Fact-checked by Grok 2 weeks ago

Tychonoff's theorem

Tychonoff's theorem states that the Cartesian product of any collection of compact topological spaces, equipped with the product topology, is itself a compact topological space.^[1] Named after the Soviet mathematician Andrey Nikolayevich Tychonoff, the theorem was first proved in its general form in his 1935 paper "Über einen Funktionenraum," building on a special case he established five years earlier for products of closed intervals.^[1]^[2] This result generalizes the Heine-Borel theorem, which characterizes compactness in Euclidean spaces, by extending the property to arbitrary infinite products rather than just finite ones.^[3] Tychonoff's theorem is a cornerstone of general topology, enabling the construction of compact spaces in infinite dimensions and playing a crucial role in areas such as functional analysis and algebraic topology.^[4] For instance, it underpins the existence of the Stone-Čech compactification of completely regular spaces and facilitates proofs of the existence of invariant means on certain function spaces.^[3] The theorem's proofs typically rely on the axiom of choice, often through tools like nets, filters, or Zorn's lemma.^[5] In set-theoretic terms, Tychonoff's theorem is logically equivalent to the axiom of choice within Zermelo-Fraenkel set theory (ZF); while the axiom of choice implies the theorem, the converse was established by John L. Kelley in 1950, showing that the theorem's validity forces the existence of choice functions for arbitrary families of nonempty sets.^[6] This equivalence highlights the theorem's deep connections to foundational mathematics.^[7]

Statement and Context

Formal Statement

Tychonoff's theorem asserts that for any index set I and any family of compact topological spaces \{X_i \mid i \in I\}, the product space \prod_{i \in I} X_i equipped with the product topology is compact.^[1] This result holds in the general setting of arbitrary products, including infinite ones, and relies on the standard definition of compactness in topological spaces. In this context, a topological space is compact if every open cover of the space admits a finite subcover.^[1] The theorem guarantees that the product inherits this property from its factors, ensuring that any collection of open sets covering the product can be reduced to finitely many that still cover it. The product topology, also known as the Tychonoff topology, is the initial topology on the product induced by the canonical projection maps \pi_j: \prod_{i \in I} X_i \to X_j for each j \in I.^[1] This topology is generated by the basic open sets of the form \prod_{i \in I} U_i, where U_i = X_i for all but finitely many i, and U_i is open in X_i for those finitely many indices. A simple instance of the theorem occurs when I has two elements, so that the product of two compact spaces X_1 \times X_2 is compact in the product topology.^[1] This finite case can be verified directly using the tube lemma or finite subcover arguments, serving as a foundational building block for the general result.

Historical Background

Andrey Nikolayevich Tychonoff first established a key case of what would become known as Tychonoff's theorem in his 1930 paper "Über die topologische Erweiterung von Räumen," published in Mathematische Annalen, where he proved that the product of arbitrarily many copies of the closed unit interval [0,1] is compact in the product topology.^[8] This result extended earlier work on compactness in product spaces; for instance, the compactness of finite products of compact spaces had been demonstrated by Pavel Alexandroff and Pavel Urysohn in their 1929 memoir "Mémoire sur les espaces topologiques compacts." Despite these partial results dating back to the late 1910s and 1920s for finite cases, the theorem is named after Tychonoff due to his pioneering generalization to infinite products. In 1935, Tychonoff published a brief note "Über einen Funktionenraum" in the same journal, extending the result to arbitrary products of compact spaces, thereby stating the full theorem as it is known today.^[9] This generalization relied on the axiom of choice, though Tychonoff did not explicitly highlight this dependency at the time. An independent and more detailed proof of the general case appeared shortly after in Eduard Čech's 1937 paper "On bicompact spaces," which provided a comprehensive verification using limits of sequences of closed sets. The historical development also includes recognition of the theorem's logical implications for set theory. In 1950, John L. Kelley demonstrated that Tychonoff's theorem for arbitrary products implies the axiom of choice.^[6] This result underscored the theorem's depth, as prior partial results for countable or finite products do not require the full axiom of choice.

Topological Prerequisites

Compactness in Topological Spaces

In topology, a space X is defined as compact if every open cover of X admits a finite subcover.^[10] This means that for any collection of open sets \{U_\alpha\}_{\alpha \in A} such that X \subseteq \bigcup_{\alpha \in A} U_\alpha, there exists a finite subcollection \{U_{\alpha_1}, \dots, U_{\alpha_n}\} with X \subseteq \bigcup_{i=1}^n U_{\alpha_i}.^[11] Several equivalent characterizations of compactness exist under specific conditions. In metric spaces, compactness is equivalent to sequential compactness, where every sequence in the space has a convergent subsequence.^[12] Limit point compactness, defined as every infinite subset having a limit point (an accumulation point), coincides with compactness in first-countable Hausdorff spaces, such as metric spaces.^[13] In Euclidean space \mathbb{R}^n with the standard topology, the Heine-Borel theorem states that a subset is compact if and only if it is closed and bounded.^[14] Examples illustrate these properties clearly. The closed interval [a, b] in \mathbb{R} is compact by the Heine-Borel theorem, as it is closed and bounded.^[15] In contrast, the open interval (a, b) is not compact, since the open cover \{(a, b - 1/n) \mid n = 2, 3, \dots\} has no finite subcover.^[16] Compactness exhibits useful properties in relation to other topological features. Every compact subset of a Hausdorff space is closed, because for any point outside the subset, disjoint open neighborhoods can be found, leading to an open cover without finite subcover otherwise.^[17] Additionally, the continuous image of a compact space is compact: if f: X \to Y is continuous and X is compact, then any open cover of f(X) pulls back to an open cover of X, which has a finite subcover whose images cover f(X).^[18] In the context of Tychonoff's theorem, compactness of each factor space X_i is a necessary condition for the product space \prod X_i to be compact in the product topology.^[19]

Product Topology and Initial Topology

The product topology on the Cartesian product \prod_{i \in I} X_i of a family of topological spaces \{X_i \mid i \in I\}, where I is an arbitrary index set, is constructed by equipping the underlying set \prod_{i \in I} X_i with a basis consisting of sets of the form \prod_{i \in I} U_i, where each U_i is open in X_i and U_i = X_i for all but finitely many i \in I.^[20] This basis generates the product topology, ensuring that open sets are arbitrary unions of such basis elements.^[20] A subbasis for the product topology is given by the collection of all sets of the form \pi_i^{-1}(U_i), where \pi_i: \prod_{j \in I} X_j \to X_i is the canonical projection map and U_i is open in X_i, for each i \in I.^[21] The finite intersections of these subbasis elements form the basis described above.^[21] The product topology is precisely the initial topology on \prod_{i \in I} X_i with respect to the family of projection maps \{\pi_i \mid i \in I\}; that is, it is the coarsest topology making each projection \pi_i continuous.^[21] By construction, each projection \pi_i is continuous and open.^[21] If each X_i is Hausdorff, then the product space \prod_{i \in I} X_i with the product topology is also Hausdorff.^[22] To see this, for distinct points x, y \in \prod_{i \in I} X_i, there exists some k \in I such that x_k \neq y_k in the Hausdorff space X_k; thus, there are disjoint open sets U_k, V_k \subset X_k containing x_k and y_k, respectively, and the preimages \pi_k^{-1}(U_k) and \pi_k^{-1}(V_k) are disjoint open neighborhoods of x and y in the product.^[22] In contrast to the box topology, which uses as a basis all products \prod_{i \in I} U_i where each U_i is open in X_i (without the finiteness condition), the product topology is strictly coarser when I is infinite.^[23] The box topology renders infinite products often pathological and unsuitable for theorems like Tychonoff's, whereas the product topology preserves continuity of projections and other desirable features for arbitrary products.^[23]

Proofs of the Theorem

Proof for Finite Products

The proof of Tychonoff's theorem for finite products proceeds by induction on the number of factors. For the base case of a single compact space X_1, the space is compact by assumption, so the result holds trivially. Assume the result holds for products of n-1 compact spaces, where n \geq 2. Consider the product X_1 \times \cdots \times X_n, where each X_i is compact. Let Z = X_1 \times \cdots \times X_{n-1}. By the induction hypothesis, Z is compact. It suffices to show that the product Z \times X_n is compact. The product topology on Z \times X_n has as a basis the sets of the form U \times V, where U is open in Z and V is open in X_n. To prove compactness of Z \times X_n, let \mathcal{A} be an open cover of Z \times X_n. Fix z_0 \in Z. The slice \{z_0\} \times X_n is homeomorphic to X_n and hence compact. Thus, there exists a finite subcollection \mathcal{A}_{z_0} = \{A_1, \dots, A_m\} [\subset](/page/Subset) \mathcal{A} such that \{z_0\} \times X_n \subset \bigcup_{i=1}^m A_i. Let N = \bigcup_{i=1}^m A_i, which is open in Z \times X_n and contains \{z_0\} \times X_n. The following tube lemma applies to extract a neighborhood of z_0 whose product with X_n is covered by \mathcal{A}_{z_0}. Tube Lemma. Let Z and X_n be topological spaces with X_n compact, and let N be an open set in Z \times X_n containing the slice \{z_0\} \times X_n. Then there exists an open neighborhood W of z_0 in Z such that W \times X_n \subset N. Proof of Tube Lemma. For each x \in X_n, the point (z_0, x) \in N lies in some basis element U_x \times V_x \subset N of the product topology, where U_x is open in Z with z_0 \in U_x and V_x is open in X_n with x \in V_x. The collection \{V_x : x \in X_n\} is an open cover of the compact space X_n, so it has a finite subcover \{V_{x_1}, \dots, V_{x_k}\}. Let W = \bigcap_{j=1}^k U_{x_j}, which is open in Z and contains z_0. For any (z, x) \in W \times X_n, there exists j such that x \in V_{x_j}, and z \in W \subset U_{x_j}, so (z, x) \in U_{x_j} \times V_{x_j} \subset N. Thus, W \times X_n \subset N. Returning to the proof, for each z \in Z, the tube lemma yields an open neighborhood W_z of z in Z such that W_z \times X_n is covered by finitely many elements of \mathcal{A}, say \mathcal{A}_z. The collection \{W_z : z \in Z\} is an open cover of the compact space Z, so there exists a finite subcollection \{W_{z_1}, \dots, W_{z_k}\} covering Z. Then Z \times X_n = \bigcup_{i=1}^k W_{z_i} \times X_n, and each W_{z_i} \times X_n is covered by the finite subcollection \mathcal{A}_{z_i} \subset \mathcal{A}. Hence, \mathcal{A} has a finite subcover, so Z \times X_n is compact. By induction, the finite product X_1 \times \cdots \times X_n is compact. This proof requires no axiom of choice beyond the standard definition of compactness, as all selections (finite subcovers for slices and for Z) involve only finite choices.

General Proof Using the Axiom of Choice

The general proof of Tychonoff's theorem establishes that the arbitrary product X = \prod_{i \in I} X_i of compact topological spaces X_i is compact in the product topology, relying on the axiom of choice through Zorn's lemma to extend the finite product case. Compactness is equivalent to the condition that every family of closed subsets with the finite intersection property (FIP) has nonempty intersection. Given such a family \mathcal{F} of closed subsets of X, consider the poset \mathcal{P} of all families of closed subsets of X that contain \mathcal{F} and have the FIP, ordered by inclusion. Any chain in \mathcal{P} has an upper bound given by its union, which preserves the FIP since finite intersections remain nonempty. By Zorn's lemma, \mathcal{P} has a maximal element \mathcal{M} \supseteq \mathcal{F}.^[24]^[25] For each i \in I, the projected family \{\pi_i(F) \mid F \in \mathcal{M}\} consists of closed subsets of the compact space X_i with the FIP, so \bigcap_{F \in \mathcal{M}} \pi_i(F) \neq \varnothing. The axiom of choice yields a selection x_i \in \bigcap_{F \in \mathcal{M}} \pi_i(F) for each i, defining the point x = (x_i)_{i \in I} \in X. To verify x \in \bigcap_{F \in \mathcal{F}} F, it suffices to show x \in \bigcap_{F \in \mathcal{M}} F by maximality. Suppose toward contradiction that x \notin G for some G \in \mathcal{M}. Then the complement X \setminus G is open and contains x, so there exists a basic open neighborhood U of x with U \cap G = \varnothing. Such a U takes the form \pi_J^{-1}(V), where J \subset I is finite, V is open in the compact finite subproduct \prod_{j \in J} X_j containing \pi_J(x), and the finite subproduct is compact by the finite product lemma.^[24]^[25]^[26] The closed set C = X \setminus U then intersects every member of \mathcal{M}: if C \cap F = \varnothing for some F \in \mathcal{M}, then F \subseteq U, so F \cap G \subseteq U \cap G = \varnothing, contradicting the FIP of \mathcal{M}. Thus, \mathcal{M} \cup \{C\} has the FIP and properly extends \mathcal{M}, contradicting maximality. Hence, no such G exists, so x \in \bigcap_{F \in \mathcal{M}} F \neq \varnothing, and thus \bigcap_{F \in \mathcal{F}} F \neq \varnothing. Any open cover of X therefore has a finite subcover, as its complements form a closed family with the FIP only if the intersection is empty, which it is not. The finite subproducts \prod_{j \in J} X_j for finite J are compact without the axiom of choice, serving as the base for projections in the construction.^[24]^[25]^[26] An alternative proof employs the well-ordering principle, equivalent to the axiom of choice, by assuming I is well-ordered as an ordinal \alpha and proceeding by transfinite induction to construct a point p \in X such that no finite subfamily of a given open cover \mathcal{U} of X covers the "initial segments" defined by p. Define subsets Z_\beta \subseteq X for \beta \leq \alpha recursively: Z_0 = X, and at successor \beta = \gamma + 1, use the tube lemma on the compact slice Z_\gamma \times X_\gamma (compact by the finite product case) to select p_\gamma \in X_\gamma such that Z_{\beta} = \{q \in X \mid q_{<\beta} = p_{<\beta}\} admits no finite subcover from \mathcal{U}. At limit ordinals \beta, finite subcovers of Z_\beta would restrict to finite subcovers on prior Z_{\gamma_j}, contradicting induction. At \beta = \alpha, Z_\alpha = \{p\} is covered by a single element of \mathcal{U}, yielding a finite subcover of X and a contradiction. This step-by-step selection uses choice for coordinates and well-ordering of I.^[26]^[27]

Applications

In Real Analysis

In real analysis, Tychonoff's theorem plays a crucial role in establishing compactness properties of function spaces and infinite products, enabling key approximation and convergence results. One prominent application is in the Stone–Weierstrass theorem, which asserts that if A is a subalgebra of the space C(K) of real-valued continuous functions on a compact Hausdorff space K, containing the constants and separating points, then the uniform closure of A is all of C(K). The proof embeds K into the compact product space [0,1]^A via evaluation maps, leveraging Tychonoff's theorem to ensure this product's compactness, which in turn implies the density of A. Another fundamental result benefiting from Tychonoff's theorem is the Ascoli–Arzelà theorem, which characterizes the relatively compact subsets of C(K), where K is a compact metric space, as those families that are equicontinuous and pointwise relatively compact in \mathbb{R}. To derive this, one considers the embedding of such a family \mathcal{F} into the product \prod_{x \in K} \overline{\mathcal{F}(x)}, where each \overline{\mathcal{F}(x)} is compact; Tychonoff's theorem guarantees the compactness of this product in the product topology, which corresponds to the topology of pointwise convergence, and equicontinuity ensures relative compactness in the uniform topology. Tychonoff's theorem also directly implies the compactness of the Hilbert cube, defined as H = \prod_{n=1}^\infty [0, 1/n] or equivalently homeomorphic to [0,1]^\mathbb{N} in the product topology. This countable infinite product of compact intervals models infinite-dimensional phenomena in analysis, such as the embedding of separable metric spaces, and its compactness facilitates the study of continuous functions on infinite-dimensional domains without relying on finite-dimensional approximations. In the theory of almost periodic functions, Tychonoff's theorem underpins the construction of the Bohr compactification of the additive group \mathbb{R}, which is the closure of the embedding of \mathbb{R} into the product \prod_{\chi} \mathbb{T} over all continuous characters \chi: \mathbb{R} \to \mathbb{T} (the unit circle group). The compactness of this uncountable product of compact groups ensures that the Bohr compactification b\mathbb{R} is a compact abelian group, and the almost periodic functions on \mathbb{R} are precisely the restrictions of continuous functions on b\mathbb{R} to \mathbb{R}, providing a uniform approximation framework for such functions. Finally, a specific instance in sequence spaces arises in Alaoglu's theorem, which proves the weak* compactness of the closed unit ball in the dual of a normed space. For the space \ell^\infty as the dual of \ell^1, this unit ball is weak* homeomorphic to the product [-1,1]^\mathbb{N} via the coordinate maps sending f to (f(n))_{n\in\mathbb{N}}, and Tychonoff's theorem establishes the compactness of this countable product of compact intervals, yielding weak* compactness essential for duality and optimization in analysis.^[28]

In Algebraic Topology and Functional Analysis

In algebraic topology, Tychonoff's theorem ensures the compactness of infinite products of Eilenberg-MacLane spaces K(G, n) equipped with certain topologies, such as the pro-finite topology when G is finite, facilitating the study of homotopy types and cohomology rings through infinite constructions.^[29] This compactness is crucial for understanding classifying spaces of infinite groups, where products model extensions and fibrations in the stable homotopy category.^[29] A key application arises in the spectrum of C*-algebras, where the Gelfand transform establishes an isometric -isomorphism between a commutative unital C-algebra A and the algebra of continuous complex-valued functions C(\Delta(A)) on its spectrum \Delta(A), a compact Hausdorff space constructed as a product of closed disks via the evaluation of characters. This representation leverages Tychonoff's theorem to guarantee the compactness of \Delta(A), enabling the duality between algebraic structures and topological spaces in noncommutative geometry. In functional analysis, Alaoglu's theorem asserts that the closed unit ball B_{X^*} = \{\mu \in X^* : \|\mu\| \leq 1\} in the dual X^* of a normed space X is compact in the weak* topology, proved by embedding B_{X^*} as a closed subset of the product \prod_{x \in X} D_x, where each D_x = \{z \in \mathbb{C} : |z| \leq \|x\|\} is compact, and applying Tychonoff's theorem to yield compactness of the product.^[28] Similarly, Pontryagin duality identifies compact abelian groups as Pontryagin duals of discrete abelian groups, with the dual \hat{G} of a discrete group G topologized as the product \prod_{g \in G} S^1 under the compact-open topology, rendering \hat{G} compact by Tychonoff's theorem.^[30] As an illustrative example in this context, profinite groups, defined as inverse limits of finite discrete groups \{G_j\} along bonding maps \phi_{ij}: G_i \to G_j, inherit compactness from the closed subspace topology within the product \prod_j G_j, which is compact by Tychonoff's theorem since each finite discrete G_j is compact.^[31] This structure underpins their role as compact totally disconnected groups in Galois theory and algebraic number theory.^[31]

Relation to the Axiom of Choice

Logical Equivalence

In Zermelo–Fraenkel set theory without the axiom of choice (ZF), Tychonoff's theorem asserting the compactness of arbitrary products of compact topological spaces is logically equivalent to the axiom of choice (AC). This equivalence highlights the foundational role of AC in general topology, as the standard proof of the theorem relies on Zorn's lemma, an equivalent of AC, while the converse—that the theorem implies AC—was rigorously established by John L. Kelley.^[32]^[33] The Boolean prime ideal theorem (BPI), which states that every Boolean algebra possesses a prime ideal, acts as an intermediate principle in this logical structure. Tychonoff's theorem implies the BPI, and the BPI is implied by AC; the reverse implications hold in the sense that AC entails Tychonoff's theorem, forming a chain of equivalences and implications central to set-theoretic topology. Specifically, the BPI is equivalent to the restricted form of Tychonoff's theorem for products of compact Hausdorff spaces.^[34]^[35] Weaker variants of Tychonoff's theorem correspond to weakened forms of AC. For instance, the theorem applied to countable products of compact spaces is equivalent to the axiom of countable choice (AC_ω). Likewise, the version for products of 2^κ many copies of the two-point discrete space {0,1} is equivalent to AC_κ, the axiom of choice for families of sets of cardinality at most κ.^[35] This equivalence was conjectured by Shizuo Kakutani and proved by John L. Kelley in 1950. This logical dependence has profound implications: Tychonoff's theorem fails in certain models of ZF negating AC, such as Fraenkel-Mostowski permutation models, where the product of an infinite collection of finite discrete spaces need not be compact.^[36]^[37]

Deriving the Axiom of Choice from the Theorem

To derive the axiom of choice from Tychonoff's theorem, consider an arbitrary family of nonempty sets \{A_i \mid i \in I\}. The proof constructs a compact product space in which the desired choice function appears as a point in a closed subset corresponding to the original product \prod_{i \in I} A_i. For each i \in I, form the enlarged space X_i = A_i \cup \{*\}, where * is a distinguished point not in A_i. Endow X_i with the topology consisting of the empty set, the singleton \{*\}, and all subsets of X_i whose complement in A_i is finite (i.e., cofinite subsets of A_i, possibly including *). This topology makes \{*\} open, so A_i is closed in X_i. Moreover, X_i is compact, as any open cover must include an open set containing *, and the cofinite opens ensure that the remaining points in A_i can be covered by finitely many opens due to their cofinite nature.^[7] The product space X = \prod_{i \in I} X_i, equipped with the product topology, is compact by Tychonoff's theorem. The projection maps \pi_i: X \to X_i are continuous, so the preimages F_i = \pi_i^{-1}(A_i) are closed subsets of X. The family \{F_i \mid i \in I\} has the finite intersection property: for any finite J \subseteq I, the intersection \bigcap_{i \in J} F_i = \left( \prod_{i \in J} A_i \right) \times \left( \prod_{i \notin J} X_i \right) is nonempty, since the finite product \prod_{i \in J} A_i is nonempty in ZF set theory (by finite induction on the number of factors, without needing choice), and the remaining product admits the constant function sending all coordinates to *, which is definable without choice.^[7] Since X is compact, every family of closed sets with the finite intersection property has nonempty intersection. Thus, \bigcap_{i \in I} F_i \neq \emptyset. But \bigcap_{i \in I} F_i = \prod_{i \in I} A_i, so this product is nonempty. Any element of \prod_{i \in I} A_i is a choice function f: I \to \bigcup_{i \in I} A_i with f(i) \in A_i for all i \in I. This establishes the axiom of choice.^[32] When each A_i is finite, the topology on X_i coincides with the discrete topology, as all subsets are cofinite on a finite set. In this case, each X_i is compact (finite discrete spaces are compact), and Tychonoff's theorem applies directly to yield the product compactness, with the same argument showing the existence of a choice function via the nonempty intersection of the F_i.^[7] For the general case with possibly infinite A_i, the cofinite topology on X_i ensures compactness even when A_i is infinite. An alternative perspective for infinite A_i embeds A_i definably into the compact space \{0,1\}^{A_i} (the generalized Cantor space with the product topology from the discrete \{0,1\}), via the injection \phi_i: a \mapsto \chi_{\{a\}}, the characteristic function of the singleton \{a\}. The product \prod_{i \in I} \{0,1\}^{A_i} is compact by Tychonoff's theorem, and the structure allows a similar finite intersection property argument on preimages to guarantee a point whose coordinates encode a choice function, avoiding the empty product.^[32]

References

[1]
https://doi.org/10.1007/BF01472255
[2]
https://doi.org/10.1007/BF01782364
[3]
[PDF] Tychonoff's theorem - Cornell Mathematics
Tychonoff's theorem asserts that the product of an arbitrary family of compact spaces is compact. This is proved in Chapter 5 of Munkres, but his proof is not.
[4]
[PDF] TYCHONOFF THEOREM AND ITS APPLICATIONS Last time
Mar 7, 2021 · 1In fact, Tychonoff defined the product topology in his 1935 paper for the first time. 1. Page 2. 2. COMPACTNESS: TYCHONOFF THEOREM AND ITS ...
[5]
[PDF] THREE PROOFS OF TYCHONOFF'S THEOREM - HAL
We give three proofs of Tychonoff's theorem on the compactness of a product of compact topological spaces. The first one proceeds “from scratch.” The second one.
[6]
The Tychonoff product theorem implies the axiom of choice - EuDML
The Tychonoff product theorem implies the axiom of choice. J. Kelley · Fundamenta Mathematicae (1950). Volume: 37, Issue: 1, page 75-76; ISSN: 0016-2736 ...Missing: paper | Show results with:paper
[7]
[PDF] Tychonoff's Theorem implies AC
Recall that Tychonoff's Theorem is the assertion that a product of compact topological spaces is compact. We will show (without using AC) that Ty- chonoff's ...
[8]
Über die topologische Erweiterung von Räumen
Cite this article. Tychonoff, A. Über die topologische Erweiterung von Räumen. Math. Ann. 102, 544–561 (1930). https://doi.org/10.1007/BF01782364. Download ...
[9]
Über einen Funktionenraum | Mathematische Annalen
Tychonoff, A. Über einen Funktionenraum. Math. Ann. 111, 762–766 (1935). https://doi.org/10.1007/BF01472255
[10]
Compact Space -- from Wolfram MathWorld
A topological space is compact if every open cover of X has a finite subcover. In other words, if X is the union of a family of open sets, there is a finite ...
[11]
[PDF] 16. Compactness
A topological space (X,T ) is said to be compact if every open cover of X has a finite subcover. We will often refer to subsets of topological spaces being ...
[12]
nLab sequentially compact metric spaces are equivalently compact ...
Jun 5, 2022 · For metric spaces, being compact and sequentially compact are equivalent. This is not true for general topological spaces.
[13]
limit point compact space in nLab
Jul 28, 2018 · A topological space is limit point compact if every infinite subset has a limit point (in the sense of Definition 2.3 there, ie, an accumulation point).
[14]
Compactness and Heine-Borel - Advanced Analysis
Jan 20, 2024 · A subset E of R n or C n is compact if and only if it is closed and bounded. We already have seen that any compact subset of a metric space must ...
[15]
examples of compact spaces - PlanetMath
Mar 22, 2013 · The unit interval [0,1] is compact. This follows from the Heine-Borel Theorem. Proving that theorem is about as hard as proving directly ...
[16]
[PDF] Topology of the Real Numbers - UC Davis Math
We will prove that a subset of R is compact if and only if it is closed and bounded. For example, every closed, bounded interval [a, b] is compact. There ...<|control11|><|separator|>
[17]
proof that a compact set in a Hausdorff space is closed - PlanetMath
Mar 22, 2013 · Let X be a Hausdorff space, and C⊆X C ⊆ X a compact subset. We are to show that C is closed. We will do so, by showing that the complement U=X∖ ...
[18]
continuous image of a compact set is compact - PlanetMath
Mar 22, 2013 · Theorem 1. The continuous image of a compact set is also compact. Proof. Let X X and Y Y be topological spaces ...
[19]
245B, Notes 10: Compactness in topological spaces - Terry Tao
Feb 9, 2009 · In these notes, we explore how compactness interacts with other key topological concepts: the Hausdorff property, bases and sub-bases, product spaces, and ...
[20]
Product Topology -- from Wolfram MathWorld
In the definition of product topology of , where is any set, the open sets are the unions of subsets , where is an open subset of with the additional condition ...
[21]
[PDF] 06. Initial and final topology - TU Graz
Note that each pi is surjective. Definition. The initial topology on X = ∏ i∈I. Xi with respect to the family {pi : i ∈ I} is called the product topology τ on X ...
[22]
product topology preserves the Hausdorff property - PlanetMath.org
Mar 22, 2013 · Then the generalized Cartesian product ∏α∈AXα ∏ α ∈ A X α equipped with the product topology is a Hausdorff space. Proof. Let Y=∏ ...
[23]
[PDF] Section 19. The Product Topology
Dec 6, 2016 · In this section we consider arbitrary products of topological spaces and give two topologies on these spaces, the box topology and the product.
[24]
[PDF] A proof of Tychonoff's theorem - OSU Math
By Zorn's lemma, there is a maximal family with the f.i.p., M ⊃ F. In the following. “construct”, “choose” etc. are just ways of speaking, as we rely on the ...
[25]
[PDF] Tychonoff's Theorem
Aug 23, 2010 · In 1950, Kelley proved that the Tychonoff theorem is equivalent to the axiom of choice [3]. In order to prove that the Tychonoff theorem ...Missing: source | Show results with:source
[26]
[PDF] Tychonoff's Theorem: The General Case
The main proof of Tychonoff's Theorem presented in Munkres' book, pages 230-235, uses Zorn's Lemma instead of transfinite induction. This alternative proof is ...<|control11|><|separator|>
[27]
What is your favorite proof of Tychonoff's Theorem? - MathOverflow
May 30, 2010 · I learned the proof from Chernoff's original paper,which was required reading in John Terilla's point set topology course.Interestingly,John ...
[28]
[PDF] Derived Stone Embedding - arXiv
Feb 22, 2025 · By Tychonoff's theorem, the product. Q Xi is compact ... which characterizes pointed connected spaces in terms of Eilenberg–MacLane spaces,.
[29]
[PDF] E.7 Alaoglu's Theorem
Even so, Alaoglu's Theorem states that the closed unit ball in X∗ is compact in the weak* topology. We will prove this theorem in this section. E.7.1 Product ...
[30]
[PDF] duality and structure of locally compact abelian groups ..... for the ...
Our aim is to describe the principal structure theorem for locally compact abelian groups, and to acquaint the reader with the Pontryagin - van Kampen duality ...
[31]
[PDF] Part III Profinite Groups - DPMMS
Feb 1, 2020 · INVERSE LIMITS. 13. By Tychonoff's Theorem, Q Gj is compact and Hausdorff. Each condition φij(gi) = gj describes a closed subset of Q Gj: the ...
[32]
[PDF] The Tychonoff product theorem implies the axiom of choice
Kuratowski, Topologie I, Monogr. Mat. 3 (1933), p. 15. Page 2. 76. J. L. Kelley. 2. Proof of the theorem. We now demonstrate the fol- lowing statement of the ...
[33]
[PDF] Equivalents of the axiom of choice - Andrés E. Caicedo
The equivalence of Tychonoff's theorem and choice is due to Kelley. That the existence of bases implies choice is due to Blass, who proved that 7 implies the.<|control11|><|separator|>
[34]
prime ideal theorem in nLab
Mar 2, 2018 · The Tychonoff theorem for compact Hausdorff spaces in turn implies that every Boolean ring B B has a maximal (and therefore prime) ideal; see ...Idea · A ladder of prime ideal theorems · BPIT implies prime ideal...
[35]
[PDF] On the roles of variants of Axiom of Choice in variations of Tychonoff ...
Dec 27, 2022 · In this expository note, we examine the roles of Axiom of Choice (AC) and its weak ... The latter can be characterized by a weakening of Tychonoff ...
[36]
[PDF] Wallman Compactifications and Tychonoff's Compactness Theorem ...
Jan 26, 2013 · CFE (Axiom of Closed Filter Extendability): For every topological space (X, T), every filter base G of closed subsets of X extends to a closed.
[37]
[PDF] Foundations of Algebraic Topology
The principal contribution of this book is an axiomatic approach to the part of algebraic topology called homology theory. It is the oldest.
[38]
Most 'unintuitive' application of the Axiom of Choice? - MathOverflow
Apr 10, 2010 · In the absence of choice, you can have an algebraic closure of Q that is a countable union of finite sets but is not itself countable.Result that follows from ZFC and not ZF but are strictly weaker than ...Unnecessary uses of the axiom of choice - MathOverflowMore results from mathoverflow.net
[39]
Tychonoff's Theorem and (the lack of) the Axiom of Choice : r/math
May 30, 2021 · I was thinking about Tychonoff's theorem and began to wonder about why it's the case that the Axiom of Choice is necessary in the proof.Some of the most mind-blowing things that are equivalent to ... - RedditOriginal uses of equivalencies of the axiom of choice : r/math - RedditMore results from www.reddit.comMissing: source | Show results with:source