Metamath Proof Explorer


Theorem metdcn

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 = MetOpen D
metdcn.2 K = TopOpen fld
Assertion metdcn D Met X D J × t J Cn K

Proof

Step Hyp Ref Expression
1 xmetdcn2.1 J = MetOpen D
2 metdcn.2 K = TopOpen fld
3 2 tgioo2 topGen ran . = K 𝑡
4 3 oveq2i J × t J Cn topGen ran . = J × t J Cn K 𝑡
5 2 cnfldtop K Top
6 cnrest2r K Top J × t J Cn K 𝑡 J × t J Cn K
7 5 6 ax-mp J × t J Cn K 𝑡 J × t J Cn K
8 4 7 eqsstri J × t J Cn topGen ran . J × t J Cn K
9 eqid topGen ran . = topGen ran .
10 1 9 metdcn2 D Met X D J × t J Cn topGen ran .
11 8 10 sselid D Met X D J × t J Cn K