Metamath Proof Explorer


Theorem kur14lem8

Description: Lemma for kur14 . Show that the set T contains at most 1 4 elements. (It could be less if some of the operators take the same value for a given set, but Kuratowski showed that this upper bound of 1 4 is tight in the sense that there exist topological spaces and subsets of these spaces for which all 1 4 generated sets are distinct, and indeed the real numbers form such a topological space.) (Contributed by Mario Carneiro, 11-Feb-2015)

Ref Expression
Hypotheses kur14lem.j J Top
kur14lem.x X = J
kur14lem.k K = cls J
kur14lem.i I = int J
kur14lem.a A X
kur14lem.b B = X K A
kur14lem.c C = K X A
kur14lem.d D = I K A
kur14lem.t T = A X A K A B C I A K B D K I A I C K D I K B K I C I K I A
Assertion kur14lem8 T Fin T 14

Proof

Step Hyp Ref Expression
1 kur14lem.j J Top
2 kur14lem.x X = J
3 kur14lem.k K = cls J
4 kur14lem.i I = int J
5 kur14lem.a A X
6 kur14lem.b B = X K A
7 kur14lem.c C = K X A
8 kur14lem.d D = I K A
9 kur14lem.t T = A X A K A B C I A K B D K I A I C K D I K B K I C I K I A
10 eqid A X A K A B C I A K B D K I A = A X A K A B C I A K B D K I A
11 eqid A X A K A B C I A = A X A K A B C I A
12 hashtplei A X A K A Fin A X A K A 3
13 hashtplei B C I A Fin B C I A 3
14 3nn0 3 0
15 3p3e6 3 + 3 = 6
16 11 12 13 14 14 15 hashunlei A X A K A B C I A Fin A X A K A B C I A 6
17 hashtplei K B D K I A Fin K B D K I A 3
18 6nn0 6 0
19 6p3e9 6 + 3 = 9
20 10 16 17 18 14 19 hashunlei A X A K A B C I A K B D K I A Fin A X A K A B C I A K B D K I A 9
21 eqid I C K D I K B K I C I K I A = I C K D I K B K I C I K I A
22 hashtplei I C K D I K B Fin I C K D I K B 3
23 hashprlei K I C I K I A Fin K I C I K I A 2
24 2nn0 2 0
25 3p2e5 3 + 2 = 5
26 21 22 23 14 24 25 hashunlei I C K D I K B K I C I K I A Fin I C K D I K B K I C I K I A 5
27 9nn0 9 0
28 5nn0 5 0
29 9p5e14 9 + 5 = 14
30 9 20 26 27 28 29 hashunlei T Fin T 14