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Theorem List for Metamath Proof Explorer - 35901-36000   *Has distinct variable group(s)
TypeLabelDescription
Statement
 
Theoremindexdom 35901* If for every element of an indexing set 𝐴 there exists a corresponding element of another set 𝐵, then there exists a subset of 𝐵 consisting only of those elements which are indexed by 𝐴, and which is dominated by the set 𝐴. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝐴𝑀 ∧ ∀𝑥𝐴𝑦𝐵 𝜑) → ∃𝑐((𝑐𝐴𝑐𝐵) ∧ (∀𝑥𝐴𝑦𝑐 𝜑 ∧ ∀𝑦𝑐𝑥𝐴 𝜑)))
 
Theoremfrinfm 35902* A subset of a well-founded set has an infimum. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑅 Fr 𝐴 ∧ (𝐵𝐶𝐵𝐴𝐵 ≠ ∅)) → ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧)))
 
Theoremwelb 35903* A nonempty subset of a well-ordered set has a lower bound. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑅 We 𝐴 ∧ (𝐵𝐶𝐵𝐴𝐵 ≠ ∅)) → (𝑅 Or 𝐵 ∧ ∃𝑥𝐵 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐵 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))))
 
Theoremsupex2g 35904 Existence of supremum. (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝐴𝐶 → sup(𝐵, 𝐴, 𝑅) ∈ V)
 
Theoremsupclt 35905* Closure of supremum. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑅 Or 𝐴 ∧ ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))) → sup(𝐵, 𝐴, 𝑅) ∈ 𝐴)
 
Theoremsupubt 35906* Upper bound property of supremum. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝑅 Or 𝐴 ∧ ∃𝑥𝐴 (∀𝑦𝐵 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦𝐴 (𝑦𝑅𝑥 → ∃𝑧𝐵 𝑦𝑅𝑧))) → (𝐶𝐵 → ¬ sup(𝐵, 𝐴, 𝑅)𝑅𝐶))
 
20.20.2  Real and complex numbers; integers
 
Theoremfilbcmb 35907* Combine a finite set of lower bounds. (Contributed by Jeff Madsen, 2-Sep-2009.)
((𝐴 ∈ Fin ∧ 𝐴 ≠ ∅ ∧ 𝐵 ⊆ ℝ) → (∀𝑥𝐴𝑦𝐵𝑧𝐵 (𝑦𝑧𝜑) → ∃𝑦𝐵𝑧𝐵 (𝑦𝑧 → ∀𝑥𝐴 𝜑)))
 
Theoremfzmul 35908 Membership of a product in a finite interval of integers. (Contributed by Jeff Madsen, 17-Jun-2010.)
((𝑀 ∈ ℤ ∧ 𝑁 ∈ ℤ ∧ 𝐾 ∈ ℕ) → (𝐽 ∈ (𝑀...𝑁) → (𝐾 · 𝐽) ∈ ((𝐾 · 𝑀)...(𝐾 · 𝑁))))
 
20.20.3  Sequences and sums
 
Theoremsdclem2 35909* Lemma for sdc 35911. (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)})    &   𝑘𝜑    &   (𝜑𝐺:𝑍𝐽)    &   (𝜑 → (𝐺𝑀):(𝑀...𝑀)⟶𝐴)    &   ((𝜑𝑤𝑍) → (𝐺‘(𝑤 + 1)) ∈ (𝑤𝐹(𝐺𝑤)))       (𝜑 → ∃𝑓(𝑓:𝑍𝐴 ∧ ∀𝑛𝑍 𝜒))
 
Theoremsdclem1 35910* Lemma for sdc 35911. (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 35911* 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 35912* 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 35913* Variant of fdc 35912 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 35914* 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 35915* 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 35916* 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 35917* 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 35918* 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 35919 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 35920 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 35921 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 35922* 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 35923 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 35924* 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 35925* 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 35926* 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 35927 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 35928* 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 35929* A constant function is a continuous function on . (Contributed by Jeff Madsen, 2-Sep-2009.) (Moved into main set.mm as cncfmptc 24084 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))
 
Theoremcnres2 35930* 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 35931 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 35932* 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 35933 Extend class notation with the class of totally bounded metric spaces.
class TotBnd
 
Syntaxcbnd 35934 Extend class notation with the class of bounded metric spaces.
class Bnd
 
Definitiondf-totbnd 35935* 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 35936* 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 35937* 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 35938* 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 35939 The predicate "totally bounded" implies 𝑀 is a metric space. (Contributed by Jeff Madsen, 2-Sep-2009.)
(𝑀 ∈ (TotBnd‘𝑋) → 𝑀 ∈ (Met‘𝑋))
 
Theorem0totbnd 35940 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 35941* 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 35942* 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 35943* 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 35944 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 35945* 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 35946* 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 35947* 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 35948 A bounded metric space is a metric space. (Contributed by Mario Carneiro, 16-Sep-2015.)
(𝑀 ∈ (Bnd‘𝑋) → 𝑀 ∈ (Met‘𝑋))
 
Theoremisbndx 35949* 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 35950* 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 35951* 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 35952* 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 35953 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 35954 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 35955* 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 35956 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 35936 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 35957* 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 35958 Lemma for equivbnd2 35959 and similar theorems. (Contributed by Jeff Madsen, 16-Sep-2015.)
𝐷 = (𝑀 ↾ (𝑌 × 𝑌))       ((𝑀 ∈ (Met‘𝑋) ∧ 𝐷 ∈ (Bnd‘𝑌)) → 𝑌𝑋)
 
Theoremequivbnd2 35959* 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 35960* 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 35961* 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 35962* 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 35963 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 35964 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 35965 Extend class notation with the class of metric space isometries.
class Ismty
 
Definitiondf-ismty 35966* 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 35967* 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 35968* 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 35969 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 35970 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 35971 Lemma for ismtyhmeo 35972. (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 35972 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 35973 Lemma for ismtybnd 35974. (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 35974 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 35975 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 35976 Lemma for heibor1 35977. 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 35977 One half of heibor 35988, that does not require any Choice. A compact metric space is complete and totally bounded. We prove completeness in cmpcmet 24492 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 35978* Lemma for heibor 35988. 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 35979* Lemma for heibor 35988. 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 35980* Lemma for heibor 35988. Using countable choice ax-cc 10200, 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 35978 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 10200 via iunctb 10339), and so we can apply ax-cc 10200 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 35981* Lemma for heibor 35988. Using the function 𝑇 constructed in heiborlem3 35980, 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 35982* Lemma for heibor 35988. The function 𝑀 is a set of point-and-radius pairs suitable for application to caubl 24481. (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 35983* Lemma for heibor 35988. 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 35984* Lemma for heibor 35988. 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 35985* Lemma for heibor 35988. 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 35986* Lemma for heibor 35988. Discharge the hypotheses of heiborlem8 35985 by applying caubl 24481 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 35987* Lemma for heibor 35988. 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 35978, 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 35988 Generalized Heine-Borel Theorem. A metric space is compact iff it is complete and totally bounded. See heibor1 35977 and heiborlem1 35978 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 35989* Lemma for bfp 35991. 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 35990* Lemma for bfp 35991. Using the point found in bfplem1 35989, 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 35991* 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 35992 Extend class notation with the n-dimensional Euclidean space.
class n
 
Definitiondf-rrn 35993* 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 35994* 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 35995* 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 35996 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 35997 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 35998* Bound on the distance between two points in Euclidean space given bounds on the distances in each coordinate. This theorem and rrndstprj1 35997 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 35999* Lemma for rrncms 36000. (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 36000 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))
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