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Theorem List for Metamath Proof Explorer - 34901-35000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremsdclem1 34901* Lemma for sdc 34902. (Contributed by Jeff Madsen, 2-Sep-2009.)
𝑍 = (ℤ𝑀)    &   (𝑔 = (𝑓 ↾ (𝑀...𝑛)) → (𝜓𝜒))    &   (𝑛 = 𝑀 → (𝜓𝜏))    &   (𝑛 = 𝑘 → (𝜓𝜃))    &   ((𝑔 = 𝑛 = (𝑘 + 1)) → (𝜓𝜎))    &   (𝜑𝐴𝑉)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑔(𝑔:{𝑀}⟶𝐴𝜏))    &   ((𝜑𝑘𝑍) → ((𝑔:(𝑀...𝑘)⟶𝐴𝜃) → ∃(:(𝑀...(𝑘 + 1))⟶𝐴𝑔 = ( ↾ (𝑀...𝑘)) ∧ 𝜎)))    &   𝐽 = {𝑔 ∣ ∃𝑛𝑍 (𝑔:(𝑀...𝑛)⟶𝐴𝜓)}    &   𝐹 = (𝑤𝑍, 𝑥𝐽 ↦ { ∣ ∃𝑘𝑍 (:(𝑀...(𝑘 + 1))⟶𝐴𝑥 = ( ↾ (𝑀...𝑘)) ∧ 𝜎)})       (𝜑 → ∃𝑓(𝑓:𝑍𝐴 ∧ ∀𝑛𝑍 𝜒))
 
Theoremsdc 34902* Strong dependent choice. Suppose we may choose an element of 𝐴 such that property 𝜓 holds, and suppose that if we have already chosen the first 𝑘 elements (represented here by a function from 1...𝑘 to 𝐴), we may choose another element so that all 𝑘 + 1 elements taken together have property 𝜓. Then there exists an infinite sequence of elements of 𝐴 such that the first 𝑛 terms of this sequence satisfy 𝜓 for all 𝑛. This theorem allows us to construct infinite seqeunces where each term depends on all the previous terms in the sequence. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 3-Jun-2014.)
𝑍 = (ℤ𝑀)    &   (𝑔 = (𝑓 ↾ (𝑀...𝑛)) → (𝜓𝜒))    &   (𝑛 = 𝑀 → (𝜓𝜏))    &   (𝑛 = 𝑘 → (𝜓𝜃))    &   ((𝑔 = 𝑛 = (𝑘 + 1)) → (𝜓𝜎))    &   (𝜑𝐴𝑉)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑 → ∃𝑔(𝑔:{𝑀}⟶𝐴𝜏))    &   ((𝜑𝑘𝑍) → ((𝑔:(𝑀...𝑘)⟶𝐴𝜃) → ∃(:(𝑀...(𝑘 + 1))⟶𝐴𝑔 = ( ↾ (𝑀...𝑘)) ∧ 𝜎)))       (𝜑 → ∃𝑓(𝑓:𝑍𝐴 ∧ ∀𝑛𝑍 𝜒))
 
Theoremfdc 34903* Finite version of dependent choice. Construct a function whose value depends on the previous function value, except at a final point at which no new value can be chosen. The final hypothesis ensures that the process will terminate. The proof does not use the Axiom of Choice. (Contributed by Jeff Madsen, 18-Jun-2010.)
𝐴 ∈ V    &   𝑀 ∈ ℤ    &   𝑍 = (ℤ𝑀)    &   𝑁 = (𝑀 + 1)    &   (𝑎 = (𝑓‘(𝑘 − 1)) → (𝜑𝜓))    &   (𝑏 = (𝑓𝑘) → (𝜓𝜒))    &   (𝑎 = (𝑓𝑛) → (𝜃𝜏))    &   (𝜂𝐶𝐴)    &   (𝜂𝑅 Fr 𝐴)    &   ((𝜂𝑎𝐴) → (𝜃 ∨ ∃𝑏𝐴 𝜑))    &   (((𝜂𝜑) ∧ (𝑎𝐴𝑏𝐴)) → 𝑏𝑅𝑎)       (𝜂 → ∃𝑛𝑍𝑓(𝑓:(𝑀...𝑛)⟶𝐴 ∧ ((𝑓𝑀) = 𝐶𝜏) ∧ ∀𝑘 ∈ (𝑁...𝑛)𝜒))
 
Theoremfdc1 34904* Variant of fdc 34903 with no specified base value. (Contributed by Jeff Madsen, 18-Jun-2010.)
𝐴 ∈ V    &   𝑀 ∈ ℤ    &   𝑍 = (ℤ𝑀)    &   𝑁 = (𝑀 + 1)    &   (𝑎 = (𝑓𝑀) → (𝜁𝜎))    &   (𝑎 = (𝑓‘(𝑘 − 1)) → (𝜑𝜓))    &   (𝑏 = (𝑓𝑘) → (𝜓𝜒))    &   (𝑎 = (𝑓𝑛) → (𝜃𝜏))    &   (𝜂 → ∃𝑎𝐴 𝜁)    &   (𝜂𝑅 Fr 𝐴)    &   ((𝜂𝑎𝐴) → (𝜃 ∨ ∃𝑏𝐴 𝜑))    &   (((𝜂𝜑) ∧ (𝑎𝐴𝑏𝐴)) → 𝑏𝑅𝑎)       (𝜂 → ∃𝑛𝑍𝑓(𝑓:(𝑀...𝑛)⟶𝐴 ∧ (𝜎𝜏) ∧ ∀𝑘 ∈ (𝑁...𝑛)𝜒))
 
Theoremseqpo 34905* Two ways to say that a sequence respects a partial order. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑅 Po 𝐴𝐹:ℕ⟶𝐴) → (∀𝑠 ∈ ℕ (𝐹𝑠)𝑅(𝐹‘(𝑠 + 1)) ↔ ∀𝑚 ∈ ℕ ∀𝑛 ∈ (ℤ‘(𝑚 + 1))(𝐹𝑚)𝑅(𝐹𝑛)))
 
Theoremincsequz 34906* An increasing sequence of positive integers takes on indefinitely large values. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝐹:ℕ⟶ℕ ∧ ∀𝑚 ∈ ℕ (𝐹𝑚) < (𝐹‘(𝑚 + 1)) ∧ 𝐴 ∈ ℕ) → ∃𝑛 ∈ ℕ (𝐹𝑛) ∈ (ℤ𝐴))
 
Theoremincsequz2 34907* An increasing sequence of positive integers takes on indefinitely large values. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝐹:ℕ⟶ℕ ∧ ∀𝑚 ∈ ℕ (𝐹𝑚) < (𝐹‘(𝑚 + 1)) ∧ 𝐴 ∈ ℕ) → ∃𝑛 ∈ ℕ ∀𝑘 ∈ (ℤ𝑛)(𝐹𝑘) ∈ (ℤ𝐴))
 
Theoremnnubfi 34908* A bounded above set of positive integers is finite. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 28-Feb-2014.)
((𝐴 ⊆ ℕ ∧ 𝐵 ∈ ℕ) → {𝑥𝐴𝑥 < 𝐵} ∈ Fin)
 
Theoremnninfnub 34909* An infinite set of positive integers is unbounded above. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 28-Feb-2014.)
((𝐴 ⊆ ℕ ∧ ¬ 𝐴 ∈ Fin ∧ 𝐵 ∈ ℕ) → {𝑥𝐴𝐵 < 𝑥} ≠ ∅)
 
20.20.4  Topology
 
Theoremsubspopn 34910 An open set is open in the subspace topology. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 15-Dec-2013.)
(((𝐽 ∈ Top ∧ 𝐴𝑉) ∧ (𝐵𝐽𝐵𝐴)) → 𝐵 ∈ (𝐽t 𝐴))
 
Theoremneificl 34911 Neighborhoods are closed under finite intersection. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 25-Nov-2013.)
(((𝐽 ∈ Top ∧ 𝑁 ⊆ ((nei‘𝐽)‘𝑆)) ∧ (𝑁 ∈ Fin ∧ 𝑁 ≠ ∅)) → 𝑁 ∈ ((nei‘𝐽)‘𝑆))
 
Theoremlpss2 34912 Limit points of a subset are limit points of the larger set. (Contributed by Jeff Madsen, 2-Sep-2009.)
𝑋 = 𝐽       ((𝐽 ∈ Top ∧ 𝐴𝑋𝐵𝐴) → ((limPt‘𝐽)‘𝐵) ⊆ ((limPt‘𝐽)‘𝐴))
 
20.20.5  Metric spaces
 
Theoremmetf1o 34913* Use a bijection with a metric space to construct a metric on a set. (Contributed by Jeff Madsen, 2-Sep-2009.)
𝑁 = (𝑥𝑌, 𝑦𝑌 ↦ ((𝐹𝑥)𝑀(𝐹𝑦)))       ((𝑌𝐴𝑀 ∈ (Met‘𝑋) ∧ 𝐹:𝑌1-1-onto𝑋) → 𝑁 ∈ (Met‘𝑌))
 
Theoremblssp 34914 A ball in the subspace metric. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 5-Jan-2014.)
𝑁 = (𝑀 ↾ (𝑆 × 𝑆))       (((𝑀 ∈ (Met‘𝑋) ∧ 𝑆𝑋) ∧ (𝑌𝑆𝑅 ∈ ℝ+)) → (𝑌(ball‘𝑁)𝑅) = ((𝑌(ball‘𝑀)𝑅) ∩ 𝑆))
 
Theoremmettrifi 34915* Generalized triangle inequality for arbitrary finite sums. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 4-Jun-2014.)
(𝜑𝐷 ∈ (Met‘𝑋))    &   (𝜑𝑁 ∈ (ℤ𝑀))    &   ((𝜑𝑘 ∈ (𝑀...𝑁)) → (𝐹𝑘) ∈ 𝑋)       (𝜑 → ((𝐹𝑀)𝐷(𝐹𝑁)) ≤ Σ𝑘 ∈ (𝑀...(𝑁 − 1))((𝐹𝑘)𝐷(𝐹‘(𝑘 + 1))))
 
Theoremlmclim2 34916* A sequence in a metric space converges to a point iff the distance between the point and the elements of the sequence converges to 0. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 5-Jun-2014.)
(𝜑𝐷 ∈ (Met‘𝑋))    &   (𝜑𝐹:ℕ⟶𝑋)    &   𝐽 = (MetOpen‘𝐷)    &   𝐺 = (𝑥 ∈ ℕ ↦ ((𝐹𝑥)𝐷𝑌))    &   (𝜑𝑌𝑋)       (𝜑 → (𝐹(⇝𝑡𝐽)𝑌𝐺 ⇝ 0))
 
Theoremgeomcau 34917* If the distance between consecutive points in a sequence is bounded by a geometric sequence, then the sequence is Cauchy. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 5-Jun-2014.)
(𝜑𝐷 ∈ (Met‘𝑋))    &   (𝜑𝐹:ℕ⟶𝑋)    &   (𝜑𝐴 ∈ ℝ)    &   (𝜑𝐵 ∈ ℝ+)    &   (𝜑𝐵 < 1)    &   ((𝜑𝑘 ∈ ℕ) → ((𝐹𝑘)𝐷(𝐹‘(𝑘 + 1))) ≤ (𝐴 · (𝐵𝑘)))       (𝜑𝐹 ∈ (Cau‘𝐷))
 
Theoremcaures 34918 The restriction of a Cauchy sequence to an upper set of integers is Cauchy. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 5-Jun-2014.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐷 ∈ (Met‘𝑋))    &   (𝜑𝐹 ∈ (𝑋pm ℂ))       (𝜑 → (𝐹 ∈ (Cau‘𝐷) ↔ (𝐹𝑍) ∈ (Cau‘𝐷)))
 
Theoremcaushft 34919* A shifted Cauchy sequence is Cauchy. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 5-Jun-2014.)
𝑍 = (ℤ𝑀)    &   (𝜑𝑀 ∈ ℤ)    &   (𝜑𝐷 ∈ (Met‘𝑋))    &   𝑊 = (ℤ‘(𝑀 + 𝑁))    &   (𝜑𝑁 ∈ ℤ)    &   ((𝜑𝑘𝑍) → (𝐹𝑘) = (𝐺‘(𝑘 + 𝑁)))    &   (𝜑𝐹 ∈ (Cau‘𝐷))    &   (𝜑𝐺:𝑊𝑋)       (𝜑𝐺 ∈ (Cau‘𝐷))
 
20.20.6  Continuous maps and homeomorphisms
 
Theoremconstcncf 34920* A constant function is a continuous function on . (Contributed by Jeff Madsen, 2-Sep-2009.) (Moved into main set.mm as cncfmptc 23448 and may be deleted by mathbox owner, JM. --MC 12-Sep-2015.) (Revised by Mario Carneiro, 12-Sep-2015.)
𝐹 = (𝑥 ∈ ℂ ↦ 𝐴)       (𝐴 ∈ ℂ → 𝐹 ∈ (ℂ–cn→ℂ))
 
Theoremidcncf 34921 The identity function is a continuous function on . (Contributed by Jeff Madsen, 11-Jun-2010.) (Moved into main set.mm as cncfmptid 23449 and may be deleted by mathbox owner, JM. --MC 12-Sep-2015.) (Revised by Mario Carneiro, 12-Sep-2015.)
𝐹 = (𝑥 ∈ ℂ ↦ 𝑥)       𝐹 ∈ (ℂ–cn→ℂ)
 
Theoremsub1cncf 34922* Subtracting a constant is a continuous function. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
𝐹 = (𝑥 ∈ ℂ ↦ (𝑥𝐴))       (𝐴 ∈ ℂ → 𝐹 ∈ (ℂ–cn→ℂ))
 
Theoremsub2cncf 34923* Subtraction from a constant is a continuous function. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
𝐹 = (𝑥 ∈ ℂ ↦ (𝐴𝑥))       (𝐴 ∈ ℂ → 𝐹 ∈ (ℂ–cn→ℂ))
 
Theoremcnres2 34924* The restriction of a continuous function to a subset is continuous. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 15-Dec-2013.)
𝑋 = 𝐽    &   𝑌 = 𝐾       (((𝐽 ∈ Top ∧ 𝐾 ∈ Top) ∧ (𝐴𝑋𝐵𝑌) ∧ (𝐹 ∈ (𝐽 Cn 𝐾) ∧ ∀𝑥𝐴 (𝐹𝑥) ∈ 𝐵)) → (𝐹𝐴) ∈ ((𝐽t 𝐴) Cn (𝐾t 𝐵)))
 
Theoremcnresima 34925 A continuous function is continuous onto its image. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 15-Dec-2013.)
((𝐽 ∈ Top ∧ 𝐾 ∈ Top ∧ 𝐹 ∈ (𝐽 Cn 𝐾)) → 𝐹 ∈ (𝐽 Cn (𝐾t ran 𝐹)))
 
Theoremcncfres 34926* A continuous function on complex numbers restricted to a subset. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Sep-2015.)
𝐴 ⊆ ℂ    &   𝐵 ⊆ ℂ    &   𝐹 = (𝑥 ∈ ℂ ↦ 𝐶)    &   𝐺 = (𝑥𝐴𝐶)    &   (𝑥𝐴𝐶𝐵)    &   𝐹 ∈ (ℂ–cn→ℂ)    &   𝐽 = (MetOpen‘((abs ∘ − ) ↾ (𝐴 × 𝐴)))    &   𝐾 = (MetOpen‘((abs ∘ − ) ↾ (𝐵 × 𝐵)))       𝐺 ∈ (𝐽 Cn 𝐾)
 
20.20.7  Boundedness
 
Syntaxctotbnd 34927 Extend class notation with the class of totally bounded metric spaces.
class TotBnd
 
Syntaxcbnd 34928 Extend class notation with the class of bounded metric spaces.
class Bnd
 
Definitiondf-totbnd 34929* Define the class of totally bounded metrics. A metric space is totally bounded iff it can be covered by a finite number of balls of any given radius. (Contributed by Jeff Madsen, 2-Sep-2009.)
TotBnd = (𝑥 ∈ V ↦ {𝑚 ∈ (Met‘𝑥) ∣ ∀𝑑 ∈ ℝ+𝑣 ∈ Fin ( 𝑣 = 𝑥 ∧ ∀𝑏𝑣𝑦𝑥 𝑏 = (𝑦(ball‘𝑚)𝑑))})
 
Theoremistotbnd 34930* The predicate "is a totally bounded metric space". (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝑀 ∈ (TotBnd‘𝑋) ↔ (𝑀 ∈ (Met‘𝑋) ∧ ∀𝑑 ∈ ℝ+𝑣 ∈ Fin ( 𝑣 = 𝑋 ∧ ∀𝑏𝑣𝑥𝑋 𝑏 = (𝑥(ball‘𝑀)𝑑))))
 
Theoremistotbnd2 34931* The predicate "is a totally bounded metric space." (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝑀 ∈ (Met‘𝑋) → (𝑀 ∈ (TotBnd‘𝑋) ↔ ∀𝑑 ∈ ℝ+𝑣 ∈ Fin ( 𝑣 = 𝑋 ∧ ∀𝑏𝑣𝑥𝑋 𝑏 = (𝑥(ball‘𝑀)𝑑))))
 
Theoremistotbnd3 34932* A metric space is totally bounded iff there is a finite ε-net for every positive ε. This differs from the definition in providing a finite set of ball centers rather than a finite set of balls. (Contributed by Mario Carneiro, 12-Sep-2015.)
(𝑀 ∈ (TotBnd‘𝑋) ↔ (𝑀 ∈ (Met‘𝑋) ∧ ∀𝑑 ∈ ℝ+𝑣 ∈ (𝒫 𝑋 ∩ Fin) 𝑥𝑣 (𝑥(ball‘𝑀)𝑑) = 𝑋))
 
Theoremtotbndmet 34933 The predicate "totally bounded" implies 𝑀 is a metric space. (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝑀 ∈ (TotBnd‘𝑋) → 𝑀 ∈ (Met‘𝑋))
 
Theorem0totbnd 34934 The metric (there is only one) on the empty set is totally bounded. (Contributed by Mario Carneiro, 16-Sep-2015.)
(𝑋 = ∅ → (𝑀 ∈ (TotBnd‘𝑋) ↔ 𝑀 ∈ (Met‘𝑋)))
 
Theoremsstotbnd2 34935* Condition for a subset of a metric space to be totally bounded. (Contributed by Mario Carneiro, 12-Sep-2015.)
𝑁 = (𝑀 ↾ (𝑌 × 𝑌))       ((𝑀 ∈ (Met‘𝑋) ∧ 𝑌𝑋) → (𝑁 ∈ (TotBnd‘𝑌) ↔ ∀𝑑 ∈ ℝ+𝑣 ∈ (𝒫 𝑋 ∩ Fin)𝑌 𝑥𝑣 (𝑥(ball‘𝑀)𝑑)))
 
Theoremsstotbnd 34936* Condition for a subset of a metric space to be totally bounded. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
𝑁 = (𝑀 ↾ (𝑌 × 𝑌))       ((𝑀 ∈ (Met‘𝑋) ∧ 𝑌𝑋) → (𝑁 ∈ (TotBnd‘𝑌) ↔ ∀𝑑 ∈ ℝ+𝑣 ∈ Fin (𝑌 𝑣 ∧ ∀𝑏𝑣𝑥𝑋 𝑏 = (𝑥(ball‘𝑀)𝑑))))
 
Theoremsstotbnd3 34937* Use a net that is not necessarily finite, but for which only finitely many balls meet the subset. (Contributed by Mario Carneiro, 14-Sep-2015.)
𝑁 = (𝑀 ↾ (𝑌 × 𝑌))       ((𝑀 ∈ (Met‘𝑋) ∧ 𝑌𝑋) → (𝑁 ∈ (TotBnd‘𝑌) ↔ ∀𝑑 ∈ ℝ+𝑣 ∈ 𝒫 𝑋(𝑌 𝑥𝑣 (𝑥(ball‘𝑀)𝑑) ∧ {𝑥𝑣 ∣ ((𝑥(ball‘𝑀)𝑑) ∩ 𝑌) ≠ ∅} ∈ Fin)))
 
Theoremtotbndss 34938 A subset of a totally bounded metric space is totally bounded. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Sep-2015.)
((𝑀 ∈ (TotBnd‘𝑋) ∧ 𝑆𝑋) → (𝑀 ↾ (𝑆 × 𝑆)) ∈ (TotBnd‘𝑆))
 
Theoremequivtotbnd 34939* If the metric 𝑀 is "strongly finer" than 𝑁 (meaning that there is a positive real constant 𝑅 such that 𝑁(𝑥, 𝑦) ≤ 𝑅 · 𝑀(𝑥, 𝑦)), then total boundedness of 𝑀 implies total boundedness of 𝑁. (Using this theorem twice in each direction states that if two metrics are strongly equivalent, then one is totally bounded iff the other is.) (Contributed by Mario Carneiro, 14-Sep-2015.)
(𝜑𝑀 ∈ (TotBnd‘𝑋))    &   (𝜑𝑁 ∈ (Met‘𝑋))    &   (𝜑𝑅 ∈ ℝ+)    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → (𝑥𝑁𝑦) ≤ (𝑅 · (𝑥𝑀𝑦)))       (𝜑𝑁 ∈ (TotBnd‘𝑋))
 
Definitiondf-bnd 34940* Define the class of bounded metrics. A metric space is bounded iff it can be covered by a single ball. (Contributed by Jeff Madsen, 2-Sep-2009.)
Bnd = (𝑥 ∈ V ↦ {𝑚 ∈ (Met‘𝑥) ∣ ∀𝑦𝑥𝑟 ∈ ℝ+ 𝑥 = (𝑦(ball‘𝑚)𝑟)})
 
Theoremisbnd 34941* The predicate "is a bounded metric space". (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Sep-2015.)
(𝑀 ∈ (Bnd‘𝑋) ↔ (𝑀 ∈ (Met‘𝑋) ∧ ∀𝑥𝑋𝑟 ∈ ℝ+ 𝑋 = (𝑥(ball‘𝑀)𝑟)))
 
Theorembndmet 34942 A bounded metric space is a metric space. (Contributed by Mario Carneiro, 16-Sep-2015.)
(𝑀 ∈ (Bnd‘𝑋) → 𝑀 ∈ (Met‘𝑋))
 
Theoremisbndx 34943* A "bounded extended metric" (meaning that it satisfies the same condition as a bounded metric, but with "metric" replaced with "extended metric") is a metric and thus is bounded in the conventional sense. (Contributed by Mario Carneiro, 12-Sep-2015.)
(𝑀 ∈ (Bnd‘𝑋) ↔ (𝑀 ∈ (∞Met‘𝑋) ∧ ∀𝑥𝑋𝑟 ∈ ℝ+ 𝑋 = (𝑥(ball‘𝑀)𝑟)))
 
Theoremisbnd2 34944* The predicate "is a bounded metric space". Uses a single point instead of an arbitrary point in the space. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑀 ∈ (Bnd‘𝑋) ∧ 𝑋 ≠ ∅) ↔ (𝑀 ∈ (∞Met‘𝑋) ∧ ∃𝑥𝑋𝑟 ∈ ℝ+ 𝑋 = (𝑥(ball‘𝑀)𝑟)))
 
Theoremisbnd3 34945* A metric space is bounded iff the metric function maps to some bounded real interval. (Contributed by Mario Carneiro, 13-Sep-2015.)
(𝑀 ∈ (Bnd‘𝑋) ↔ (𝑀 ∈ (Met‘𝑋) ∧ ∃𝑥 ∈ ℝ 𝑀:(𝑋 × 𝑋)⟶(0[,]𝑥)))
 
Theoremisbnd3b 34946* A metric space is bounded iff the metric function maps to some bounded real interval. (Contributed by Mario Carneiro, 22-Sep-2015.)
(𝑀 ∈ (Bnd‘𝑋) ↔ (𝑀 ∈ (Met‘𝑋) ∧ ∃𝑥 ∈ ℝ ∀𝑦𝑋𝑧𝑋 (𝑦𝑀𝑧) ≤ 𝑥))
 
Theorembndss 34947 A subset of a bounded metric space is bounded. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑀 ∈ (Bnd‘𝑋) ∧ 𝑆𝑋) → (𝑀 ↾ (𝑆 × 𝑆)) ∈ (Bnd‘𝑆))
 
Theoremblbnd 34948 A ball is bounded. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 15-Jan-2014.)
((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑌𝑋𝑅 ∈ ℝ) → (𝑀 ↾ ((𝑌(ball‘𝑀)𝑅) × (𝑌(ball‘𝑀)𝑅))) ∈ (Bnd‘(𝑌(ball‘𝑀)𝑅)))
 
Theoremssbnd 34949* A subset of a metric space is bounded iff it is contained in a ball around 𝑃, for any 𝑃 in the larger space. (Contributed by Mario Carneiro, 14-Sep-2015.)
𝑁 = (𝑀 ↾ (𝑌 × 𝑌))       ((𝑀 ∈ (Met‘𝑋) ∧ 𝑃𝑋) → (𝑁 ∈ (Bnd‘𝑌) ↔ ∃𝑑 ∈ ℝ 𝑌 ⊆ (𝑃(ball‘𝑀)𝑑)))
 
Theoremtotbndbnd 34950 A totally bounded metric space is bounded. This theorem fails for extended metrics - a bounded extended metric is a metric, but there are totally bounded extended metrics that are not metrics (if we were to weaken istotbnd 34930 to only require that 𝑀 be an extended metric). A counterexample is the discrete extended metric (assigning distinct points distance +∞) on a finite set. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 12-Sep-2015.)
(𝑀 ∈ (TotBnd‘𝑋) → 𝑀 ∈ (Bnd‘𝑋))
 
Theoremequivbnd 34951* If the metric 𝑀 is "strongly finer" than 𝑁 (meaning that there is a positive real constant 𝑅 such that 𝑁(𝑥, 𝑦) ≤ 𝑅 · 𝑀(𝑥, 𝑦)), then boundedness of 𝑀 implies boundedness of 𝑁. (Using this theorem twice in each direction states that if two metrics are strongly equivalent, then one is bounded iff the other is.) (Contributed by Mario Carneiro, 14-Sep-2015.)
(𝜑𝑀 ∈ (Bnd‘𝑋))    &   (𝜑𝑁 ∈ (Met‘𝑋))    &   (𝜑𝑅 ∈ ℝ+)    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → (𝑥𝑁𝑦) ≤ (𝑅 · (𝑥𝑀𝑦)))       (𝜑𝑁 ∈ (Bnd‘𝑋))
 
Theorembnd2lem 34952 Lemma for equivbnd2 34953 and similar theorems. (Contributed by Jeff Madsen, 16-Sep-2015.)
𝐷 = (𝑀 ↾ (𝑌 × 𝑌))       ((𝑀 ∈ (Met‘𝑋) ∧ 𝐷 ∈ (Bnd‘𝑌)) → 𝑌𝑋)
 
Theoremequivbnd2 34953* If balls are totally bounded in the metric 𝑀, then balls are totally bounded in the equivalent metric 𝑁. (Contributed by Mario Carneiro, 15-Sep-2015.)
(𝜑𝑀 ∈ (Met‘𝑋))    &   (𝜑𝑁 ∈ (Met‘𝑋))    &   (𝜑𝑅 ∈ ℝ+)    &   (𝜑𝑆 ∈ ℝ+)    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → (𝑥𝑁𝑦) ≤ (𝑅 · (𝑥𝑀𝑦)))    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → (𝑥𝑀𝑦) ≤ (𝑆 · (𝑥𝑁𝑦)))    &   𝐶 = (𝑀 ↾ (𝑌 × 𝑌))    &   𝐷 = (𝑁 ↾ (𝑌 × 𝑌))    &   (𝜑 → (𝐶 ∈ (TotBnd‘𝑌) ↔ 𝐶 ∈ (Bnd‘𝑌)))       (𝜑 → (𝐷 ∈ (TotBnd‘𝑌) ↔ 𝐷 ∈ (Bnd‘𝑌)))
 
Theoremprdsbnd 34954* The product metric over finite index set is bounded if all the factors are bounded. (Contributed by Mario Carneiro, 13-Sep-2015.)
𝑌 = (𝑆Xs𝑅)    &   𝐵 = (Base‘𝑌)    &   𝑉 = (Base‘(𝑅𝑥))    &   𝐸 = ((dist‘(𝑅𝑥)) ↾ (𝑉 × 𝑉))    &   𝐷 = (dist‘𝑌)    &   (𝜑𝑆𝑊)    &   (𝜑𝐼 ∈ Fin)    &   (𝜑𝑅 Fn 𝐼)    &   ((𝜑𝑥𝐼) → 𝐸 ∈ (Bnd‘𝑉))       (𝜑𝐷 ∈ (Bnd‘𝐵))
 
Theoremprdstotbnd 34955* The product metric over finite index set is totally bounded if all the factors are totally bounded. (Contributed by Mario Carneiro, 20-Sep-2015.)
𝑌 = (𝑆Xs𝑅)    &   𝐵 = (Base‘𝑌)    &   𝑉 = (Base‘(𝑅𝑥))    &   𝐸 = ((dist‘(𝑅𝑥)) ↾ (𝑉 × 𝑉))    &   𝐷 = (dist‘𝑌)    &   (𝜑𝑆𝑊)    &   (𝜑𝐼 ∈ Fin)    &   (𝜑𝑅 Fn 𝐼)    &   ((𝜑𝑥𝐼) → 𝐸 ∈ (TotBnd‘𝑉))       (𝜑𝐷 ∈ (TotBnd‘𝐵))
 
Theoremprdsbnd2 34956* If balls are totally bounded in each factor, then balls are bounded in a metric product. (Contributed by Mario Carneiro, 16-Sep-2015.)
𝑌 = (𝑆Xs𝑅)    &   𝐵 = (Base‘𝑌)    &   𝑉 = (Base‘(𝑅𝑥))    &   𝐸 = ((dist‘(𝑅𝑥)) ↾ (𝑉 × 𝑉))    &   𝐷 = (dist‘𝑌)    &   (𝜑𝑆𝑊)    &   (𝜑𝐼 ∈ Fin)    &   (𝜑𝑅 Fn 𝐼)    &   𝐶 = (𝐷 ↾ (𝐴 × 𝐴))    &   ((𝜑𝑥𝐼) → 𝐸 ∈ (Met‘𝑉))    &   ((𝜑𝑥𝐼) → ((𝐸 ↾ (𝑦 × 𝑦)) ∈ (TotBnd‘𝑦) ↔ (𝐸 ↾ (𝑦 × 𝑦)) ∈ (Bnd‘𝑦)))       (𝜑 → (𝐶 ∈ (TotBnd‘𝐴) ↔ 𝐶 ∈ (Bnd‘𝐴)))
 
Theoremcntotbnd 34957 A subset of the complex numbers is totally bounded iff it is bounded. (Contributed by Mario Carneiro, 14-Sep-2015.)
𝐷 = ((abs ∘ − ) ↾ (𝑋 × 𝑋))       (𝐷 ∈ (TotBnd‘𝑋) ↔ 𝐷 ∈ (Bnd‘𝑋))
 
Theoremcnpwstotbnd 34958 A subset of 𝐴𝐼, where 𝐴 ⊆ ℂ, is totally bounded iff it is bounded. (Contributed by Mario Carneiro, 14-Sep-2015.)
𝑌 = ((ℂflds 𝐴) ↑s 𝐼)    &   𝐷 = ((dist‘𝑌) ↾ (𝑋 × 𝑋))       ((𝐴 ⊆ ℂ ∧ 𝐼 ∈ Fin) → (𝐷 ∈ (TotBnd‘𝑋) ↔ 𝐷 ∈ (Bnd‘𝑋)))
 
20.20.8  Isometries
 
Syntaxcismty 34959 Extend class notation with the class of metric space isometries.
class Ismty
 
Definitiondf-ismty 34960* Define a function which takes two metric spaces and returns the set of isometries between the spaces. An isometry is a bijection which preserves distance. (Contributed by Jeff Madsen, 2-Sep-2009.)
Ismty = (𝑚 ran ∞Met, 𝑛 ran ∞Met ↦ {𝑓 ∣ (𝑓:dom dom 𝑚1-1-onto→dom dom 𝑛 ∧ ∀𝑥 ∈ dom dom 𝑚𝑦 ∈ dom dom 𝑚(𝑥𝑚𝑦) = ((𝑓𝑥)𝑛(𝑓𝑦)))})
 
Theoremismtyval 34961* The set of isometries between two metric spaces. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌)) → (𝑀 Ismty 𝑁) = {𝑓 ∣ (𝑓:𝑋1-1-onto𝑌 ∧ ∀𝑥𝑋𝑦𝑋 (𝑥𝑀𝑦) = ((𝑓𝑥)𝑁(𝑓𝑦)))})
 
Theoremisismty 34962* The condition "is an isometry". (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌)) → (𝐹 ∈ (𝑀 Ismty 𝑁) ↔ (𝐹:𝑋1-1-onto𝑌 ∧ ∀𝑥𝑋𝑦𝑋 (𝑥𝑀𝑦) = ((𝐹𝑥)𝑁(𝐹𝑦)))))
 
Theoremismtycnv 34963 The inverse of an isometry is an isometry. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌)) → (𝐹 ∈ (𝑀 Ismty 𝑁) → 𝐹 ∈ (𝑁 Ismty 𝑀)))
 
Theoremismtyima 34964 The image of a ball under an isometry is another ball. (Contributed by Jeff Madsen, 31-Jan-2014.)
(((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌) ∧ 𝐹 ∈ (𝑀 Ismty 𝑁)) ∧ (𝑃𝑋𝑅 ∈ ℝ*)) → (𝐹 “ (𝑃(ball‘𝑀)𝑅)) = ((𝐹𝑃)(ball‘𝑁)𝑅))
 
Theoremismtyhmeolem 34965 Lemma for ismtyhmeo 34966. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Sep-2015.)
𝐽 = (MetOpen‘𝑀)    &   𝐾 = (MetOpen‘𝑁)    &   (𝜑𝑀 ∈ (∞Met‘𝑋))    &   (𝜑𝑁 ∈ (∞Met‘𝑌))    &   (𝜑𝐹 ∈ (𝑀 Ismty 𝑁))       (𝜑𝐹 ∈ (𝐽 Cn 𝐾))
 
Theoremismtyhmeo 34966 An isometry is a homeomorphism on the induced topology. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Sep-2015.)
𝐽 = (MetOpen‘𝑀)    &   𝐾 = (MetOpen‘𝑁)       ((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌)) → (𝑀 Ismty 𝑁) ⊆ (𝐽Homeo𝐾))
 
Theoremismtybndlem 34967 Lemma for ismtybnd 34968. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 19-Jan-2014.)
((𝑁 ∈ (∞Met‘𝑌) ∧ 𝐹 ∈ (𝑀 Ismty 𝑁)) → (𝑀 ∈ (Bnd‘𝑋) → 𝑁 ∈ (Bnd‘𝑌)))
 
Theoremismtybnd 34968 Isometries preserve boundedness. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 19-Jan-2014.)
((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌) ∧ 𝐹 ∈ (𝑀 Ismty 𝑁)) → (𝑀 ∈ (Bnd‘𝑋) ↔ 𝑁 ∈ (Bnd‘𝑌)))
 
Theoremismtyres 34969 A restriction of an isometry is an isometry. The condition 𝐴𝑋 is not necessary but makes the proof easier. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 12-Sep-2015.)
𝐵 = (𝐹𝐴)    &   𝑆 = (𝑀 ↾ (𝐴 × 𝐴))    &   𝑇 = (𝑁 ↾ (𝐵 × 𝐵))       (((𝑀 ∈ (∞Met‘𝑋) ∧ 𝑁 ∈ (∞Met‘𝑌)) ∧ (𝐹 ∈ (𝑀 Ismty 𝑁) ∧ 𝐴𝑋)) → (𝐹𝐴) ∈ (𝑆 Ismty 𝑇))
 
20.20.9  Heine-Borel Theorem
 
Theoremheibor1lem 34970 Lemma for heibor1 34971. A compact metric space is complete. This proof works by considering the collection cls(𝐹 “ (ℤ𝑛)) for each 𝑛 ∈ ℕ, which has the finite intersection property because any finite intersection of upper integer sets is another upper integer set, so any finite intersection of the image closures will contain (𝐹 “ (ℤ𝑚)) for some 𝑚. Thus, by compactness, the intersection contains a point 𝑦, which must then be the convergent point of 𝐹. (Contributed by Jeff Madsen, 17-Jan-2014.) (Revised by Mario Carneiro, 5-Jun-2014.)
𝐽 = (MetOpen‘𝐷)    &   (𝜑𝐷 ∈ (Met‘𝑋))    &   (𝜑𝐽 ∈ Comp)    &   (𝜑𝐹 ∈ (Cau‘𝐷))    &   (𝜑𝐹:ℕ⟶𝑋)       (𝜑𝐹 ∈ dom (⇝𝑡𝐽))
 
Theoremheibor1 34971 One half of heibor 34982, that does not require any Choice. A compact metric space is complete and totally bounded. We prove completeness in cmpcmet 23851 and total boundedness here, which follows trivially from the fact that the set of all 𝑟-balls is an open cover of 𝑋, so finitely many cover 𝑋. (Contributed by Jeff Madsen, 16-Jan-2014.)
𝐽 = (MetOpen‘𝐷)       ((𝐷 ∈ (Met‘𝑋) ∧ 𝐽 ∈ Comp) → (𝐷 ∈ (CMet‘𝑋) ∧ 𝐷 ∈ (TotBnd‘𝑋)))
 
Theoremheiborlem1 34972* Lemma for heibor 34982. We work with a fixed open cover 𝑈 throughout. The set 𝐾 is the set of all subsets of 𝑋 that admit no finite subcover of 𝑈. (We wish to prove that 𝐾 is empty.) If a set 𝐶 has no finite subcover, then any finite cover of 𝐶 must contain a set that also has no finite subcover. (Contributed by Jeff Madsen, 23-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐵 ∈ V       ((𝐴 ∈ Fin ∧ 𝐶 𝑥𝐴 𝐵𝐶𝐾) → ∃𝑥𝐴 𝐵𝐾)
 
Theoremheiborlem2 34973* Lemma for heibor 34982. Substitutions for the set 𝐺. (Contributed by Jeff Madsen, 23-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐴 ∈ V    &   𝐶 ∈ V       (𝐴𝐺𝐶 ↔ (𝐶 ∈ ℕ0𝐴 ∈ (𝐹𝐶) ∧ (𝐴𝐵𝐶) ∈ 𝐾))
 
Theoremheiborlem3 34974* Lemma for heibor 34982. Using countable choice ax-cc 9846, we have fixed in advance a collection of finite 2↑-𝑛 nets (𝐹𝑛) for 𝑋 (note that an 𝑟-net is a set of points in 𝑋 whose 𝑟 -balls cover 𝑋). The set 𝐺 is the subset of these points whose corresponding balls have no finite subcover (i.e. in the set 𝐾). If the theorem was false, then 𝑋 would be in 𝐾, and so some ball at each level would also be in 𝐾. But we can say more than this; given a ball (𝑦𝐵𝑛) on level 𝑛, since level 𝑛 + 1 covers the space and thus also (𝑦𝐵𝑛), using heiborlem1 34972 there is a ball on the next level whose intersection with (𝑦𝐵𝑛) also has no finite subcover. Now since the set 𝐺 is a countable union of finite sets, it is countable (which needs ax-cc 9846 via iunctb 9985), and so we can apply ax-cc 9846 to 𝐺 directly to get a function from 𝐺 to itself, which points from each ball in 𝐾 to a ball on the next level in 𝐾, and such that the intersection between these balls is also in 𝐾. (Contributed by Jeff Madsen, 18-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))       (𝜑 → ∃𝑔𝑥𝐺 ((𝑔𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑔𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))
 
Theoremheiborlem4 34975* Lemma for heibor 34982. Using the function 𝑇 constructed in heiborlem3 34974, construct an infinite path in 𝐺. (Contributed by Jeff Madsen, 23-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))    &   (𝜑 → ∀𝑥𝐺 ((𝑇𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑇𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))    &   (𝜑𝐶𝐺0)    &   𝑆 = seq0(𝑇, (𝑚 ∈ ℕ0 ↦ if(𝑚 = 0, 𝐶, (𝑚 − 1))))       ((𝜑𝐴 ∈ ℕ0) → (𝑆𝐴)𝐺𝐴)
 
Theoremheiborlem5 34976* Lemma for heibor 34982. The function 𝑀 is a set of point-and-radius pairs suitable for application to caubl 23840. (Contributed by Jeff Madsen, 23-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))    &   (𝜑 → ∀𝑥𝐺 ((𝑇𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑇𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))    &   (𝜑𝐶𝐺0)    &   𝑆 = seq0(𝑇, (𝑚 ∈ ℕ0 ↦ if(𝑚 = 0, 𝐶, (𝑚 − 1))))    &   𝑀 = (𝑛 ∈ ℕ ↦ ⟨(𝑆𝑛), (3 / (2↑𝑛))⟩)       (𝜑𝑀:ℕ⟶(𝑋 × ℝ+))
 
Theoremheiborlem6 34977* Lemma for heibor 34982. Since the sequence of balls connected by the function 𝑇 ensures that each ball nontrivially intersects with the next (since the empty set has a finite subcover, the intersection of any two successive balls in the sequence is nonempty), and each ball is half the size of the previous one, the distance between the centers is at most 3 / 2 times the size of the larger, and so if we expand each ball by a factor of 3 we get a nested sequence of balls. (Contributed by Jeff Madsen, 23-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))    &   (𝜑 → ∀𝑥𝐺 ((𝑇𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑇𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))    &   (𝜑𝐶𝐺0)    &   𝑆 = seq0(𝑇, (𝑚 ∈ ℕ0 ↦ if(𝑚 = 0, 𝐶, (𝑚 − 1))))    &   𝑀 = (𝑛 ∈ ℕ ↦ ⟨(𝑆𝑛), (3 / (2↑𝑛))⟩)       (𝜑 → ∀𝑘 ∈ ℕ ((ball‘𝐷)‘(𝑀‘(𝑘 + 1))) ⊆ ((ball‘𝐷)‘(𝑀𝑘)))
 
Theoremheiborlem7 34978* Lemma for heibor 34982. Since the sizes of the balls decrease exponentially, the sequence converges to zero. (Contributed by Jeff Madsen, 23-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))    &   (𝜑 → ∀𝑥𝐺 ((𝑇𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑇𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))    &   (𝜑𝐶𝐺0)    &   𝑆 = seq0(𝑇, (𝑚 ∈ ℕ0 ↦ if(𝑚 = 0, 𝐶, (𝑚 − 1))))    &   𝑀 = (𝑛 ∈ ℕ ↦ ⟨(𝑆𝑛), (3 / (2↑𝑛))⟩)       𝑟 ∈ ℝ+𝑘 ∈ ℕ (2nd ‘(𝑀𝑘)) < 𝑟
 
Theoremheiborlem8 34979* Lemma for heibor 34982. The previous lemmas establish that the sequence 𝑀 is Cauchy, so using completeness we now consider the convergent point 𝑌. By assumption, 𝑈 is an open cover, so 𝑌 is an element of some 𝑍𝑈, and some ball centered at 𝑌 is contained in 𝑍. But the sequence contains arbitrarily small balls close to 𝑌, so some element ball(𝑀𝑛) of the sequence is contained in 𝑍. And finally we arrive at a contradiction, because {𝑍} is a finite subcover of 𝑈 that covers ball(𝑀𝑛), yet ball(𝑀𝑛) ∈ 𝐾. For convenience, we write this contradiction as 𝜑𝜓 where 𝜑 is all the accumulated hypotheses and 𝜓 is anything at all. (Contributed by Jeff Madsen, 22-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))    &   (𝜑 → ∀𝑥𝐺 ((𝑇𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑇𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))    &   (𝜑𝐶𝐺0)    &   𝑆 = seq0(𝑇, (𝑚 ∈ ℕ0 ↦ if(𝑚 = 0, 𝐶, (𝑚 − 1))))    &   𝑀 = (𝑛 ∈ ℕ ↦ ⟨(𝑆𝑛), (3 / (2↑𝑛))⟩)    &   (𝜑𝑈𝐽)    &   𝑌 ∈ V    &   (𝜑𝑌𝑍)    &   (𝜑𝑍𝑈)    &   (𝜑 → (1st𝑀)(⇝𝑡𝐽)𝑌)       (𝜑𝜓)
 
Theoremheiborlem9 34980* Lemma for heibor 34982. Discharge the hypotheses of heiborlem8 34979 by applying caubl 23840 to get a convergent point and adding the open cover assumption. (Contributed by Jeff Madsen, 20-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))    &   (𝜑 → ∀𝑥𝐺 ((𝑇𝑥)𝐺((2nd𝑥) + 1) ∧ ((𝐵𝑥) ∩ ((𝑇𝑥)𝐵((2nd𝑥) + 1))) ∈ 𝐾))    &   (𝜑𝐶𝐺0)    &   𝑆 = seq0(𝑇, (𝑚 ∈ ℕ0 ↦ if(𝑚 = 0, 𝐶, (𝑚 − 1))))    &   𝑀 = (𝑛 ∈ ℕ ↦ ⟨(𝑆𝑛), (3 / (2↑𝑛))⟩)    &   (𝜑𝑈𝐽)    &   (𝜑 𝑈 = 𝑋)       (𝜑𝜓)
 
Theoremheiborlem10 34981* Lemma for heibor 34982. The last remaining piece of the proof is to find an element 𝐶 such that 𝐶𝐺0, i.e. 𝐶 is an element of (𝐹‘0) that has no finite subcover, which is true by heiborlem1 34972, since (𝐹‘0) is a finite cover of 𝑋, which has no finite subcover. Thus, the rest of the proof follows to a contradiction, and thus there must be a finite subcover of 𝑈 that covers 𝑋, i.e. 𝑋 is compact. (Contributed by Jeff Madsen, 22-Jan-2014.)
𝐽 = (MetOpen‘𝐷)    &   𝐾 = {𝑢 ∣ ¬ ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin)𝑢 𝑣}    &   𝐺 = {⟨𝑦, 𝑛⟩ ∣ (𝑛 ∈ ℕ0𝑦 ∈ (𝐹𝑛) ∧ (𝑦𝐵𝑛) ∈ 𝐾)}    &   𝐵 = (𝑧𝑋, 𝑚 ∈ ℕ0 ↦ (𝑧(ball‘𝐷)(1 / (2↑𝑚))))    &   (𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝐹:ℕ0⟶(𝒫 𝑋 ∩ Fin))    &   (𝜑 → ∀𝑛 ∈ ℕ0 𝑋 = 𝑦 ∈ (𝐹𝑛)(𝑦𝐵𝑛))       ((𝜑 ∧ (𝑈𝐽 𝐽 = 𝑈)) → ∃𝑣 ∈ (𝒫 𝑈 ∩ Fin) 𝐽 = 𝑣)
 
Theoremheibor 34982 Generalized Heine-Borel Theorem. A metric space is compact iff it is complete and totally bounded. See heibor1 34971 and heiborlem1 34972 for a description of the proof. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 28-Jan-2014.)
𝐽 = (MetOpen‘𝐷)       ((𝐷 ∈ (Met‘𝑋) ∧ 𝐽 ∈ Comp) ↔ (𝐷 ∈ (CMet‘𝑋) ∧ 𝐷 ∈ (TotBnd‘𝑋)))
 
20.20.10  Banach Fixed Point Theorem
 
Theorembfplem1 34983* Lemma for bfp 34985. The sequence 𝐺, which simply starts from any point in the space and iterates 𝐹, satisfies the property that the distance from 𝐺(𝑛) to 𝐺(𝑛 + 1) decreases by at least 𝐾 after each step. Thus, the total distance from any 𝐺(𝑖) to 𝐺(𝑗) is bounded by a geometric series, and the sequence is Cauchy. Therefore, it converges to a point ((⇝𝑡𝐽)‘𝐺) since the space is complete. (Contributed by Jeff Madsen, 17-Jun-2014.)
(𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝑋 ≠ ∅)    &   (𝜑𝐾 ∈ ℝ+)    &   (𝜑𝐾 < 1)    &   (𝜑𝐹:𝑋𝑋)    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → ((𝐹𝑥)𝐷(𝐹𝑦)) ≤ (𝐾 · (𝑥𝐷𝑦)))    &   𝐽 = (MetOpen‘𝐷)    &   (𝜑𝐴𝑋)    &   𝐺 = seq1((𝐹 ∘ 1st ), (ℕ × {𝐴}))       (𝜑𝐺(⇝𝑡𝐽)((⇝𝑡𝐽)‘𝐺))
 
Theorembfplem2 34984* Lemma for bfp 34985. Using the point found in bfplem1 34983, we show that this convergent point is a fixed point of 𝐹. Since for any positive 𝑥, the sequence 𝐺 is in 𝐵(𝑥 / 2, 𝑃) for all 𝑘 ∈ (ℤ𝑗) (where 𝑃 = ((⇝𝑡𝐽)‘𝐺)), we have 𝐷(𝐺(𝑗 + 1), 𝐹(𝑃)) ≤ 𝐷(𝐺(𝑗), 𝑃) < 𝑥 / 2 and 𝐷(𝐺(𝑗 + 1), 𝑃) < 𝑥 / 2, so 𝐹(𝑃) is in every neighborhood of 𝑃 and 𝑃 is a fixed point of 𝐹. (Contributed by Jeff Madsen, 5-Jun-2014.)
(𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝑋 ≠ ∅)    &   (𝜑𝐾 ∈ ℝ+)    &   (𝜑𝐾 < 1)    &   (𝜑𝐹:𝑋𝑋)    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → ((𝐹𝑥)𝐷(𝐹𝑦)) ≤ (𝐾 · (𝑥𝐷𝑦)))    &   𝐽 = (MetOpen‘𝐷)    &   (𝜑𝐴𝑋)    &   𝐺 = seq1((𝐹 ∘ 1st ), (ℕ × {𝐴}))       (𝜑 → ∃𝑧𝑋 (𝐹𝑧) = 𝑧)
 
Theorembfp 34985* Banach fixed point theorem, also known as contraction mapping theorem. A contraction on a complete metric space has a unique fixed point. We show existence in the lemmas, and uniqueness here - if 𝐹 has two fixed points, then the distance between them is less than 𝐾 times itself, a contradiction. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 5-Jun-2014.)
(𝜑𝐷 ∈ (CMet‘𝑋))    &   (𝜑𝑋 ≠ ∅)    &   (𝜑𝐾 ∈ ℝ+)    &   (𝜑𝐾 < 1)    &   (𝜑𝐹:𝑋𝑋)    &   ((𝜑 ∧ (𝑥𝑋𝑦𝑋)) → ((𝐹𝑥)𝐷(𝐹𝑦)) ≤ (𝐾 · (𝑥𝐷𝑦)))       (𝜑 → ∃!𝑧𝑋 (𝐹𝑧) = 𝑧)
 
20.20.11  Euclidean space
 
Syntaxcrrn 34986 Extend class notation with the n-dimensional Euclidean space.
class n
 
Definitiondf-rrn 34987* Define n-dimensional Euclidean space as a metric space with the standard Euclidean norm given by the quadratic mean. (Contributed by Jeff Madsen, 2-Sep-2009.)
n = (𝑖 ∈ Fin ↦ (𝑥 ∈ (ℝ ↑m 𝑖), 𝑦 ∈ (ℝ ↑m 𝑖) ↦ (√‘Σ𝑘𝑖 (((𝑥𝑘) − (𝑦𝑘))↑2))))
 
Theoremrrnval 34988* The n-dimensional Euclidean space. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 13-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)       (𝐼 ∈ Fin → (ℝn𝐼) = (𝑥𝑋, 𝑦𝑋 ↦ (√‘Σ𝑘𝐼 (((𝑥𝑘) − (𝑦𝑘))↑2))))
 
Theoremrrnmval 34989* The value of the Euclidean metric. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 13-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)       ((𝐼 ∈ Fin ∧ 𝐹𝑋𝐺𝑋) → (𝐹(ℝn𝐼)𝐺) = (√‘Σ𝑘𝐼 (((𝐹𝑘) − (𝐺𝑘))↑2)))
 
Theoremrrnmet 34990 Euclidean space is a metric space. (Contributed by Jeff Madsen, 2-Sep-2009.) (Proof shortened by Mario Carneiro, 5-Jun-2014.)
𝑋 = (ℝ ↑m 𝐼)       (𝐼 ∈ Fin → (ℝn𝐼) ∈ (Met‘𝑋))
 
Theoremrrndstprj1 34991 The distance between two points in Euclidean space is greater than the distance between the projections onto one coordinate. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 13-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)    &   𝑀 = ((abs ∘ − ) ↾ (ℝ × ℝ))       (((𝐼 ∈ Fin ∧ 𝐴𝐼) ∧ (𝐹𝑋𝐺𝑋)) → ((𝐹𝐴)𝑀(𝐺𝐴)) ≤ (𝐹(ℝn𝐼)𝐺))
 
Theoremrrndstprj2 34992* Bound on the distance between two points in Euclidean space given bounds on the distances in each coordinate. This theorem and rrndstprj1 34991 can be used to show that the supremum norm and Euclidean norm are equivalent. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 13-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)    &   𝑀 = ((abs ∘ − ) ↾ (ℝ × ℝ))       (((𝐼 ∈ (Fin ∖ {∅}) ∧ 𝐹𝑋𝐺𝑋) ∧ (𝑅 ∈ ℝ+ ∧ ∀𝑛𝐼 ((𝐹𝑛)𝑀(𝐺𝑛)) < 𝑅)) → (𝐹(ℝn𝐼)𝐺) < (𝑅 · (√‘(♯‘𝐼))))
 
Theoremrrncmslem 34993* Lemma for rrncms 34994. (Contributed by Jeff Madsen, 6-Jun-2014.) (Revised by Mario Carneiro, 13-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)    &   𝑀 = ((abs ∘ − ) ↾ (ℝ × ℝ))    &   𝐽 = (MetOpen‘(ℝn𝐼))    &   (𝜑𝐼 ∈ Fin)    &   (𝜑𝐹 ∈ (Cau‘(ℝn𝐼)))    &   (𝜑𝐹:ℕ⟶𝑋)    &   𝑃 = (𝑚𝐼 ↦ ( ⇝ ‘(𝑡 ∈ ℕ ↦ ((𝐹𝑡)‘𝑚))))       (𝜑𝐹 ∈ dom (⇝𝑡𝐽))
 
Theoremrrncms 34994 Euclidean space is complete. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 13-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)       (𝐼 ∈ Fin → (ℝn𝐼) ∈ (CMet‘𝑋))
 
Theoremrepwsmet 34995 The supremum metric on ℝ↑𝐼 is a metric. (Contributed by Jeff Madsen, 15-Sep-2015.)
𝑌 = ((ℂflds ℝ) ↑s 𝐼)    &   𝐷 = (dist‘𝑌)    &   𝑋 = (ℝ ↑m 𝐼)       (𝐼 ∈ Fin → 𝐷 ∈ (Met‘𝑋))
 
Theoremrrnequiv 34996 The supremum metric on ℝ↑𝐼 is equivalent to the n metric. (Contributed by Jeff Madsen, 15-Sep-2015.)
𝑌 = ((ℂflds ℝ) ↑s 𝐼)    &   𝐷 = (dist‘𝑌)    &   𝑋 = (ℝ ↑m 𝐼)    &   (𝜑𝐼 ∈ Fin)       ((𝜑 ∧ (𝐹𝑋𝐺𝑋)) → ((𝐹𝐷𝐺) ≤ (𝐹(ℝn𝐼)𝐺) ∧ (𝐹(ℝn𝐼)𝐺) ≤ ((√‘(♯‘𝐼)) · (𝐹𝐷𝐺))))
 
Theoremrrntotbnd 34997 A set in Euclidean space is totally bounded iff its is bounded. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 16-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)    &   𝑀 = ((ℝn𝐼) ↾ (𝑌 × 𝑌))       (𝐼 ∈ Fin → (𝑀 ∈ (TotBnd‘𝑌) ↔ 𝑀 ∈ (Bnd‘𝑌)))
 
Theoremrrnheibor 34998 Heine-Borel theorem for Euclidean space. A subset of Euclidean space is compact iff it is closed and bounded. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 22-Sep-2015.)
𝑋 = (ℝ ↑m 𝐼)    &   𝑀 = ((ℝn𝐼) ↾ (𝑌 × 𝑌))    &   𝑇 = (MetOpen‘𝑀)    &   𝑈 = (MetOpen‘(ℝn𝐼))       ((𝐼 ∈ Fin ∧ 𝑌𝑋) → (𝑇 ∈ Comp ↔ (𝑌 ∈ (Clsd‘𝑈) ∧ 𝑀 ∈ (Bnd‘𝑌))))
 
20.20.12  Intervals (continued)
 
Theoremismrer1 34999* An isometry between and ℝ↑1. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 22-Sep-2015.)
𝑅 = ((abs ∘ − ) ↾ (ℝ × ℝ))    &   𝐹 = (𝑥 ∈ ℝ ↦ ({𝐴} × {𝑥}))       (𝐴𝑉𝐹 ∈ (𝑅 Ismty (ℝn‘{𝐴})))
 
Theoremreheibor 35000 Heine-Borel theorem for real numbers. A subset of is compact iff it is closed and bounded. (Contributed by Jeff Madsen, 2-Sep-2009.) (Revised by Mario Carneiro, 22-Sep-2015.)
𝑀 = ((abs ∘ − ) ↾ (𝑌 × 𝑌))    &   𝑇 = (MetOpen‘𝑀)    &   𝑈 = (topGen‘ran (,))       (𝑌 ⊆ ℝ → (𝑇 ∈ Comp ↔ (𝑌 ∈ (Clsd‘𝑈) ∧ 𝑀 ∈ (Bnd‘𝑌))))
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144 14301-14400 145 14401-14500 146 14501-14600 147 14601-14700 148 14701-14800 149 14801-14900 150 14901-15000 151 15001-15100 152 15101-15200 153 15201-15300 154 15301-15400 155 15401-15500 156 15501-15600 157 15601-15700 158 15701-15800 159 15801-15900 160 15901-16000 161 16001-16100 162 16101-16200 163 16201-16300 164 16301-16400 165 16401-16500 166 16501-16600 167 16601-16700 168 16701-16800 169 16801-16900 170 16901-17000 171 17001-17100 172 17101-17200 173 17201-17300 174 17301-17400 175 17401-17500 176 17501-17600 177 17601-17700 178 17701-17800 179 17801-17900 180 17901-18000 181 18001-18100 182 18101-18200 183 18201-18300 184 18301-18400 185 18401-18500 186 18501-18600 187 18601-18700 188 18701-18800 189 18801-18900 190 18901-19000 191 19001-19100 192 19101-19200 193 19201-19300 194 19301-19400 195 19401-19500 196 19501-19600 197 19601-19700 198 19701-19800 199 19801-19900 200 19901-20000 201 20001-20100 202 20101-20200 203 20201-20300 204 20301-20400 205 20401-20500 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268 26701-26800 269 26801-26900 270 26901-27000 271 27001-27100 272 27101-27200 273 27201-27300 274 27301-27400 275 27401-27500 276 27501-27600 277 27601-27700 278 27701-27800 279 27801-27900 280 27901-28000 281 28001-28100 282 28101-28200 283 28201-28300 284 28301-28400 285 28401-28500 286 28501-28600 287 28601-28700 288 28701-28800 289 28801-28900 290 28901-29000 291 29001-29100 292 29101-29200 293 29201-29300 294 29301-29400 295 29401-29500 296 29501-29600 297 29601-29700 298 29701-29800 299 29801-29900 300 29901-30000 301 30001-30100 302 30101-30200 303 30201-30300 304 30301-30400 305 30401-30500 306 30501-30600 307 30601-30700 308 30701-30800 309 30801-30900 310 30901-31000 311 31001-31100 312 31101-31200 313 31201-31300 314 31301-31400 315 31401-31500 316 31501-31600 317 31601-31700 318 31701-31800 319 31801-31900 320 31901-32000 321 32001-32100 322 32101-32200 323 32201-32300 324 32301-32400 325 32401-32500 326 32501-32600 327 32601-32700 328 32701-32800 329 32801-32900 330 32901-33000 331 33001-33100 332 33101-33200 333 33201-33300 334 33301-33400 335 33401-33500 336 33501-33600 337 33601-33700 338 33701-33800 339 33801-33900 340 33901-34000 341 34001-34100 342 34101-34200 343 34201-34300 344 34301-34400 345 34401-34500 346 34501-34600 347 34601-34700 348 34701-34800 349 34801-34900 350 34901-35000 351 35001-35100 352 35101-35200 353 35201-35300 354 35301-35400 355 35401-35500 356 35501-35600 357 35601-35700 358 35701-35800 359 35801-35900 360 35901-36000 361 36001-36100 362 36101-36200 363 36201-36300 364 36301-36400 365 36401-36500 366 36501-36600 367 36601-36700 368 36701-36800 369 36801-36900 370 36901-37000 371 37001-37100 372 37101-37200 373 37201-37300 374 37301-37400 375 37401-37500 376 37501-37600 377 37601-37700 378 37701-37800 379 37801-37900 380 37901-38000 381 38001-38100 382 38101-38200 383 38201-38300 384 38301-38400 385 38401-38500 386 38501-38600 387 38601-38700 388 38701-38800 389 38801-38900 390 38901-39000 391 39001-39100 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