HomeHome Metamath Proof Explorer < Previous   Next >
Browser slow? Try the
Unicode version.

Jump to page: Contents + 1 1-100 2 101-200 3 201-300 4 301-400 5 401-500 6 501-600 7 601-700 8 701-800 9 801-900 10 901-1000 11 1001-1100 12 1101-1200 13 1201-1300 14 1301-1400 15 1401-1500 16 1501-1600 17 1601-1700 18 1701-1800 19 1801-1900 20 1901-2000 21 2001-2100 22 2101-2200 23 2201-2300 24 2301-2400 25 2401-2500 26 2501-2600 27 2601-2700 28 2701-2800 29 2801-2900 30 2901-3000 31 3001-3100 32 3101-3200 33 3201-3300 34 3301-3400 35 3401-3500 36 3501-3600 37 3601-3700 38 3701-3800 39 3801-3900 40 3901-4000 41 4001-4100 42 4101-4200 43 4201-4300 44 4301-4400 45 4401-4500 46 4501-4600 47 4601-4700 48 4701-4800 49 4801-4900 50 4901-5000 51 5001-5100 52 5101-5200 53 5201-5300 54 5301-5400 55 5401-5500 56 5501-5600 57 5601-5700 58 5701-5800 59 5801-5900 60 5901-6000 61 6001-6100 62 6101-6200 63 6201-6300 64 6301-6400 65 6401-6500 66 6501-6600 67 6601-6700 68 6701-6800 69 6801-6900 70 6901-7000 71 7001-7100 72 7101-7200 73 7201-7300 74 7301-7400 75 7401-7500 76 7501-7600 77 7601-7700 78 7701-7800 79 7801-7900 80 7901-8000 81 8001-8100 82 8101-8200 83 8201-8300 84 8301-8400 85 8401-8500 86 8501-8600 87 8601-8700 88 8701-8800 89 8801-8900 90 8901-9000 91 9001-9100 92 9101-9200 93 9201-9300 94 9301-9400 95 9401-9500 96 9501-9600 97 9601-9700 98 9701-9800 99 9801-9900 100 9901-10000 101 10001-10100 102 10101-10200 103 10201-10300 104 10301-10400 105 10401-10500 106 10501-10600 107 10601-10700 108 10701-10800 109 10801-10900 110 10901-11000 111 11001-11100 112 11101-11200 113 11201-11300 114 11301-11400 115 11401-11500 116 11501-11600 117 11601-11700 118 11701-11800 119 11801-11900 120 11901-12000 121 12001-12100 122 12101-12200 123 12201-12229

Color key:    Metamath Proof Explorer  Metamath Proof Explorer
(1-9062)
  Hilbert Space Explorer  Hilbert Space Explorer
(9063-10650)
  Users' Mathboxes  Users' Mathboxes
(10651-12229)
 

Statement List for Metamath Proof Explorer - 4601-4700 - Page 47 of 123
TypeLabelDescription
Statement
 
Theoremsbthlem10 4601 Lemma for sbth 4602.
 
Theoremsbth 4602 Schroeder-Bernstein Theorem. Theorem 18 of [Suppes] p. 95. This theorem states that if set A is smaller (has lower cardinality) than B and vice-versa, then A and B are equinumerous (have the same cardinality). The interesting thing is that this can be proved without invoking the Axiom of Choice, as we do here, but the proof as you can see is quite difficult. (The theorem can be proved more easily if we allow AC.) The main proof consists of lemmas sbthlem1 4592 through sbthlem10 4601; this final piece mainly changes bound variables to eliminate the hypotheses of sbthlem10 4601. We follow closely the proof in Suppes, which you should consult to understand our proof at a higher level.
|- ((A ~<_ B /\ B ~<_ A) -> A ~~ B)
 
Theoremsbthbg 4603 Schroeder-Bernstein Theorem and its converse.
|- (B e. C -> ((A ~<_ B /\ B ~<_ A) <-> A ~~ B))
 
Theoremsbthcl 4604 Schroeder-Bernstein Theorem in class form.
|- ~~ = ( ~<_ i^i `' ~<_ )
 
Theoremdfsdom2 4605 Alternate definition of strict dominance. Compare Definition 3 of [Suppes] p. 97.
|- ~< = ( ~<_ \ `' ~<_ )
 
Theorembrsdom2 4606 Alternate definition of strict dominance. Definition 3 of [Suppes] p. 97.
|- A e. V   &   |- B e. V   =>   |- (A ~< B <-> (A ~<_ B /\ -. B ~<_ A))
 
Theoremsdomnsym 4607 Strict dominance is not symmetric. Theorem 21(ii) of [Suppes] p. 97.
|- (A ~< B -> -. B ~< A)
 
Theoremdomnsym 4608 Theorem 22(i) of [Suppes] p. 97.
|- (A ~<_ B -> -. B ~< A)
 
Theorem0dom 4609 Any set dominates the empty set.
|- (/) ~<_ A
 
Theoremdom0 4610 A set dominated by the empty set is empty.
|- (A ~<_ (/) <-> A = (/))
 
Theorem0sdomg 4611 A set strictly dominates the empty set iff it is not empty.
|- (A e. B -> ((/) ~< A <-> A =/= (/)))
 
Theorem0sdom 4612 A set strictly dominates the empty set iff it is not empty.
|- A e. V   =>   |- ((/) ~< A <-> A =/= (/))
 
Theoremsdom0 4613 The empty set does not strictly dominate any set.
|- -. A ~< (/)
 
Theoremsdomdomtr 4614 Transitivity of strict dominance and dominance. Theorem 22(iii) of [Suppes] p. 97.
|- (C e. D -> ((A ~< B /\ B ~<_ C) -> A ~< C))
 
Theoremsdomentr 4615 Transitivity of strict dominance and equinumerosity. Exercise 11 of [Suppes] p. 98.
|- (C e. D -> ((A ~< B /\ B ~~ C) -> A ~< C))
 
Theoremensdomtr 4616 Transitivity of equinumerosity and strict dominance.
|- ((A ~~ B /\ B ~< C) -> A ~< C)
 
Theoremsdomirr 4617 Strict dominance is irreflexive. Theorem 21(i) of [Suppes] p. 97.
|- -. A ~< A
 
Theoremsdomex 4618 Technical lemma for simplifying proofs involving strict dominance.
|- (A ~< B -> (A e. V /\ B e. V))
 
Theoremsdomtr 4619 Strict dominance is transitive. Theorem 21(iii) of [Suppes] p. 97.
|- ((A ~< B /\ B ~< C) -> A ~< C)
 
Theoremsdomn2lp 4620 Strict dominance has no 2-cycle loops.
|- -. (A ~< B /\ B ~< A)
 
Theoremdomsdomtr 4621 Transitivity of dominance and strict dominance. Theorem 22(ii) of [Suppes] p. 97.
|- ((A ~<_ B /\ B ~< C) -> A ~< C)
 
Theoremenen1 4622 Equality-like theorem for equinumerosity.
|- ((B e. D /\ A ~~ B) -> (A ~~ C <-> B ~~ C))
 
Theoremenen2 4623 Equality-like theorem for equinumerosity.
|- ((B e. D /\ A ~~ B) -> (C ~~ A <-> C ~~ B))
 
Theoremdomen1 4624 Equality-like theorem for equinumerosity and dominance.
|- ((B e. D /\ A ~~ B) -> (A ~<_ C <-> B ~<_ C))
 
Theoremdomen2 4625 Equality-like theorem for equinumerosity and dominance.
|- ((B e. D /\ A ~~ B) -> (C ~<_ A <-> C ~<_ B))
 
Theoremsdomen1 4626 Equality-like theorem for equinumerosity and strict dominance.
|- ((B e. D /\ A ~~ B) -> (A ~< C <-> B ~< C))
 
Theoremsdomen2 4627 Equality-like theorem for equinumerosity and strict dominance.
|- ((B e. D /\ A ~~ B) -> (C ~< A <-> C ~< B))
 
Theoremfodomr 4628 There exists a mapping from a set onto any (non-empty) set that it dominates.
|- ((A e. C /\ (/) ~< B /\ B ~<_ A) -> E.f f:A-onto->B)
 
Theoremcanth2 4629 Cantor's Theorem. No set is equinumerous to its power set. Specifically, any set has a cardinality (size) strictly less than the cardinality of its power set. For example, the cardinality of real numbers is the same as the cardinality of the power set of integers, so real numbers cannot be put into a one-to-one correspondence with integers. Theorem 23 of [Suppes] p. 97. For the function version, see canth 4205.
|- A e. V   =>   |- A ~< P~A
 
Theoremcanth2g 4630 Cantor's theorem with the sethood requirement expressed as an antecedent. Theorem 23 of [Suppes] p. 97.
|- (A e. B -> A ~< P~A)
 
Theorempwuninel 4631 The power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set.
|- -. P~U.A e. A
 
Theorem2pwuninel 4632 The power set of the power set of the union of a set does not belong to the set. This theorem provides a way of constructing a new set that doesn't belong to a given set.
|- -. P~P~U.A e. A
 
Theorempwne 4633 No set equals its power set.
|- (A e. B -> P~A =/= A)
 
Theorem2pwne 4634 No set equals the power set of its power set.
|- (A e. B -> P~P~A =/= A)
 
Theoremxpen 4635 Equinumerosity law for cross product. Proposition 4.22(b) of [Mendelson] p. 254.
|- A e. V   &   |- B e. V   &   |- C e. V   &   |- D e. V   =>   |- ((A ~~ B /\ C ~~ D) -> (A X. C) ~~ (B X. D))
 
Theoremmapenlem1 4636 Lemma for mapen 4638.
 
Theoremmapenlem2 4637 Lemma for mapen 4638.
 
Theoremmapen 4638 Two set exponentiations are equinumerous when their bases and exponents are equinumerous. Theorem 6H(c) of [Enderton] p. 139.
|- A e. V   &   |- B e. V   &   |- C e. V   &   |- D e. V   =>   |- ((A ~~ B /\ C ~~ D) -> (A ^m C) ~~ (B ^m D))
 
Theoremmapdom1 4639 Order-preserving property of set exponentiation. Theorem 6L(c) of [Enderton] p. 149.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- (A ~<_ B -> (A ^m C) ~<_ (B ^m C))
 
Theoremmapdom2lem 4640 Lemma for mapdom2 4641.
 
Theoremmapdom2 4641 Order-preserving property of set exponentiation. Theorem 6L(d) of [Enderton] p. 149.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- ((A ~<_ B /\ -. (A = (/) /\ C = (/))) -> (C ^m A) ~<_ (C ^m B))
 
Theoremmapxpen 4642 Equinumerosity law for double set exponentiation. Proposition 10.45 of [TakeutiZaring] p. 96.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- ((A ^m B) ^m C) ~~ (A ^m (B X. C))
 
Theoremxpmapenlem1 4643 Lemma for xpmapen 4648.
 
Theoremxpmapenlem2 4644 Lemma for xpmapen 4648.
 
Theoremxpmapenlem3 4645 Lemma for xpmapen 4648.
 
Theoremxpmapenlem4 4646 Lemma for xpmapen 4648.
 
Theoremxpmapenlem5 4647 Lemma for xpmapen 4648.
 
Theoremxpmapen 4648 Equinumerosity law for set exponentiation of a cross product. Exercise 4.47 of [Mendelson] p. 255.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- ((A X. B) ^m C) ~~ ((A ^m C) X. (B ^m C))
 
Theoremmapunen 4649 Equinumerosity law for set exponentiation of a disjoint union. Exercise 4.45 of [Mendelson] p. 255.
|- A e. V   &   |- B e. V   &   |- C e. V   =>   |- ((A i^i B) = (/) -> (C ^m (A u. B)) ~~ ((C ^m A) X. (C ^m B)))
 
Theorempwen 4650 If two sets are equinumerous, then their power sets are equinumerous. Proposition 10.15 of [TakeutiZaring] p. 87.
|- B e. V   =>   |- (A ~~ B -> P~A ~~ P~B)
 
Theoremssenen 4651 Equinumerosity of equinumerous subsets of a set.
|- A e. V   &   |- B e. V   =>   |- (A ~~ B -> {x | (x (_ A /\ x ~~ C)} ~~ {x | (x (_ B /\ x ~~ C)})
 
Theoremlimenpsi 4652 A limit ordinal is equinumerous to a proper subset of itself.
|- Lim A   =>   |- (A e. B -> A ~~ (A \ {(/)}))
 
Theoremlimensuci 4653 A limit ordinal is equinumerous to its successor.
|- Lim A   =>   |- (A e. B -> A ~~ suc A)
 
Theoremlimensuc 4654 A limit ordinal is equinumerous to its successor.
|- ((A e. B /\ Lim A) -> A ~~ suc A)
 
Pigeonhole Principle
 
Theoremphplem1 4655 Lemma for Pigeonhole Principle. If we join a natural number to itself minus an element, we end up with its successor minus the same element.
 
Theoremphplem2 4656 Lemma for Pigeonhole Principle. A natural number is equinumerous to its successor minus one of its elements.
 
Theoremphplem3 4657 Lemma for Pigeonhole Principle. A natural number is equinumerous to its successor minus any element of the successor.
 
Theoremphplem4 4658 Lemma for Pigeonhole Principle. Equinumerosity of successors implies equinumerosity of the original natural numbers.
 
Theoremnneneq 4659 Two equinumerous natural numbers are equal. Proposition 10.20 of [TakeutiZaring] p. 90 and its converse. Also compare Corollary 6E of [Enderton] p. 136.
|- ((A e. om /\ B e. om) -> (A ~~ B <-> A = B))
 
Theoremphp 4660 Pigeonhole Principle. A natural number is not equinumerous to a proper subset of itself. Theorem (Pigeonhole Principle) of [Enderton] p. 134. The theorem is so-called because you can't put n + 1 pigeons into n holes (if each hole holds only one pigeon). The proof consists of lemmas phplem1 4655 through phplem4 4658, nneneq 4659, and this final piece of the proof.
|- ((A e. om /\ B (. A) -> -. A ~~ B)
 
Theoremphp2 4661 Corollary of Pigeonhole Principle.
|- ((A e. om /\ B (. A) -> B ~< A)
 
Theoremphp3 4662 Corollary of Pigeonhole Principle. If A is finite and B is a proper subset of A, the B is strictly less numerous than A. Stronger version of Corollary 6C of [Enderton] p. 135.
|- ((A e. Fin /\ B (. A) -> B ~< A)
 
Theoremphp4 4663 Corollary of the Pigeonhole Principle php 4660: a natural number is strictly dominated by its successor.
|- (A e. om -> A ~< suc A)
 
Theoremphp5 4664 Corollary of the Pigeonhole Principle php 4660: a natural number is not equinumerous to its successor. Corollary 10.21(1) of [TakeutiZaring] p. 90.
|- (A e. om -> -. A ~~ suc A)
 
Finite sets
 
Theoremonomeneq 4665 An ordinal number equinumerous to a natural number is equal to it. Proposition 10.22 of [TakeutiZaring] p. 90 and its converse.
|- ((A e. On /\ B e. om) -> (A ~~ B <-> A = B))
 
Theoremonfin 4666 An ordinal number is finite iff it is a natural number. Proposition 10.32 of [TakeutiZaring] p. 92.
|- (A e. On -> (A e. Fin <-> A e. om))
 
Theoremnndomo 4667 Cardinal ordering agrees with natural number ordering. Example 3 of [Enderton] p. 146.
|- ((A e. om /\ B e. om) -> (A ~<_ B <-> A (_ B))
 
Theoremnnsdomo 4668 Cardinal ordering agrees with natural number ordering.
|- ((A e. om /\ B e. om) -> (A ~< B <-> A (. B))
 
Theoremomsucdom 4669 Strict dominance of natural numbers is the same as dominance over the successor of the smaller.
|- ((A e. om /\ B e. om) -> (A ~< B <-> suc A ~<_ B))
 
Theoremsucdomi 4670 Dominance of a set over a successor of a natural number implies strict dominance over the number. For the converse, see sucdom 4992.
|- ((A e. om /\ B e. C) -> (suc A ~<_ B -> A ~< B))
 
Theorem0sdom1dom 4671 Strict dominance over zero is the same as dominance over one.
|- A e. V   =>   |- ((/) ~< A <-> 1o ~<_ A)
 
Theorem1sdom2 4672 Ordinal 1 is strictly dominated by ordinal 2.
|- 1o ~< 2o
 
Theoremfinsucdom 4673 Strict dominance of a finite set over a natural number is the same as dominance over its successor.
|- ((A e. om /\ B e. Fin) -> (A ~< B <-> suc A ~<_ B))
 
Theorempssinf 4674 A set equinumerous to a proper subset of itself is infinite. Corollary 6D(a) of [Enderton] p. 136.
|- ((A (. B /\ A ~~ B) -> -. B e. Fin)
 
Theoremominf 4675 The set of natural numbers is infinite. Corollary 6D(b) of [Enderton] p. 136.
|- -. om e. Fin
 
Theoremomsdomnn 4676 Omega strictly dominates a natural number. Example 3 of [Enderton] p. 146. Here we use A ~<_ om /\ -. om ~~ A instead of A ~< om because, due to a peculiarity ultimately caused our ordered pair definition, we would need the Axiom of infinity (which we have avoided up to now) in order to prove the latter.
|- (A e. om -> (A ~<_ om /\ -. om ~~ A))
 
Theoremisfinite1 4677 Omega strictly dominates a finite set. See comment in omsdomnn 4676.
|- (A e. Fin -> (A ~<_ om /\ -. om ~~ A))
 
Theoreminfsdomnn 4678 An infinite set strictly dominates a natural number.
|- A e. V   =>   |- ((om ~<_ A /\ B e. om) -> B ~< A)
 
Theoreminfn0 4679 An infinite set is not empty.
|- A e. V   =>   |- (om ~<_ A -> A =/= (/))
 
Theoremenfi 4680 Equinmerous sets have the same finiteness.
|- ((B e. C /\ A ~~ B) -> (A e. Fin <-> B e. Fin))
 
Theorempssnn 4681 A proper subset of a natural number is equinumerous to some smaller number. Lemma 6F of [Enderton] p. 137.
|- ((A e. om /\ B (. A) -> E.x e. A B ~~ x)
 
Theoremssnnfi 4682 A subset of a natural number is finite.
|- ((A e. om /\ B (_ A) -> B e. Fin)
 
Theoremssfi 4683 A subset of a finite set is finite. Corollary 6G of [Enderton] p. 138.
|- ((A e. Fin /\ B (_ A) -> B e. Fin)
 
Theoremdomfi 4684 A set dominated by a finite set is finite.
|- ((A e. Fin /\ B ~<_ A) -> B e. Fin)
 
Theoremxpfi 4685 The components of a non-empty finite cross product are finite. (Contributed by Paul Chapman, 11-Apr-2009.)
|- (((A X. B) e. Fin /\ (A X. B) =/= (/)) -> (A e. Fin /\ B e. Fin))
 
Theoremunblem1 4686 Lemma for unbnn 4690. After removing the successor of an element from an unbounded set of natural numbers, the intersection of the result belongs to the original unbounded set.
 
Theoremunblem2 4687 Lemma for unbnn 4690. The value of the function F belongs to the unbounded set of natural numbers A.
 
Theoremunblem3 4688 Lemma for unbnn 4690. The value of the function F is less than its value at a successor.
 
Theoremunblem4 4689 Lemma for unbnn 4690. The function F maps the set of natural numbers one-to-one to the set of unbounded natural numbers A.
 
Theoremunbnn 4690 Any unbounded subset of natural numbers is equinumerous to the set of all natural numbers. Part of the proof of Theorem 42 of [Suppes] p. 151. See unbnn3 4785 for a stronger version without the hypothesis.
|- A e. V   =>   |- ((A (_ om /\ A.x e. om E.y e. A x e. y) -> A ~~ om)
 
Theoremunbnn2 4691 Version of unbnn 4690 that does not require a strict upper bound.
|- A e. V   =>   |- ((A (_ om /\ A.x e. om E.y e. A x (_ y) -> A ~~ om)
 
Theoremisfinite2 4692 Any set strictly dominated by the class of natural numbers is finite. Sufficiency part of Theorem 42 of [Suppes] p. 151. This theorem does not require the Axiom of Infinity.
|- (A ~< om -> A e. Fin)
 
Theoremfin2inf 4693 This (useless) theorem, which was proved without the Axiom of Infinity, demonstrates an artifact of our definition of strict dominance, which is meaningful only when its arguments exist. In particular, the antecedent cannot be satisfied unless om exists.
|- (A ~< om -> om e. V)
 
Theoremunfilem1 4694 Lemma for proving that the union of two finite sets is finite.
 
Theoremunfilem2 4695 Lemma for proving that the union of two finite sets is finite.
 
Theoremunfilem3 4696 Lemma for proving that the union of two finite sets is finite.
 
Theoremunfi 4697 The union of two finite sets is finite. Part of Corollary 6K of [Enderton] p. 144.
|- ((A e. Fin /\ B e. Fin) -> (A u. B) e. Fin)
 
Theoremunfi2 4698 The union of two finite sets is finite. Part of Corollary 6K of [Enderton] p. 144. This version of unfi 4697 is useful only if we assume the Axiom of Infinity (see comments in fin2inf 4693).
|- ((A ~< om /\ B ~< om) -> (A u. B) ~< om)
 
Theoreminfcntss 4699 Every infinite set has a denumerable subset. Similar to Exercise 8 of [TakeutiZaring] p. 91. (However, we need neither AC nor the Axiom of Infinity because of the way we express "infinite" in the antecedent.)
|- A e. V   =>   |- (om ~<_ A -> E.x(x (_ A /\ x ~~ om))
 
Theoremprfi 4700 An unordered pair is finite.
|- {A, B} e. Fin

MPE Home   Contents Copyright terms: Public domain < Previous  Next >