Metamath Proof Explorer


Theorem ressms

Description: The restriction of a metric space is a metric space. (Contributed by Mario Carneiro, 24-Aug-2015)

Ref Expression
Assertion ressms K MetSp A V K 𝑠 A MetSp

Proof

Step Hyp Ref Expression
1 msxms K MetSp K ∞MetSp
2 ressxms K ∞MetSp A V K 𝑠 A ∞MetSp
3 1 2 sylan K MetSp A V K 𝑠 A ∞MetSp
4 eqid Base K = Base K
5 eqid dist K Base K × Base K = dist K Base K × Base K
6 4 5 msmet K MetSp dist K Base K × Base K Met Base K
7 6 adantr K MetSp A V dist K Base K × Base K Met Base K
8 metres dist K Base K × Base K Met Base K dist K Base K × Base K A × A Met Base K A
9 7 8 syl K MetSp A V dist K Base K × Base K A × A Met Base K A
10 resres dist K Base K × Base K A × A = dist K Base K × Base K A × A
11 inxp Base K × Base K A × A = Base K A × Base K A
12 11 reseq2i dist K Base K × Base K A × A = dist K Base K A × Base K A
13 10 12 eqtri dist K Base K × Base K A × A = dist K Base K A × Base K A
14 eqid K 𝑠 A = K 𝑠 A
15 eqid dist K = dist K
16 14 15 ressds A V dist K = dist K 𝑠 A
17 16 adantl K MetSp A V dist K = dist K 𝑠 A
18 incom Base K A = A Base K
19 14 4 ressbas A V A Base K = Base K 𝑠 A
20 19 adantl K MetSp A V A Base K = Base K 𝑠 A
21 18 20 syl5eq K MetSp A V Base K A = Base K 𝑠 A
22 21 sqxpeqd K MetSp A V Base K A × Base K A = Base K 𝑠 A × Base K 𝑠 A
23 17 22 reseq12d K MetSp A V dist K Base K A × Base K A = dist K 𝑠 A Base K 𝑠 A × Base K 𝑠 A
24 13 23 syl5eq K MetSp A V dist K Base K × Base K A × A = dist K 𝑠 A Base K 𝑠 A × Base K 𝑠 A
25 21 fveq2d K MetSp A V Met Base K A = Met Base K 𝑠 A
26 9 24 25 3eltr3d K MetSp A V dist K 𝑠 A Base K 𝑠 A × Base K 𝑠 A Met Base K 𝑠 A
27 eqid TopOpen K = TopOpen K
28 14 27 resstopn TopOpen K 𝑡 A = TopOpen K 𝑠 A
29 eqid Base K 𝑠 A = Base K 𝑠 A
30 eqid dist K 𝑠 A Base K 𝑠 A × Base K 𝑠 A = dist K 𝑠 A Base K 𝑠 A × Base K 𝑠 A
31 28 29 30 isms K 𝑠 A MetSp K 𝑠 A ∞MetSp dist K 𝑠 A Base K 𝑠 A × Base K 𝑠 A Met Base K 𝑠 A
32 3 26 31 sylanbrc K MetSp A V K 𝑠 A MetSp