hcju/proofwikiqa huggingface dataset:
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$. Thus it follows immediately from the fact that $S$ is a ring with unity, that $\struct {S_R, *}$ is a unitary algebra. {{qed}} Category:Algebras cpt3ip7bivyes7cyh5urvhe82e5sjko
By definition, the multiplication of $\struct {S_R, *}$ is the ring product of $S$. Thus it follows immediately from the fact that $S$ is a ring, that $\struct {S_R, *}$ is an associative algebra. {{qed}}
Note that by Center of Commutative Ring, the image of $f$ is indeed a subset of the center of $S$. By definition, the multiplication of $(S_R, *)$ is the ring product of $S$. Thus it follows immediately from the fact that $S$ is a ring, that $(S_R, *)$ is an commutative algebra. {{qed}}
{{begin-eqn}} {{eqn | n = 1       | l = 207 \, 622 \, 273       | r = 4 \times 51 \, 905 \, 568 + 1       | c = and is the $11 \, 477 \, 482$nd prime }} {{eqn | n = 2       | l = 207 \, 622 \, 297       | r = 4 \times 51 \, 905 \, 574 + 1       | c = and is the $11 \, 477 \, 483$rd prime }} {{eqn | n = 3       | l = 207 \, 622 \, 301       | r = 4 \times 51 \, 905 \, 575 + 1       | c = and is the $11 \, 477 \, 484$th prime }} {{eqn | n = 4       | l = 207 \, 622 \, 313       | r = 4 \times 51 \, 905 \, 578 + 1       | c = and is the $11 \, 477 \, 485$th prime }} {{eqn | n = 5       | l = 207 \, 622 \, 321       | r = 4 \times 51 \, 905 \, 580 + 1       | c = and is the $11 \, 477 \, 486$th prime }} {{eqn | n = 6       | l = 207 \, 622 \, 381       | r = 4 \times 51 \, 905 \, 595 + 1       | c = and is the $11 \, 477 \, 487$th prime }} {{eqn | n = 7       | l = 207 \, 622 \, 409       | r = 4 \times 51 \, 905 \, 602 + 1       | c = and is the $11 \, 477 \, 488$th prime }} {{eqn | n = 8       | l = 207 \, 622 \, 417       | r = 4 \times 51 \, 905 \, 604 + 1       | c = and is the $11 \, 477 \, 489$th prime }} {{eqn | n = 9       | l = 207 \, 622 \, 421       | r = 4 \times 51 \, 905 \, 605 + 1       | c = and is the $11 \, 477 \, 490$th prime }} {{eqn | n = 10       | l = 207 \, 622 \, 489       | r = 4 \times 51 \, 905 \, 622 + 1       | c = and is the $11 \, 477 \, 491$st prime }} {{eqn | n = 11       | l = 207 \, 622 \, 501       | r = 4 \times 51 \, 905 \, 625 + 1       | c = and is the $11 \, 477 \, 492$nd prime }} {{eqn | n = 12       | l = 207 \, 622 \, 517       | r = 4 \times 51 \, 905 \, 629 + 1       | c = and is the $11 \, 477 \, 493$rd prime }} {{eqn | n = 13       | l = 207 \, 622 \, 537       | r = 4 \times 51 \, 905 \, 634 + 1       | c = and is the $11 \, 477 \, 494$th prime }} {{eqn | n = 14       | l = 207 \, 622 \, 549       | r = 4 \times 51 \, 905 \, 637 + 1       | c = and is the $11 \, 477 \, 495$th prime }} {{eqn | n = 15       | l = 207 \, 622 \, 553       | r = 4 \times 51 \, 905 \, 638 + 1       | c = and is the $11 \, 477 \, 496$th prime }} {{eqn | n = 16       | l = 207 \, 622 \, 561       | r = 4 \times 51 \, 905 \, 640 + 1       | c = and is the $11 \, 477 \, 497$th prime }} {{end-eqn}} Note that the $11 \, 477 \, 481$st prime: :$207 \, 622 \, 271 = 4 \times 51 \, 905 \, 568 - 1$ and the $11 \, 477 \, 498$th prime: :$207 \, 622 \, 567 = 4 \times 51 \, 905 \, 642 - 1$ and so are not of the form $4 n + 1$. {{qed}}
The abelian cases are handled by the corollary to Abelian Group Factored by Prime. {{qed|lemma}} Let $G$ be non-abelian. By Lagrange's theorem the order of non-identity elements in $G$ is either $2$, $4$ or $8$. {{AimForCont}} that there exists an order $8$ element. Then $G$ is generated by this element. So $G$ is by definition cyclic. But Cyclic Group is Abelian, contradicting the assumption that $G$ is non-abelian. So there is no order $8$ element. By Non-Abelian Order 8 Group has Order 4 Element, there exists at least one order $4$ element in $G$. Let it be denoted by $a$. Let $A$ denote the subgroup generated by $a$. By Lagrange's theorem there are two cosets in $G$: $A$ and $G \setminus A$. Let $b \in G \setminus A$. Then $\set {a, b}$ is a generator of $G$. Now we consider how $a$ and $b$ interact with each other. Consider the element $x = b a b^{-1}$. By Subgroup of Index 2 is Normal, $b A b^{-1} = A$. So $x \in A$. By Order of Conjugate Element equals Order of Element, the only possible choices are $x = a$ or $x = a^3$. If $x = a$, then $a$ and $b$ commute. Since $\set {a, b}$ generates $G$, this makes $G$ an abelian group, which is a contradiction. So $b a b^{-1} = a^3$. It suffices to consider the order of $b$: :If $\order b = 2$, then $G \cong D_4$. :If $\order b = 4$, then $G \cong \Dic 2$. {{qed}}
{{begin-eqn}} {{eqn | l = T_3       | r = \frac {3 \left({3 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = 6       | c =  }} {{eqn | r = 1 \times 2 \times 3       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = T_{15}       | r = \frac {15 \left({15 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = 120       | c =  }} {{eqn | r = 4 \times 5 \times 6       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = T_{20}       | r = \frac {20 \left({20 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = 210       | c =  }} {{eqn | r = 5 \times 6 \times 7       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = T_{44}       | r = \frac {44 \left({44 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = 990       | c =  }} {{eqn | r = 9 \times 10 \times 11       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = T_{608}       | r = \frac {608 \left({608 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = 185 \, 136       | c =  }} {{eqn | r = 56 \times 57 \times 58       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = T_{22 \, 736}       | r = \frac {22 \, 736 \left({22 \, 736 + 1}\right)} 2       | c = Closed Form for Triangular Numbers }} {{eqn | r = 258 \, 474 \, 216       | c =  }} {{eqn | r = 2^3 \times 3 \times 7^2 \times 11 \times 13 \times 29 \times 53       | c =  }} {{eqn | r = \left({2^2 \times 3 \times 53}\right) \times \left({7^2 \times 13}\right) \times \left({2 \times 11 \times 29}\right)       | c =  }} {{eqn | r = 636 \times 637 \times 638       | c =  }} {{end-eqn}} {{ProofWanted|It remains to be shown these are the only ones}}
Let $\operatorname{id} : R \to R$ be the identity mapping. Let $1$ be the unity of $R$. By Identity Mapping is Ring Automorphism, $\operatorname{id}$ is a ring homomorphism. By Universal Property of Polynomial Ring, there exists a ring homomorphism $h : R[X] \to R$ with $h\circ \iota = \operatorname{id}$. By Identity Mapping is Injection, $\operatorname{id}$ is an injection. By Injection if Composite is Injection, $\iota$ is an injection. Thus $\iota$ is a ring monomorphism. {{qed}} Category:Polynomial Theory 9v6ubd9bbba5c4eqfgu96fv9lmn3i0o
Use Universal Property of Polynomial Ring and Universal Poperty of Field of Fractions. {{ProofWanted}} Category:Polynomial Theory Category:Universal Properties 6grwged78wrdv6kuqtrvbjrflq9g02z
First it is demonstrated that indeed $381 \, 654 \, 729$ has this property: {{begin-eqn}} {{eqn | l = 3       | r = 1 \times 3 }} {{eqn | l = 38       | r = 2 \times 19 }} {{eqn | l = 381       | r = 3 \times 127 }} {{eqn | l = 3816       | r = 4 \times 954 }} {{eqn | l = 38 \, 165       | r = 5 \times 7633 }} {{eqn | l = 381 \, 654       | r = 6 \times 63 \, 609 }} {{eqn | l = 3 \, 816 \, 547       | r = 7 \times 545 \, 221 }} {{eqn | l = 38 \, 165 \, 472       | r = 8 \times 4 \, 770 \, 684 }} {{eqn | l = 381 \, 654 \, 729       | r = 9 \times 42 \, 406 \, 081 }} {{end-eqn}} {{qed|lemma}} It remains to be demonstrated that it is the only such number. Let $\sqbrk {abcdefghi}$ be a polydivisible number which is pandigital. By Divisibility by 5: :$e = 5$ By Divisibility by 2: :$b, d, f, h$ are even. Hence: :$a, c, g, i$ are odd. By Divisibility by 3: :$a + b + c$, $d + 5 + f$, $g + h + i$ are divisible by $3$. By Divisibility by 8: :$\sqbrk {fgh}$ is divisible by $8$. Because $f$ is even: :$\sqbrk {f00}$ is divisible by $8$. We must therefore have: :$\sqbrk {gh}$ divisible by $8$. Thus: :$\sqbrk {gh}$ must be one of $16, 32, 72, 96$. Because $i$ is odd and $g + h + i$ are divisible by $3$: :$\sqbrk {ghi}$ must be one of $321, 327, 723, 729, 963$. By Divisibility by 4: :$\sqbrk {cd}$ is divisible by $4$. Because $c$ is odd: :$d$ must be $2$ or $6$. Suppose $h = 6$. Then $\sqbrk {ghi} = 963$ and $d = 2$. Since $d + 5 + f$ is divisible by $3$ and $f$ is even: :$f = 8$ The remaining even number $b$ must be $4$. So: :$\sqbrk {abcdefghi} = \sqbrk {a4c258963}$ Thus the only remaining options for $a$ and $c$ are $1$ and $7$. However, we find that neither $1472589$ nor $7412689$ is divisible by $7$. Therefore we must have: :$h = 2$ It follow therefore that: :$d = 6$ Because $d + 5 + f$ is divisible by $3$, and $f$ is even: :$f = 4$ The remaining even number $b$ must be $8$. So: :$\sqbrk {abcdefghi} = \sqbrk {a8c654g2i}$ Since $a + 8 + c$ is divisible by $3$: :$\sqbrk {a8c}$ must be one of $183, 189, 381, 387, 783, 789, 981, 987$. Suppose $g = 3$. Then: :$\sqbrk {a8c}$ must be one of $189, 789, 981, 987$. Inspection of the possible numbers reveals that none of $1896543$, $7896543$, $9816543$, $9876543$ is divisible by $7$. Suppose $g = 7$. Then: :$\sqbrk {a8c}$ must be one of $183, 189, 381, 981$. Among $1836547$, $1896547$, $3816547$, $9816547$, only $3816547$ is divisible by $7$. The results follow. {{qed}}
Consider the numbers $\sqbrk {999 \, 999 \, 9ab}$. Since $999 \, 999 \, 000$ is divisible by $2, 3, 5, 7, 11, 13$, if $\sqbrk {9ab}$ is divisible by these primes, so is $\sqbrk {999 \, 999 \, 9ab}$. After this elimination the only $\sqbrk {ab} > 37$ that remains are: :$41, 43, 47, 53, 61, 67, 71, 77, 83, 89, 91, 97$ We have: {{begin-eqn}} {{eqn | l = 999 \, 999 \, 941       | r = 113 \times 149 \times 59 \, 393 }} {{eqn | l = 999 \, 999 \, 943       | r = 5 \, 623 \times 177 \, 841 }} {{eqn | l = 999 \, 999 \, 947       | r = 163 \times 6 \, 134 \, 969 }} {{eqn | l = 999 \, 999 \, 953       | r = 29 \times 31 \times 773 \times 1 \, 439 }} {{eqn | l = 999 \, 999 \, 961       | r = 3 \, 673 \times 272 \, 257 }} {{eqn | l = 999 \, 999 \, 967       | r = 3 \, 257 \times 307 \, 031 }} {{eqn | l = 999 \, 999 \, 971       | r = 193 \times 5 \, 181 \, 347 }} {{eqn | l = 999 \, 999 \, 977       | r = 2 \, 971 \times 336 \, 587 }} {{eqn | l = 999 \, 999 \, 983       | r = 337 \times 2 \, 967 \, 359 }} {{eqn | l = 999 \, 999 \, 989       | r = 4 \, 327 \times 231 \, 107 }} {{eqn | l = 999 \, 999 \, 991       | r = 67 \times 14 \, 925 \, 373 }} {{eqn | l = 999 \, 999 \, 997       | r = 71 \times 2 \, 251 \times 6 \, 257 }} {{end-eqn}} And we do have $999 \, 999 \, 937$ is prime. {{qed}}
By assumption, some logical formula $\phi$ is not an $\mathscr M$-tautology. Since $\mathscr P$ is sound for $\mathscr M$, $\phi$ is also not a $\mathscr P$-theorem. But then by definition $\mathscr P$ is consistent. {{qed}}
=== Definition 1 implies Definition 2 === Suppose that $\neg \vdash_{\mathscr P} \phi$. Suppose additionally that there is some logical formula $\psi$ such that: :$\vdash_{\mathscr P} \psi, \neg \psi$ By the Rule of Explosion: :$\psi, \neg \psi \vdash_{\mathscr P} \phi$ By Provable Consequence of Theorems is Theorem, we conclude: :$\vdash_{\mathscr P} \phi$ in contradiction to our assumption. {{qed|lemma}} === Definition 2 implies Definition 1 === Suppose either $\phi$ or $\neg \phi$ is not a theorem of $\mathscr P$. The implication follows trivially. {{qed}}
{{BeginTableau|p, \neg p \vdash q|Instance 2 of the Hilbert-style systems}} {{Assumption|1|p}} {{Assumption|2|\neg p}} {{TableauLine  |n = 3  |f = q \implies (p \lor q)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A2$ }} {{TableauLine  |n = 4  |f = \neg p \implies (q \lor \neg p)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 1$  |dep = 3  |c = $\neg p \, / \, q$, $q \, / \, p$ }} {{TableauLine  |n = 5  |f = q \lor \neg p  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep = 2, 4 }} {{TableauLine  |n = 6  |f = (q \lor \neg p) \implies (\neg p \lor q)  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Axiom $A3$, Rule $RST \, 1$  |c = $\neg p \, / \, q$, $q \, / \, p$ }} {{TableauLine  |n = 7  |f = \neg p \lor q  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep = 5, 6 }} {{TableauLine  |n = 8  |f = p \implies q  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 2 \, (2)$ }} {{TableauLine  |n = 9  |f = q  |rlnk = Definition:Hilbert Proof System/Instance 2  |rtxt = Rule $RST \, 3$  |dep = 1, 8 }} {{EndTableau}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = 1 \, 787 \, 109 \, 376^2       | r = \enspace 3 \, 193 \, 759 \, 92 \mathbf {1 \, 787 \, 109 \, 376} }} {{eqn | l = 8 \, 212 \, 890 \, 625^2       | r = 67 \, 451 \, 572 \, 41 \mathbf {8 \, 212 \, 890 \, 625} }} {{end-eqn}} thus demonstrating they are automorphic. By Automorphic Numbers in Base 10, there are no others. {{qed}}
Let $n$ be an automorphic number of $d$ digits, expressed in base $b$. By {{Defof|Automorphic Number}}, we have: :$n^2 \equiv n \pmod {b^d}$ Let some digits be removed from the left-hand end of $n$, so that only $d'$ digits remain. This only makes sense when $d' < d$. Define this new number as $n'$. Then we have: :$n \equiv n' \pmod {b^{d'} }$ Thus we have: {{begin-eqn}} {{eqn | l = n^2       | o = \equiv       | r = n       | rr = \pmod {b^{d'} }       | c = Congruence by Divisor of Modulus: from $n^2 \equiv n \pmod {b^d}$ and $b^{d'} \divides b^d$ }} {{eqn | l = n^2       | o = \equiv       | r = \paren {n'}^2       | rr = \pmod {b^{d'} }       | c = Congruence of Powers: from $n \equiv n' \pmod {b^{d'} }$ }} {{eqn | ll = \leadsto       | l = \paren {n'}^2       | o = \equiv       | r = n'       | rr = \pmod {b^{d'} } }} {{end-eqn}} Hence $n'$ is an automorphic number of $d'$ digits in base $b$. {{qed}}
We have that: {{begin-eqn}} {{eqn | o =        | r = 1 \, 979 \, 339 \, 333       | c = is prime }} {{eqn | o =        | r = 1 \, 979 \, 339 \, 339       | c = is prime }} {{end-eqn}} For both, the truncation process is the same: {{begin-eqn}} {{eqn | o =        | r = 197 \, 933 \, 933       | c = is the $10 \, 970 \, 817$th prime }} {{eqn | o =        | r = 19 \, 793 \, 393       | c = is the $1 \, 252 \, 285$th prime }} {{eqn | o =        | r = 1 \, 979 \, 339       | c = is the $147 \, 488$th prime }} {{eqn | o =        | r = 197 \, 933       | c = is the $17 \, 815$th prime }} {{eqn | o =        | r = 19 \, 793       | c = is the $2240$th prime }} {{eqn | o =        | r = 1979       | c = is the $299$th prime }} {{eqn | o =        | r = 197       | c = is the $45$th prime }} {{eqn | o =        | r = 19       | c = is the $8$th prime }} {{eqn | o =        | r = 1       | c = has been defined temporarily to be prime }} {{end-eqn}} {{qed}}
By definition of complete multiplicativity: :$\forall m, n \in \Z: \map f {m n} = \map f m \map f n$ Hence by True Statement is implied by Every Statement: :$\forall m, n \in \Z: m \perp n \implies \map f {m n} = \map f m \map f n$ So $f$ is multiplicative. {{qed}} Category:Number Theory Category:Completely Multiplicative Functions Category:Multiplicative Functions st2el7aotybal8kzslr6xuxckfq4fqy
We have: :$5 \, 391 \, 411 \, 025 = 5^2 \times 7 \times 11 \times 13 \times 17 \times 19 \times 23 \times 29$ showing it is not divisible by $3$. Then from {{SigmaLink|5,391,411,025|5 \, 391 \, 411 \, 025}} we have: :$\map \sigma {5 \, 391 \, 411 \, 025} = 10 \, 799 \, 308 \, 800 = 2 \times 5 \, 391 \, 411 \, 025 + 16 \, 486 \, 750$ demonstrating that $5 \, 391 \, 411 \, 025$ is abundant. {{ProofWanted|It is still to be shown it is the smallest such.}}
=== $(1)$ implies $(2)$ === Let $T$ be an irreducible space by Definition 1. That is, $T$ is not the union of any two proper closed sets. {{AimForCont}} $T$ admits a finite cover by proper closed sets. By the Well-Ordering Principle, there exists a minimal natural number $n \in \N$ such that $T$ has a cover by $n$ proper closed sets, say $F_1, \ldots, F_n$. By definition of proper subset, it must be the case that $n > 1$. By Finite Union of Closed Sets is Closed in Topological Space, $F_{n - 1} \cup F_n$ is closed. By Union is Associative: :$S = F_1 \cup \cdots \cup \paren {F_{n - 1} \cup F_n}$ By minimality of $n$, $F_{n - 1} \cup F_n$ is not a proper subset. Thus $F_{n - 1} \cup F_n = S$. This contradicts the definition of an irreducible space by Definition 1. Thus $T$ has no finite cover by proper closed sets. That is, $T$ is an irreducible space by Definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $T$ be an irreducible space by Definition 2. Then $T$ has no finite cover by proper closed sets. A cover by two sets is a finite cover. Hence it immediately follows that $T$ has no finite cover by $2$ proper closed sets. That is, $T$ is an irreducible space by Definition 1. {{qed|lemma}} === $(1)$ implies $(3)$ === Let $T$ be an irreducible space by Definition 1. That is, $T$ is not the union of any two proper closed sets. Let $U$ and $V$ be two arbitrary non-empty open sets of $T$. {{AimForCont}} $U$ and $V$ have an empty intersection. By De Morgan's Laws, their complements are proper closed sets whose union is $T$. This contradicts the definition of an irreducible space by Definition 1. Thus: :$U \cap V \ne \O$ As $U$ and $V$ are arbitrary, it follows that $T$ is an irreducible space by Definition 3. {{qed|lemma}} === $(3)$ implies $(1)$ === Let $T$ be an irreducible space by Definition 3. Let $F$ and $G$ be two arbitrary proper closed sets of $T$ {{AimForCont}} their union is $T$. By De Morgan's Laws, their complements are non-empty open sets whose intersection is empty. This contradicts the definition of an irreducible space by Definition 3. Thus $F \cup G \ne T$. As $F$ and $G$ are arbitrary, it follows that $T$ is an irreducible space by Definition 1. {{qed|lemma}} === $(3)$ iff $(4)$ === By definition of everywhere dense subset, an open set is dense {{iff}} it has non-empty intersection with any other open set. {{qed|lemma}} === $(4)$ iff $(5)$ === Follows from Everywhere Dense iff Interior of Complement is Empty. {{qed|lemma}} === $(3)$ iff $(6)$ === {{:Equivalence of Definitions of Irreducible Space/3 iff 6}}{{qed|lemma}} === $(3)$ iff $(7)$ === {{:Equivalence of Definitions of Irreducible Space/3 iff 7}}{{qed}} Category:Irreducible Spaces 7xdaqgcr1cbcv1zvtjlp566gsliu1ru
By the definition of an irreducible subset, $Y\subset X$ is irreducible if and only if any nonempty open subset in $Y$ intersects.  Since the open subsets in $\bar{Y}$ is the same as the open subsets in $Y$, any two of them still trivially intersects in $\bar{Y}$, showing that $\bar{Y}$ is also irreducible. More generally, we can also show that if $\bar{Y}$ is irreducible for a subset $Y\subset X$, then $Y$ is also irreducible in $X$.  We prove it by contradiction.  Assume $Y$ is not irreducible, then there exist two closed proper subsets $Y_1$, $Y_2$ such that $Y=Y_1\cup Y_2$, then $\bar{Y}=\bar{Y_1}\cup \bar{Y_2}$, which contradicts with the assumption.
Because Trivial Topological Space is Irreducible, this is a special case of Irreducible Subspace is Contained in Irreducible Component. {{qed}} Category:Irreducible Spaces 0c7d4q4hjsg3o0hxvk280dbnknp8qme
By definition, an irreducible component of $X$ is an irreducible subspace that is maximal among the irreducible subspaces, ordered by inclusion. {{ProofWanted}} {{AoC|Zorn's Lemma}}
By Closure of Irreducible Subspace is Irreducible, the closure $Y^-$ of $Y$ is irreducible. By Set is Subset of its Topological Closure, $Y \subseteq Y^-$. Because $Y$ is an irreducible component, we must have $Y =  Y^-$. By Set is Closed iff Equals Topological Closure, $Y$ is closed in $T$. {{qed}} Category:Irreducible Spaces o1xyk5a7bvm1x768k3roqodm8ipyohl
Follows from: :Trivial Topological Space is Indiscrete :Indiscrete Space is Irreducible {{qed}} Category:Irreducible Spaces Category:Trivial Topological Space j50l4l5dfsvui33i4zypvnnzjtiwbcx
By Subspace of Hausdorff Space is Hausdorff, the irreducible components of $X$ are also Hausdorff. By Irreducible Hausdorff Space is Singleton, they can only be singletons. By Trivial Topological Space is Irreducible, every singleton of $X$ is indeed irreducible. {{qed}} Category:Irreducible Spaces 8ylvofgxxdjnjln8zk4c1fjg1geuhjw
=== Open Sets Intersect implies Open Sets are Connected === Let $T = \struct {S, \tau}$ be irreducible in the sense that: :no two non-empty open sets of $T$ are disjoint. Let $U \subseteq S$ be an open set of $T$. {{AimForCont}} $U$ is not connected. Then there exist non-empty open sets $V, W$ of $U$ that are disjoint and whose union is $U$. By Open Set in Open Subspace, $V$ and $W$ are open sets of $T$. Because $V \cap W = \O$, $T$ is not irreducible. This is a contradiction. Thus $U$ is connected. Thus $T = \struct {S, \tau}$ is irreducible in the sense that: :every open set of $T$ is connected.. {{qed|lemma}} === Open Sets are Connected implies Open Sets Intersect === Let $T = \struct {S, \tau}$ be irreducible in the sense that: :every open set of $T$ is connected. Let $V$ and $W$ be open sets of $T$. By definition of topology, their union $V \cup W$ is open in $T$. By hypothesis, $V \cup W$ is connected. By Open Set in Open Subspace, $V$ and $W$ are open sets of $V \cup W$. Because $V \cup W$ is connected, $V \cap W$ is non-empty. Because $V$ and $W$ were arbitrary, $T$ is irreducible. Thus $T = \struct {S, \tau}$ is irreducible in the sense that: :no two non-empty open sets of $T$ are disjoint.
We check all the square of numbers from $99 \, 067$ up to $\floor {9876543210} = 99 \, 380$, with the following constraints: Since all these squares has $9$ as its leftmost digit, the number cannot end with $3$ or $7$. The number cannot end with $0$ since its square will end in $00$. A pandigital number is divisible by $9$, so our number must be divisible by $3$. These constraints leaves us with the around $60$ candidates: :$99069, 99072, 99075, 99078, 99081, 99084, 99096, 99099, 99102, 99105, 99108, 99111, 99114, 99126, 99129, 99132, \dots, 99378$ {{finish}}
We have that: :$10 \, 662 \, 526 \, 601 = 2201^3$ There are no others whose cube root is below $10^{15}$.
Let $T = \struct {S, \tau}$ be an irreducible topological space. Let $U$ be a non-empty open set of $T$. {{AimForCont}} $U$ is not irreducible in $T$. Then $U = V_1 \cup V_2$ for some closed sets $V_1$ and $V_2$ of $\struct {U, \tau_U}$. By definition of subspace topology: :$V_1 = U \cap W_1$ and: :$V_2 = U \cap W_2$ for some closed sets $W_1$ and $W_2$ of $T$. Because $W_1 = S \implies V_1 = U \cap W_1 = U$, it follows that: :$W_1 \ne S$ This is a contradiction, because $V_1$ is a proper subset of $U$. Now $U \ne \O$ implies that $S \setminus U$ is a proper subset of $S$ which is a closed set of $T$. Let $W_3 := W_2 \cup \paren {S \setminus U}$. Then $W_3$ is a closed set of $T$. Also, $W_3 \ne S$, because otherwise $V_2 = U \cap W_2 = U \cap W_3 = U$. Thus $S = W_1 \cup W_3$ shows that $T$ is not irreducible. The result follows by Proof by Contradiction. {{qed}}