Metamath Proof Explorer


Theorem imasf1obl

Description: The image of a metric space ball. (Contributed by Mario Carneiro, 28-Aug-2015)

Ref Expression
Hypotheses imasf1obl.u φ U = F 𝑠 R
imasf1obl.v φ V = Base R
imasf1obl.f φ F : V 1-1 onto B
imasf1obl.r φ R Z
imasf1obl.e E = dist R V × V
imasf1obl.d D = dist U
imasf1obl.m φ E ∞Met V
imasf1obl.x φ P V
imasf1obl.s φ S *
Assertion imasf1obl φ F P ball D S = F P ball E S

Proof

Step Hyp Ref Expression
1 imasf1obl.u φ U = F 𝑠 R
2 imasf1obl.v φ V = Base R
3 imasf1obl.f φ F : V 1-1 onto B
4 imasf1obl.r φ R Z
5 imasf1obl.e E = dist R V × V
6 imasf1obl.d D = dist U
7 imasf1obl.m φ E ∞Met V
8 imasf1obl.x φ P V
9 imasf1obl.s φ S *
10 f1ocnvfv2 F : V 1-1 onto B x B F F -1 x = x
11 3 10 sylan φ x B F F -1 x = x
12 11 oveq2d φ x B F P D F F -1 x = F P D x
13 1 adantr φ x B U = F 𝑠 R
14 2 adantr φ x B V = Base R
15 3 adantr φ x B F : V 1-1 onto B
16 4 adantr φ x B R Z
17 7 adantr φ x B E ∞Met V
18 8 adantr φ x B P V
19 f1ocnv F : V 1-1 onto B F -1 : B 1-1 onto V
20 3 19 syl φ F -1 : B 1-1 onto V
21 f1of F -1 : B 1-1 onto V F -1 : B V
22 20 21 syl φ F -1 : B V
23 22 ffvelrnda φ x B F -1 x V
24 13 14 15 16 5 6 17 18 23 imasdsf1o φ x B F P D F F -1 x = P E F -1 x
25 12 24 eqtr3d φ x B F P D x = P E F -1 x
26 25 breq1d φ x B F P D x < S P E F -1 x < S
27 9 adantr φ x B S *
28 elbl2 E ∞Met V S * P V F -1 x V F -1 x P ball E S P E F -1 x < S
29 17 27 18 23 28 syl22anc φ x B F -1 x P ball E S P E F -1 x < S
30 26 29 bitr4d φ x B F P D x < S F -1 x P ball E S
31 30 pm5.32da φ x B F P D x < S x B F -1 x P ball E S
32 1 2 3 4 5 6 7 imasf1oxmet φ D ∞Met B
33 f1of F : V 1-1 onto B F : V B
34 3 33 syl φ F : V B
35 34 8 ffvelrnd φ F P B
36 elbl D ∞Met B F P B S * x F P ball D S x B F P D x < S
37 32 35 9 36 syl3anc φ x F P ball D S x B F P D x < S
38 f1ofn F -1 : B 1-1 onto V F -1 Fn B
39 elpreima F -1 Fn B x F -1 -1 P ball E S x B F -1 x P ball E S
40 20 38 39 3syl φ x F -1 -1 P ball E S x B F -1 x P ball E S
41 31 37 40 3bitr4d φ x F P ball D S x F -1 -1 P ball E S
42 41 eqrdv φ F P ball D S = F -1 -1 P ball E S
43 imacnvcnv F -1 -1 P ball E S = F P ball E S
44 42 43 eqtrdi φ F P ball D S = F P ball E S