)\) restricted to finitely many coordinates and \(\mathbb{R}\) elsewhere has no finite subcover, since any finite subfamily leaves some coordinate unbounded.

\(X\) carries the product topology, not the box topology. Take \(X_n=[0,1]\) for all \(n\in\mathbb{N}\), each individually compact, but give \(X=\prod_n[0,1]\) the box topology (arbitrary products of open sets are open). The cover by boxes \(V_x=\prod_n(x_n-\tfrac1n\cdot\tfrac12,\,x_n+\tfrac1n\cdot\tfrac12)\cap[0,1]^{\mathbb{N}}\), one for every \(x\in X\), has no finite subcover — see Worked Example 2/Problem 2 for the explicit diagonal argument.

Axiom of Choice (equivalently Zorn's Lemma / the Ultrafilter Lemma at minimum). Kelley (1950) proved that "the product of compact spaces is compact" (for arbitrary families, without assuming Hausdorff) is logically equivalent to AC. Without any choice principle one cannot even guarantee \(\prod_{i\in I}X_i\) is nonempty when each \(X_i\) is a nonempty compact space, let alone that it is compact.

The index set \(I\) may be infinite and even uncountable — the proof must not proceed "coordinate by coordinate". For finite \(I\) the theorem is elementary (induction plus the Tube Lemma, Problem 1) and needs no choice. For countably infinite \(I\) a sequential/diagonal argument works if the factors are metrizable, but this technique breaks down completely once \(I\) is uncountable or the factors are not first countable — one must argue via open covers (Alexander's Lemma), not sequences.

Proof

X=\prod_{i\in I}X_i \text{ carries the topology with subbase } \mathcal{S}=\{\pi_i^{-1}(U): i\in I,\ U\text{ open in }X_i\}.

Definition of the product topology: a basic open set is a finite intersection \(\pi_{i_1}^{-1}(U_1)\cap\cdots\cap\pi_{i_n}^{-1}(U_n)\), and \(\mathcal{S}\) generates exactly these. A

\textbf{Alexander Subbase Lemma. } \text{If } \mathcal{S} \text{ is a subbase for a space } Y \text{ and every cover of } Y \text{ by elements of } \mathcal{S} \text{ has a finite subcover, then } Y \text{ is compact.}

This is the key reduction: it lets us test compactness only on subbasic covers, which for a product are indexed by single coordinates. We prove it in Steps 3–5 rather than merely citing it, since it is the technical heart of the theorem. B

\text{Suppose, for contradiction, } Y \text{ is not compact. Fix an open cover } \mathcal{U}_0 \text{ of } Y \text{ with no finite subcover.}

Consider the poset \(\mathcal{P}\) of all open covers of \(Y\) that admit no finite subcover, containing \(\mathcal{U}_0\), ordered by inclusion. Any chain \(\{\mathcal{C}_\alpha\}\) in \(\mathcal{P}\) has upper bound \(\bigcup_\alpha \mathcal{C}_\alpha\): it covers \(Y\), and if it had a finite subcover, that finite subcover — being finite — would already lie in a single \(\mathcal{C}_\alpha\) (chains are totally ordered by inclusion), contradicting \(\mathcal{C}_\alpha\in\mathcal{P}\). By Zorn's Lemma \(\mathcal{P}\) has a maximal element \(\mathcal{C}\): a cover of \(Y\) with no finite subcover, maximal with this property. C

\text{Maximality of } \mathcal{C}\text{: if } V \text{ is open and } V\notin\mathcal{C}, \text{ then } \exists\, C_1,\dots,C_n\in\mathcal{C} \text{ with } V\cup C_1\cup\cdots\cup C_n=Y.

If no such finite set existed, \(\mathcal{C}\cup\{V\}\) would still be a cover of \(Y\) with no finite subcover, properly containing \(\mathcal{C}\) — contradicting maximality of \(\mathcal{C}\) in \(\mathcal{P}\). C

\mathcal{C}\cap\mathcal{S} \text{ covers } Y.

Let \(x\in Y\); since \(\mathcal{C}\) covers \(Y\), pick \(A\in\mathcal{C}\) with \(x\in A\). As \(\mathcal{S}\) is a subbase, \(A\) is a union of finite intersections of subbasic sets, so there are \(S_1,\dots,S_n\in\mathcal{S}\) with \(x\in S_1\cap\cdots\cap S_n\subseteq A\). If every \(S_k\in\mathcal{C}\) we are done (\(x\) is covered by a subbasic member of \(\mathcal{C}\)). Otherwise suppose (toward a contradiction) that for every \(x\) some \(S_k\notin\mathcal{C}\); apply Step 4 to each such \(S_k\): there is a finite \(\mathcal{F}_k\subseteq\mathcal{C}\) with \(S_k\cup\bigcup\mathcal{F}_k=Y\), i.e. \(Y\setminus\bigcup\mathcal{F}_k\subseteq S_k\). Intersecting over \(k=1,\dots,n\): \(Y\setminus\bigcup_k\bigcup\mathcal{F}_k\subseteq S_1\cap\cdots\cap S_n\subseteq A\). Hence \(\{A\}\cup\bigcup_k\mathcal{F}_k\) is a finite subfamily of \(\mathcal{C}\) covering \(Y\) — contradicting that \(\mathcal{C}\in\mathcal{P}\) has no finite subcover. So in fact every \(x\) admits a subbasic \(S_k\in\mathcal{C}\) with \(x\in S_k\), proving \(\mathcal{C}\cap\mathcal{S}\) covers \(Y\). C

\mathcal{C}\cap\mathcal{S} \text{ is a cover of } Y \text{ by subbasic sets with no finite subcover} \Rightarrow\Leftarrow \text{hypothesis. Hence } Y \text{ is compact, proving the Lemma.}

A finite subcover of \(\mathcal{C}\cap\mathcal{S}\) would in particular be a finite subcover drawn from \(\mathcal{C}\), which we assumed does not exist; this contradicts the standing hypothesis of Alexander's Lemma that every subbasic cover has a finite subcover. The contradiction refutes the Step 3 assumption. B

\text{Now apply Step 2 to } Y=X=\prod_{i\in I}X_i \text{ with } \mathcal{S}=\{\pi_i^{-1}(U)\}. \text{ Let } \mathcal{U}\subseteq\mathcal{S} \text{ be any cover of } X. \text{ For each } i\in I \text{ set } \mathcal{U}_i=\{U \text{ open in } X_i : \pi_i^{-1}(U)\in\mathcal{U}\}.

Just sorting the subbasic members of \(\mathcal{U}\) by which coordinate they depend on; this is the step where the arbitrary product genuinely reduces to statements about individual factors. A

\exists\, i_0\in I \text{ such that } \mathcal{U}_{i_0} \text{ covers } X_{i_0}.

Suppose not: for every \(i\in I\), \(\mathcal{U}_i\) fails to cover \(X_i\), so the set \(X_i\setminus\bigcup\mathcal{U}_i\) is nonempty. By the Axiom of Choice applied to this \(I\)-indexed family of nonempty sets, choose \(x_i\in X_i\setminus\bigcup\mathcal{U}_i\) for every \(i\), giving a point \(x=(x_i)_{i\in I}\in X\). If some \(\pi_i^{-1}(U)\in\mathcal{U}\) contained \(x\), then \(x_i\in U\in\mathcal{U}_i\), contradicting the choice of \(x_i\). So \(x\) lies in no member of \(\mathcal{U}\), contradicting that \(\mathcal{U}\) covers \(X\). This is precisely the point at which the Axiom of Choice is invoked for arbitrary \(I\). B

\text{Since } X_{i_0} \text{ is compact and } \mathcal{U}_{i_0} \text{ is an open cover of it, } \exists\, U_1,\dots,U_n\in\mathcal{U}_{i_0} \text{ with } X_{i_0}=U_1\cup\cdots\cup U_n.

Direct use of the hypothesis that each factor \(X_i\) is compact — applied here to \(X_{i_0}\). A

X=\pi_{i_0}^{-1}(U_1)\cup\cdots\cup\pi_{i_0}^{-1}(U_n), \text{ a finite subfamily of } \mathcal{U}.

\(\pi_{i_0}^{-1}\) preserves unions, and \(\pi_{i_0}^{-1}(X_{i_0})=X\); each \(\pi_{i_0}^{-1}(U_k)\) lies in \(\mathcal{U}\) by definition of \(\mathcal{U}_{i_0}\). So every subbasic cover \(\mathcal{U}\) of \(X\) has a finite subcover; by the Alexander Subbase Lemma (Step 2/6), \(X\) is compact. \(\blacksquare\) A

Result

\prod_{i\in I}X_i \text{ compact} \iff \text{each } X_i \text{ compact (and nonempty), for the product topology, } \forall\, i\in I

Reading. No matter how many factors you multiply together — even uncountably many — and no matter how "thin" or wild the index set is, gluing compact spaces together with the product topology never destroys compactness. The product topology is exactly calibrated (via its subbase of single-coordinate conditions) to make this true; the box topology is not.

Scope. Holds for arbitrary topological spaces (no separation axioms needed) and arbitrary index sets, provided the product carries the product topology and is nonempty by choice of a point in each factor. Hausdorffness of the factors is not required for compactness of the product (though it is required, together with compactness, for the product to be compact Hausdorff, the setting used in Stone–Čech theory).

Corollaries & converses

Finite products. The case \(|I\vert \lt \infty\) is a corollary (and can be proved without any choice, by induction using the Tube Lemma — Problem 1).
Cubes. \([0,1]^I\) is compact for every set \(I\); every compact Hausdorff space embeds in some such cube (Tychonoff/Urysohn embedding), which underlies the Stone–Čech compactification.
Banach–Alaoglu. The closed unit ball of the dual \(X^*\) of a normed space, in the weak\(^*\) topology, is compact: it embeds as a closed subset of \(\prod_{x\in X}[-\Vert x\Vert,\Vert x\Vert]\), which is compact by Tychonoff.
Converse, set-theoretic. The statement "an arbitrary product of compact spaces is compact" is equivalent to the Axiom of Choice (Kelley, 1950): assuming Tychonoff's theorem for all families of (possibly non-Hausdorff) compact spaces, one can recover AC. So there is no strictly weaker hypothesis that still yields the fully general theorem.
Converse, "does compactness of the product force compactness of each factor" — yes, and this direction needs no choice: each \(X_i=\pi_i(X)\) is a continuous image of a compact space (provided \(X\neq\varnothing\)), hence compact.

Fails without

Compactness of a factor dropped. \(\mathbb{R}\times[0,1]^{\mathbb{N}}\) (with \(\mathbb{R}\) non-compact) is not compact: projection onto the \(\mathbb{R}\) factor is continuous and surjective, and a continuous image of a compact space would have to be compact, but \(\mathbb{R}\) is not — so the whole product cannot be compact.
Box topology instead of product topology. \([0,1]^{\mathbb{N}}\) with the box topology is not compact: the cover \(\{V_x\}_{x\in[0,1]^{\mathbb{N}}}\), \(V_x=\prod_n\big(x_n-\tfrac{1}{n+2},x_n+\tfrac{1}{n+2}\big)\cap[0,1]^{\mathbb{N}}\), has no finite subcover, because finitely many \(V_x\)'s can only "control" finitely many coordinates tightly and one can always build a point escaping all of them in some coordinate (Problem 2 works this out in full).
No Axiom of Choice at all. In a model of set theory without AC it is consistent that a countable product of nonempty finite sets (regarded as compact discrete spaces) is empty, so the product cannot even be a topological space with points, let alone compact; the theorem genuinely requires choice once \(I\) is infinite.

Common errors

Trying to prove the infinite case by a diagonal/sequential-compactness argument (as one does for \([0,1]^{\mathbb{N}}\)): this only shows sequential compactness, and fails to even make sense for uncountable \(I\); worse, sequential compactness and compactness diverge for non-metrizable products (see Discussion).
Silently using the box topology's "obvious" basis (products of arbitrarily many opens) instead of the product topology's basis (all but finitely many factors equal to the whole space) — this is the single most common error, and immediately invalidates any subsequent finite-subcover argument.
Believing the theorem needs the factors to be Hausdorff; compactness of the product holds regardless, though most applications (embedding into cubes, uniqueness of limits) implicitly also want Hausdorff for other reasons.
Applying Alexander's Subbase Lemma to a cover by basic (finite-intersection) open sets rather than genuinely subbasic ones, then wrongly concluding compactness — the lemma's hypothesis is about the subbase specifically, and checking it on a larger collection (the basis) is a strictly stronger requirement that does not reduce to single coordinates.
Forgetting that nonemptiness of the infinite product itself already uses choice, and treating "the theorem is about compactness" and "the theorem needs AC just to have a point to work with" as unrelated issues.

Discussion

Tychonoff proved the theorem in 1930 for products of copies of \([0,1]\), in the course of constructing what is now called the Tychonoff cube and showing every completely regular space embeds in one — this is what makes the compactification of such a space (the Stone–Čech compactification) possible: take the closure of the image inside a compact cube. Čech extended the embedding technique in 1937, and the fully general statement — arbitrary products of arbitrary compact spaces — together with its proof via the Alexander Subbase Lemma (1939, using Zorn's Lemma) is the form now taught universally.

The proof given above is a genuine "compactness is a covering property" argument: it never looks at sequences or nets, only at open covers, which is exactly why it survives arbitrary — including wildly uncountable — index sets. An equivalent proof phrases the same idea using ultrafilters: an ultrafilter on \(X\) converges coordinatewise (since each \(X_i\) is compact, the pushed-forward ultrafilter \((\pi_i)_*\mathcal{F}\) converges in \(X_i\)), and one checks the resulting limit point works simultaneously in every coordinate; this is the same "sort by coordinate, use compactness one factor at a time, glue" idea repackaged.

John Kelley's 1950 theorem that Tychonoff's theorem (in the generality of arbitrary, not-necessarily-Hausdorff spaces) is logically equivalent to the Axiom of Choice is one of the cleanest topology-to-set-theory equivalences known. Weaker fragments correspond to weaker choice principles: Tychonoff's theorem restricted to compact Hausdorff spaces is equivalent to the Boolean Prime Ideal Theorem / Ultrafilter Lemma, which is strictly weaker than full AC — a fact used, for instance, to run compactness-theorem arguments in logic in models where full AC may fail but the Ultrafilter Lemma holds.

Common misconception. Students often assume "compact" and "sequentially compact" are interchangeable, because they coincide for metric spaces and for \([0,1]^{\mathbb{N}}\). They are not equivalent for general products: \([0,1]^{[0,1]}\) (functions from \([0,1]\) to \([0,1]\), pointwise convergence topology) is compact by Tychonoff, yet it is not sequentially compact — there exist sequences of functions with no pointwise-convergent subsequence (a diagonal/back-and-forth cardinality argument on an uncountable index set defeats the usual diagonal trick). Compactness, proved via open covers, is the robust notion; sequential compactness is not.

Worked examples

\text{Show the Hilbert cube } H=[0,1]^{\mathbb{N}} \text{ (product topology) is compact.}

Each factor \([0,1]\subset\mathbb{R}\) is compact by the Heine–Borel theorem. A

I=\mathbb{N} \text{ is a (countably infinite) index set; the family } \{[0,1]\}_{n\in\mathbb{N}} \text{ satisfies the hypotheses of Tychonoff's theorem verbatim.}

No further work is needed beyond checking the hypotheses — each factor compact, product topology used. A

H=[0,1]^{\mathbb{N}}\text{ is compact.}

Reading. \(H\) is additionally metrizable (via \(d(x,y)=\sum_n 2^{-n}\vert x_n-y_n\vert\)), so on \(H\) compactness, sequential compactness, and completeness-plus-total-boundedness all coincide — this example is exactly the boundary case where the "naive" diagonal argument for sequential compactness also works, because \(\mathbb{N}\) is countable.

\text{Deduce the Compactness Theorem for propositional logic from Tychonoff: a set } \Sigma \text{ of propositional formulas over variables } \{p_i\}_{i\in I} \text{ is satisfiable if every finite subset of } \Sigma \text{ is satisfiable.}

Model the space of truth assignments as \(X=\prod_{i\in I}\{0,1\}\) with each \(\{0,1\}\) discrete, hence compact. B

X=\prod_{i\in I}\{0,1\} \text{ is compact by Tychonoff's theorem (each factor is finite, hence compact).}

Direct application of the theorem to a family of two-point discrete spaces. A

\text{For each } \varphi\in\Sigma, \text{ let } C_\varphi=\{x\in X : x \text{ satisfies } \varphi\}. \text{ Each } C_\varphi \text{ is closed in } X.

\(\varphi\) depends on only finitely many variables \(p_{i_1},\dots,p_{i_k}\); the truth value of \(\varphi\) is a continuous (indeed locally constant) function of \((x_{i_1},\dots,x_{i_k})\), so \(C_\varphi\) is the preimage of a closed (here: clopen) set under a continuous coordinate projection, hence closed. B

\{C_\varphi\}_{\varphi\in\Sigma} \text{ has the finite intersection property: every finite subfamily has nonempty intersection.}

This is exactly the hypothesis "every finite subset of \(\Sigma\) is satisfiable": a satisfying assignment for \(\{\varphi_1,\dots,\varphi_n\}\) is a point in \(C_{\varphi_1}\cap\cdots\cap C_{\varphi_n}\). A

\bigcap_{\varphi\in\Sigma}C_\varphi\neq\varnothing.

In a compact space, a family of closed sets with the finite intersection property has nonempty total intersection (the contrapositive of the open-cover definition of compactness, applied to the complements \(X\setminus C_\varphi\)). This is where compactness of \(X\) — i.e. Tychonoff's theorem — is used. B

\text{Any } x\in\bigcap_{\varphi\in\Sigma}C_\varphi \text{ satisfies every formula in } \Sigma, \text{ so } \Sigma \text{ is satisfiable.}

Reading. Local, finite consistency (every finite piece is satisfiable) forces global consistency, purely because the space of all candidate truth assignments is compact. This is the standard topological proof of the propositional Compactness Theorem, and the same scheme (Stone space of a Boolean algebra, or the space of first-order structures via the Löwenheim–Skolem apparatus) generalizes to first-order logic.

Problems

Prove directly, without invoking the Alexander Subbase Lemma, that if \(X\) and \(Y\) are compact then \(X\times Y\) is compact. (Use the Tube Lemma: if \(N\) is open in \(X\times Y\) and contains the slice \(\{x_0\}\times Y\), then \(N\) contains a tube \(W\times Y\) for some open \(W\ni x_0\).)
Solution
Let \(\mathcal{U}\) be an open cover of \(X\times Y\). Fix \(x_0\in X\); the slice \(\{x_0\}\times Y\) is homeomorphic to the compact space \(Y\), so finitely many members \(U_1,\dots,U_n\in\mathcal{U}\) cover it. Let \(N=U_1\cup\cdots\cup U_n\), an open set containing \(\{x_0\}\times Y\). By the Tube Lemma there is an open \(W_{x_0}\ni x_0\) with \(W_{x_0}\times Y\subseteq N\subseteq U_1\cup\cdots\cup U_n\). Doing this for every \(x_0\in X\) gives an open cover \(\{W_{x_0}\}_{x_0\in X}\) of \(X\); by compactness of \(X\) finitely many \(W_{x_1},\dots,W_{x_m}\) cover \(X\). Each tube \(W_{x_j}\times Y\) is covered by a finite subfamily \(\mathcal{U}_j\subset\mathcal{U}\) of size \(n_j\); the union \(\mathcal{U}_1\cup\cdots\cup\mathcal{U}_m\) is then a finite subfamily of \(\mathcal{U}\) covering \(\bigcup_j W_{x_j}\times Y=X\times Y\). Hence \(X\times Y\) is compact. (This is the \(n=2\) base case of the finite-product corollary; induction gives all finite products without choice.)
Show explicitly that \([0,1]^{\mathbb{N}}\) with the box topology is not compact, by exhibiting an open cover with no finite subcover.
Solution
For \(x=(x_n)_{n\in\mathbb{N}}\in[0,1]^{\mathbb{N}}\) let \(V_x=\prod_{n}\big(x_n-\tfrac{1}{n+2},\,x_n+\tfrac{1}{n+2}\big)\cap[0,1]^{\mathbb{N}}\), a box-open neighbourhood of \(x\) (an arbitrary product of opens is open in the box topology). Then \(\mathcal{V}=\{V_x : x\in[0,1]^{\mathbb{N}}\}\) covers \([0,1]^{\mathbb{N}}\) (each \(x\in V_x\)). Suppose \(V_{x^{(1)}},\dots,V_{x^{(m)}}\) is a finite subfamily. Define \(y\in[0,1]^{\mathbb{N}}\) coordinatewise: for each \(n\), choose \(y_n\in[0,1]\) with \(\vert y_n-x^{(k)}_n\vert\geq\tfrac{1}{n+2}\) for every \(k=1,\dots,m\) — possible because \([0,1]\) is uncountable while we only need to avoid \(m\) small intervals of total length \(\leq \tfrac{2m}{n+2}\), which is \(\lt 1\) once \(n\) is large; for small \(n\) simply enlarge using that \([0,1]\) minus finitely many open intervals of controllable length is still nonempty (choose \(n\) large enough first, or note we only need the escape to work for at least one coordinate). Concretely: pick \(n^\*\) with \(\tfrac{2m}{n^\*+2}\lt 1\), and set \(y_{n^\*}\) to avoid all \(m\) forbidden intervals in that coordinate (possible since their total measure is \(\lt 1\)), and \(y_n=0\) for \(n\neq n^\*\). Then for every \(k\), \(y\notin V_{x^{(k)}}\) because the \(n^\*\)-th coordinate condition fails. Hence \(y\) is not covered by \(V_{x^{(1)}},\dots,V_{x^{(m)}}\), so no finite subfamily of \(\mathcal{V}\) covers \([0,1]^{\mathbb{N}}\), and the box-topology cube is not compact.
Use Tychonoff's theorem to prove the Banach–Alaoglu theorem for a normed space \(X\): the closed unit ball \(B^\*=\{f\in X^\* : \Vert f\Vert\leq 1\}\) of the dual, with the weak\(^*\) topology, is compact.
Solution
For each \(x\in X\) let \(I_x=[-\Vert x\Vert,\Vert x\Vert]\subset\mathbb{R}\), compact by Heine–Borel. By Tychonoff, \(P=\prod_{x\in X}I_x\) (product topology) is compact. Define \(\Phi:B^\*\to P\) by \(\Phi(f)=(f(x))_{x\in X}\); this is well-defined since \(\vert f(x)\vert\leq\Vert f\Vert\Vert x\Vert\leq\Vert x\Vert\), injective (a functional is determined by its values on all of \(X\)), and a homeomorphism onto its image when \(B^\*\) carries the weak\(^*\) topology and \(P\) the product topology, since both are exactly "convergence/closeness coordinatewise on points of \(X\)". It remains to show \(\Phi(B^\*)\) is closed in \(P\): the conditions "linearity", \(f(x+y)=f(x)+f(y)\) and \(f(\lambda x)=\lambda f(x)\), pin down, for each fixed \(x,y,\lambda\), a closed condition on coordinates \((\cdot)_x,(\cdot)_y,(\cdot)_{x+y},(\cdot)_{\lambda x}\) of \(P\) (preimage of \(\{0\}\) under a continuous map \(P\to\mathbb{R}\) built from finitely many coordinate projections), so \(\Phi(B^\*)\) is an intersection of closed sets, hence closed in the compact space \(P\), hence compact. Since \(\Phi\) is a homeomorphism onto its image, \(B^\*\) is compact in the weak\(^*\) topology.
Let \(I\) be uncountable and \(X_i=\{0,1\}\) (discrete) for each \(i\in I\). Show \(X=\prod_{i\in I}\{0,1\}\) is compact but is not first countable at any point (hence not metrizable), illustrating why sequence-based arguments cannot prove Tychonoff's theorem in general.
Solution
Compactness is immediate from Tychonoff, since each two-point discrete space is (finite, hence) compact. For non-first-countability, fix \(x\in X\) and suppose \(\{N_k\}_{k\in\mathbb{N}}\) were a countable neighbourhood base at \(x\). Each basic open neighbourhood of \(x\) constrains only finitely many coordinates, so each \(N_k\) (containing some basic neighbourhood) constrains at most a countable union of finitely many coordinates across all \(k\) — i.e. the set \(J=\bigcup_k F_k\) of coordinates ever constrained by some basic set inside some \(N_k\) is countable, while \(I\) is uncountable. Pick \(i_0\in I\setminus J\) and let \(U=\pi_{i_0}^{-1}(\{1-x_{i_0}\})^c=\pi_{i_0}^{-1}(\{x_{i_0}\})\), a basic open neighbourhood of \(x\) constraining only coordinate \(i_0\). No \(N_k\) can be contained in \(U\), since every \(N_k\) contains points differing from \(x\) at coordinate \(i_0\) (as \(i_0\notin J\), no \(N_k\) restricts that coordinate). So \(\{N_k\}\) is not a neighbourhood base — contradiction. Since \(X\) is not first countable, sequential arguments (which only "see" countably many constraints) cannot certify compactness here; only the open-cover argument of the theorem works.
(Converse direction, no choice needed.) Show, without using Tychonoff's theorem, that if \(X=\prod_{i\in I}X_i\) is compact and nonempty then every factor \(X_i\) is compact.
Solution
Fix \(i_0\in I\). Since \(X\neq\varnothing\), the projection \(\pi_{i_0}:X\to X_{i_0}\) is a well-defined, continuous (by definition of the product topology) and surjective map (surjective because for any \(y\in X_{i_0}\), any point \(x\in X\) can be modified only at coordinate \(i_0\) to have \(x_{i_0}=y\) — no choice is needed here since we only alter a single coordinate of an already-existing point \(x\in X\)). The continuous image of a compact space is compact: given an open cover \(\{U_\alpha\}\) of \(X_{i_0}\), the sets \(\pi_{i_0}^{-1}(U_\alpha)\) form an open cover of \(X\); by compactness of \(X\) finitely many \(\pi_{i_0}^{-1}(U_{\alpha_1}),\dots,\pi_{i_0}^{-1}(U_{\alpha_n})\) cover \(X\); applying the surjection \(\pi_{i_0}\), \(U_{\alpha_1},\dots,U_{\alpha_n}\) cover \(X_{i_0}\). Hence \(X_{i_0}\) is compact. As \(i_0\) was arbitrary, every factor is compact.