Metamath Proof Explorer


Theorem nlpineqsn

Description: For every point p of a subset A of X with no limit points, there exists an open set n that intersects A only at p . (Contributed by ML, 23-Mar-2021)

Ref Expression
Hypothesis nlpineqsn.x X = J
Assertion nlpineqsn J Top A X limPt J A = p A n J p n n A = p

Proof

Step Hyp Ref Expression
1 nlpineqsn.x X = J
2 simp1 J Top A X p A J Top
3 simp2 J Top A X p A A X
4 ssel2 A X p A p X
5 4 3adant1 J Top A X p A p X
6 2 3 5 3jca J Top A X p A J Top A X p X
7 noel ¬ p
8 eleq2 limPt J A = p limPt J A p
9 7 8 mtbiri limPt J A = ¬ p limPt J A
10 9 adantl J Top A X p X limPt J A = ¬ p limPt J A
11 1 islp3 J Top A X p X p limPt J A n J p n n A p
12 11 adantr J Top A X p X limPt J A = p limPt J A n J p n n A p
13 10 12 mtbid J Top A X p X limPt J A = ¬ n J p n n A p
14 nne ¬ n A p n A p =
15 14 anbi2i p n ¬ n A p p n n A p =
16 annim p n ¬ n A p ¬ p n n A p
17 15 16 bitr3i p n n A p = ¬ p n n A p
18 17 rexbii n J p n n A p = n J ¬ p n n A p
19 rexnal n J ¬ p n n A p ¬ n J p n n A p
20 18 19 bitri n J p n n A p = ¬ n J p n n A p
21 13 20 sylibr J Top A X p X limPt J A = n J p n n A p =
22 6 21 sylan J Top A X p A limPt J A = n J p n n A p =
23 indif2 n A p = n A p
24 23 eqeq1i n A p = n A p =
25 ssdif0 n A p n A p =
26 24 25 bitr4i n A p = n A p
27 elin p n A p n p A
28 n0i p n A ¬ n A =
29 biorf ¬ n A = n A = p n A = n A = p
30 28 29 syl p n A n A = p n A = n A = p
31 sssn n A p n A = n A = p
32 30 31 syl6rbbr p n A n A p n A = p
33 27 32 sylbir p n p A n A p n A = p
34 26 33 syl5bb p n p A n A p = n A = p
35 34 ancoms p A p n n A p = n A = p
36 35 pm5.32da p A p n n A p = p n n A = p
37 36 rexbidv p A n J p n n A p = n J p n n A = p
38 37 3ad2ant3 J Top A X p A n J p n n A p = n J p n n A = p
39 38 adantr J Top A X p A limPt J A = n J p n n A p = n J p n n A = p
40 22 39 mpbid J Top A X p A limPt J A = n J p n n A = p
41 40 3an1rs J Top A X limPt J A = p A n J p n n A = p
42 41 ralrimiva J Top A X limPt J A = p A n J p n n A = p