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=BaseM
msdcn.d D=distM
msdcn.j J=TopOpenM
msdcn.2 K=topGenran.
Assertion msdcn MMetSpDX×XJ×tJCnK

Proof

Step Hyp Ref Expression
1 msdcn.x X=BaseM
2 msdcn.d D=distM
3 msdcn.j J=TopOpenM
4 msdcn.2 K=topGenran.
5 1 2 msmet2 MMetSpDX×XMetX
6 eqid MetOpenDX×X=MetOpenDX×X
7 6 4 metdcn2 DX×XMetXDX×XMetOpenDX×X×tMetOpenDX×XCnK
8 5 7 syl MMetSpDX×XMetOpenDX×X×tMetOpenDX×XCnK
9 2 reseq1i DX×X=distMX×X
10 3 1 9 mstopn MMetSpJ=MetOpenDX×X
11 10 10 oveq12d MMetSpJ×tJ=MetOpenDX×X×tMetOpenDX×X
12 11 oveq1d MMetSpJ×tJCnK=MetOpenDX×X×tMetOpenDX×XCnK
13 8 12 eleqtrrd MMetSpDX×XJ×tJCnK