Metamath Proof Explorer


Theorem dnsconst

Description: If a continuous mapping to a T_1 space is constant on a dense subset, it is constant on the entire space. Note that ( ( clsJ )A ) = X means " A is dense in X " and A C_ (`' F " { P } ) means " F is constant on A ` " (see funconstss ). (Contributed by NM, 15-Mar-2007) (Proof shortened by Mario Carneiro, 21-Aug-2015)

Ref Expression
Hypotheses dnsconst.1 X = J
dnsconst.2 Y = K
Assertion dnsconst K Fre F J Cn K P Y A F -1 P cls J A = X F : X P

Proof

Step Hyp Ref Expression
1 dnsconst.1 X = J
2 dnsconst.2 Y = K
3 simplr K Fre F J Cn K P Y A F -1 P cls J A = X F J Cn K
4 1 2 cnf F J Cn K F : X Y
5 ffn F : X Y F Fn X
6 3 4 5 3syl K Fre F J Cn K P Y A F -1 P cls J A = X F Fn X
7 simpr3 K Fre F J Cn K P Y A F -1 P cls J A = X cls J A = X
8 simpll K Fre F J Cn K P Y A F -1 P cls J A = X K Fre
9 simpr1 K Fre F J Cn K P Y A F -1 P cls J A = X P Y
10 2 t1sncld K Fre P Y P Clsd K
11 8 9 10 syl2anc K Fre F J Cn K P Y A F -1 P cls J A = X P Clsd K
12 cnclima F J Cn K P Clsd K F -1 P Clsd J
13 3 11 12 syl2anc K Fre F J Cn K P Y A F -1 P cls J A = X F -1 P Clsd J
14 simpr2 K Fre F J Cn K P Y A F -1 P cls J A = X A F -1 P
15 1 clsss2 F -1 P Clsd J A F -1 P cls J A F -1 P
16 13 14 15 syl2anc K Fre F J Cn K P Y A F -1 P cls J A = X cls J A F -1 P
17 7 16 eqsstrrd K Fre F J Cn K P Y A F -1 P cls J A = X X F -1 P
18 fconst3 F : X P F Fn X X F -1 P
19 6 17 18 sylanbrc K Fre F J Cn K P Y A F -1 P cls J A = X F : X P