Metamath Proof Explorer


Theorem msdcn

Description: The metric function of a metric space is always continuous in the topology generated by it. (Contributed by Mario Carneiro, 5-May-2014) (Revised by Mario Carneiro, 5-Oct-2015)

Ref Expression
Hypotheses msdcn.x X = Base M
msdcn.d D = dist M
msdcn.j J = TopOpen M
msdcn.2 K = topGen ran .
Assertion msdcn M MetSp D X × X J × t J Cn K

Proof

Step Hyp Ref Expression
1 msdcn.x X = Base M
2 msdcn.d D = dist M
3 msdcn.j J = TopOpen M
4 msdcn.2 K = topGen ran .
5 1 2 msmet2 M MetSp D X × X Met X
6 eqid MetOpen D X × X = MetOpen D X × X
7 6 4 metdcn2 D X × X Met X D X × X MetOpen D X × X × t MetOpen D X × X Cn K
8 5 7 syl M MetSp D X × X MetOpen D X × X × t MetOpen D X × X Cn K
9 2 reseq1i D X × X = dist M X × X
10 3 1 9 mstopn M MetSp J = MetOpen D X × X
11 10 10 oveq12d M MetSp J × t J = MetOpen D X × X × t MetOpen D X × X
12 11 oveq1d M MetSp J × t J Cn K = MetOpen D X × X × t MetOpen D X × X Cn K
13 8 12 eleqtrrd M MetSp D X × X J × t J Cn K