Metamath Proof Explorer


Theorem conndisj

Description: If a topology is connected, its underlying set can't be partitioned into two nonempty non-overlapping open sets. (Contributed by FL, 16-Nov-2008) (Proof shortened by Mario Carneiro, 10-Mar-2015)

Ref Expression
Hypotheses isconn.1 X=J
connclo.1 φJConn
connclo.2 φAJ
connclo.3 φA
conndisj.4 φBJ
conndisj.5 φB
conndisj.6 φAB=
Assertion conndisj φABX

Proof

Step Hyp Ref Expression
1 isconn.1 X=J
2 connclo.1 φJConn
3 connclo.2 φAJ
4 connclo.3 φA
5 conndisj.4 φBJ
6 conndisj.5 φB
7 conndisj.6 φAB=
8 elssuni AJAJ
9 3 8 syl φAJ
10 9 1 sseqtrrdi φAX
11 uneqdifeq AXAB=AB=XXA=B
12 10 7 11 syl2anc φAB=XXA=B
13 simpr φXA=BXA=B
14 13 difeq2d φXA=BXXA=XB
15 dfss4 AXXXA=A
16 10 15 sylib φXXA=A
17 16 adantr φXA=BXXA=A
18 2 adantr φXA=BJConn
19 5 adantr φXA=BBJ
20 6 adantr φXA=BB
21 1 isconn JConnJTopJClsdJ=X
22 21 simplbi JConnJTop
23 2 22 syl φJTop
24 1 opncld JTopAJXAClsdJ
25 23 3 24 syl2anc φXAClsdJ
26 25 adantr φXA=BXAClsdJ
27 13 26 eqeltrrd φXA=BBClsdJ
28 1 18 19 20 27 connclo φXA=BB=X
29 28 difeq2d φXA=BXB=XX
30 difid XX=
31 29 30 eqtrdi φXA=BXB=
32 14 17 31 3eqtr3d φXA=BA=
33 32 ex φXA=BA=
34 12 33 sylbid φAB=XA=
35 34 necon3d φAABX
36 4 35 mpd φABX