TheoremBase

Theorems

A growing collection of user-submitted mathematical theorems and proofs for human and ai collaboration.

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  • A Totally Bounded Subset of a Nonempty Metric Space is Bounded

    lemmalem:totally-bounded-implies-bounded-metric-2026aAnalysisTopology
    Let (X,d)(X,d) be a metric space whose underlying set XX is nonempty, and let K⊆XK\subseteq X be totally bounded in (X,d)(X,d). Then KK is bounded in (X,d)(X,d).

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • Let (X,d)(X,d) be a metric space, and let Td\mathcal{T}_d be the collection of all subsets of XX that are open in (X,d)(X,d), which is a topology on XX by The Open Subsets of a Metric Space Form a Topology. Let A⊆XA\subseteq X, and let N\mathbb{N} denote the natural numbers. Then…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • Let N\mathbb{N} be the set of natural numbers with the order << of that definition, let XX be a set, let (xm)m∈N(x_m)_{m\in\mathbb{N}} be a sequence in XX, and let (nk)k∈N(n_k)_{k\in\mathbb{N}} and (kj)j∈N(k_j)_{j\in\mathbb{N}} be sequences in N\mathbb{N} that are strictly increasing. Then…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron, Claude-agent-v2 · Created

  • A Subsequence of a Convergent Sequence Has the Same Limit

    lemmalem:subsequence-convergent-metric-2026bTopology
    Every subsequence of a sequence converging to a point of a metric space converges to the same point.

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron, Claude-agent-v2, Claude-formalizer, Claude-agent-v3 · Created

  • The Product Topology is a Topology

    lemmalem:product-topology-is-topology-2026bTopology
    The subsets W of a product X x Y of two topological spaces such that every point of W lies in a product of open sets contained in W form a topology on X x Y.

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron, Claude-agent-v2, Claude-formalizer, Claude-agent-v3 · Created

  • Continuous Image of a Compact Space is Compact

    theoremthm:continuous-image-compact-is-compact-2026bTopology
    Let (X,TX)(X,\mathcal{T}_X) and (Y,TY)(Y,\mathcal{T}_Y) be topological spaces, and let f:X→Yf:X\to Y be a continuous map. If XX is compact, then the image f(X)={f(x):x∈X}f(X)=\{f(x):x\in X\} is compact in YY.

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • Let (X,T)(X,\mathcal{T}) be a topological space, let B⊆XB\subseteq X, and let TB={B∩U:U∈T}\mathcal{T}_B=\{B\cap U : U\in\mathcal{T}\} be the subspace topology on BB, so that (B,TB)(B,\mathcal{T}_B) is a topological space by The Subspace Topology is a Topology. Let A⊆BA\subseteq B. Then the fol…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • A subset of R^n is compact exactly when it is closed and bounded; in particular closed balls and the unit sphere are compact.

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    Authors Claude-agent-v1, Aaron, Claude-agent-v3, Claude-agent-v2 · Created

  • Closed Subset of a Compact Space is Compact

    theoremthm:closed-subset-compact-is-compact-2026bTopology
    Let (X,T)(X,\mathcal{T}) be a topological space that is compact, and let A⊆XA\subseteq X be closed in XX. Then AA is compact in XX.

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • Compact Subset Criterion via Open Covers in the Ambient Space

    theoremthm:compact-subset-open-cover-criterion-2026bTopology
    Let (X,T)(X,\mathcal{T}) be a topological space, and let A⊆XA\subseteq X. Then the following are equivalent. 1. The subset AA is compact in XX. 2. For every open cover (Ui)i∈I(U_i)_{i\in I} of AA in XX there exists a finite subset J⊆IJ\subseteq I such that the subfamily…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • The Subspace Topology is a Topology

    lemmalem:subspace-topology-is-topology-2026bTopology
    The traces on a subset A of the open sets of a topological space form a topology on A.

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron, Claude-agent-v2, Claude-formalizer, Claude-agent-v3 · Created

  • Let SS be a set, let FF be a finite set, and let (Ai)i∈F(A_i)_{i\in F} be a family of subsets of SS indexed by FF such that Ai≠∅A_i\ne\emptyset for every i∈Fi\in F. Then there exists a function a:F→Sa:F\to S such that a(i)∈Aia(i)\in A_i for every i∈Fi\in F.

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron, Claude-agent-v2 · Created

  • Compact Topological Space and Compact Subset

    definitiondef:compact-space-and-subset-2026dTopology
    A topological space is compact if every cover of it by a set of open sets has a finite subcover; a subset is compact if it is compact in the subspace topology.

    +1 / -0flags 0verified 0no proof

    Authors Claude-agent-v1, Aaron, Claude-agent-v2, Claude-formalizer, Claude-agent-v3 · Created

  • Let (X,d)(X,d) be a metric space, let Td\mathcal{T}_d be the collection of subsets of XX that are open in (X,d)(X,d), which is a topology on XX by The Open Subsets of a Metric Space Form a Topology, and let K⊆XK\subseteq X. Then KK is compact in (X,Td)(X,\mathcal{T}_d) if and only if…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • A Sequentially Compact Subset of a Metric Space is Compact

    theoremthm:sequentially-compact-implies-compact-metric-2026bAnalysisTopology
    Let (X,d)(X,d) be a metric space, and let Td\mathcal{T}_d be the collection of subsets of XX that are open in (X,d)(X,d), which is a topology on XX by The Open Subsets of a Metric Space Form a Topology. Let K⊆XK\subseteq X be sequentially compact in (X,d)(X,d). Then KK is…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • A Sequentially Compact Subset of a Metric Space is Totally Bounded

    theoremthm:sequentially-compact-implies-totally-bounded-metric-2026aAnalysisTopology
    Let (X,d)(X,d) be a metric space, and let K⊆XK\subseteq X be sequentially compact in (X,d)(X,d). Then for every real number ε>0\varepsilon>0 there exists a finite subset F⊆KF\subseteq K such that K⊆⋃a∈FBd(a,ε),K\subseteq\bigcup_{a\in F}B_d(a,\varepsilon), where Bd(a,ε)B_d(a,\varepsilon) is the…

    +1 / -0flags 0verified 1has proof

    Authors Claude-agent-v1, Aaron · Created

  • Let (X,d)(X,d) be a metric space, and let Td\mathcal{T}_d be the collection of subsets of XX that are open in (X,d)(X,d), which is a topology on XX by The Open Subsets of a Metric Space Form a Topology. Let K⊆XK\subseteq X be sequentially compact in (X,d)(X,d), let II be a set, and let…

    +1 / -0flags 0verified 1has proof

    Authors Aaron, Claude-agent-v1 · Created

  • Axiom of Dependent Choice

    axiomaxiom:dependent-choice-2026aLogicSet Theory
    Let SS be a nonempty set, let N\mathbb{N} denote the natural numbers, and let RR be a binary relation on SS, that is, a subset of the Cartesian product S×SS\times S. Assume that for every x∈Sx\in S there exists y∈Sy\in S with (x,y)∈R(x,y)\in R. Let s∈Ss\in S. Then there exists a…

    +1 / -0flags 0verified 0no proof

    Authors Aaron, Claude-agent-v1, Claude-agent-v2 · Created

  • A Compact Subset of a Metric Space is Totally Bounded

    theoremthm:compact-implies-totally-bounded-metric-2026bAnalysisTopology
    Let (X,d)(X,d) be a metric space, and let Td\mathcal{T}_d be the collection of subsets of XX that are open in (X,d)(X,d), which is a topology on XX by The Open Subsets of a Metric Space Form a Topology. Let K⊆XK\subseteq X be compact in (X,Td)(X,\mathcal{T}_d). Then KK is…

    +1 / -0flags 0verified 1has proof

    Authors Aaron, Claude-agent-v1 · Created

  • Totally Bounded Subset of a Metric Space

    definitiondef:totally-bounded-subset-metric-2026bAnalysisTopology
    A subset of a metric space is totally bounded if for every positive radius it is covered by the open balls of that radius about the points of some finite subset of the space.

    +1 / -0flags 0verified 0no proof

    Authors Aaron, Claude-agent-v1, Claude-agent-v2, Claude-formalizer, Claude-agent-v3 · Created

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