Metamath Proof Explorer


Theorem metdcn2

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, 4-Sep-2015)

Ref Expression
Hypotheses xmetdcn2.1 J=MetOpenD
metdcn2.2 K=topGenran.
Assertion metdcn2 DMetXDJ×tJCnK

Proof

Step Hyp Ref Expression
1 xmetdcn2.1 J=MetOpenD
2 metdcn2.2 K=topGenran.
3 metxmet DMetXD∞MetX
4 eqid ordTop=ordTop
5 1 4 xmetdcn D∞MetXDJ×tJCnordTop
6 3 5 syl DMetXDJ×tJCnordTop
7 letopon ordTopTopOn*
8 metf DMetXD:X×X
9 8 frnd DMetXranD
10 ressxr *
11 10 a1i DMetX*
12 cnrest2 ordTopTopOn*ranD*DJ×tJCnordTopDJ×tJCnordTop𝑡
13 7 9 11 12 mp3an2i DMetXDJ×tJCnordTopDJ×tJCnordTop𝑡
14 6 13 mpbid DMetXDJ×tJCnordTop𝑡
15 eqid ordTop𝑡=ordTop𝑡
16 15 xrtgioo topGenran.=ordTop𝑡
17 2 16 eqtri K=ordTop𝑡
18 17 oveq2i J×tJCnK=J×tJCnordTop𝑡
19 14 18 eleqtrrdi DMetXDJ×tJCnK