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 JTopAXlimPtJA=pAnJpnnA=p

Proof

Step Hyp Ref Expression
1 nlpineqsn.x X=J
2 simp1 JTopAXpAJTop
3 simp2 JTopAXpAAX
4 ssel2 AXpApX
5 4 3adant1 JTopAXpApX
6 2 3 5 3jca JTopAXpAJTopAXpX
7 noel ¬p
8 eleq2 limPtJA=plimPtJAp
9 7 8 mtbiri limPtJA=¬plimPtJA
10 9 adantl JTopAXpXlimPtJA=¬plimPtJA
11 1 islp3 JTopAXpXplimPtJAnJpnnAp
12 11 adantr JTopAXpXlimPtJA=plimPtJAnJpnnAp
13 10 12 mtbid JTopAXpXlimPtJA=¬nJpnnAp
14 nne ¬nApnAp=
15 14 anbi2i pn¬nAppnnAp=
16 annim pn¬nAp¬pnnAp
17 15 16 bitr3i pnnAp=¬pnnAp
18 17 rexbii nJpnnAp=nJ¬pnnAp
19 rexnal nJ¬pnnAp¬nJpnnAp
20 18 19 bitri nJpnnAp=¬nJpnnAp
21 13 20 sylibr JTopAXpXlimPtJA=nJpnnAp=
22 6 21 sylan JTopAXpAlimPtJA=nJpnnAp=
23 indif2 nAp=nAp
24 23 eqeq1i nAp=nAp=
25 ssdif0 nApnAp=
26 24 25 bitr4i nAp=nAp
27 elin pnApnpA
28 sssn nApnA=nA=p
29 n0i pnA¬nA=
30 biorf ¬nA=nA=pnA=nA=p
31 29 30 syl pnAnA=pnA=nA=p
32 28 31 bitr4id pnAnApnA=p
33 27 32 sylbir pnpAnApnA=p
34 26 33 bitrid pnpAnAp=nA=p
35 34 ancoms pApnnAp=nA=p
36 35 pm5.32da pApnnAp=pnnA=p
37 36 rexbidv pAnJpnnAp=nJpnnA=p
38 37 3ad2ant3 JTopAXpAnJpnnAp=nJpnnA=p
39 38 adantr JTopAXpAlimPtJA=nJpnnAp=nJpnnA=p
40 22 39 mpbid JTopAXpAlimPtJA=nJpnnA=p
41 40 3an1rs JTopAXlimPtJA=pAnJpnnA=p
42 41 ralrimiva JTopAXlimPtJA=pAnJpnnA=p