Metamath Proof Explorer


Theorem xpstps

Description: A binary product of topologies is a topological space. (Contributed by Mario Carneiro, 27-Aug-2015)

Ref Expression
Hypothesis xpstps.t ⊢ T = R × 𝑠 S
Assertion xpstps ⊢ R ∈ TopSp ∧ S ∈ TopSp → T ∈ TopSp

Proof

Step Hyp Ref Expression
1 xpstps.t ⊢ T = R × 𝑠 S
2 eqid ⊢ Base R = Base R
3 eqid ⊢ Base S = Base S
4 simpl ⊢ R ∈ TopSp ∧ S ∈ TopSp → R ∈ TopSp
5 simpr ⊢ R ∈ TopSp ∧ S ∈ TopSp → S ∈ TopSp
6 eqid ⊢ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y = x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y
7 eqid ⊢ Scalar ⁡ R = Scalar ⁡ R
8 eqid ⊢ Scalar ⁡ R ⨉ 𝑠 ∅ R 1 𝑜 S = Scalar ⁡ R ⨉ 𝑠 ∅ R 1 𝑜 S
9 1 2 3 4 5 6 7 8 xpsval ⊢ R ∈ TopSp ∧ S ∈ TopSp → T = x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y -1 “ 𝑠 Scalar ⁡ R ⨉ 𝑠 ∅ R 1 𝑜 S
10 1 2 3 4 5 6 7 8 xpsrnbas ⊢ R ∈ TopSp ∧ S ∈ TopSp → ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y = Base Scalar ⁡ R ⨉ 𝑠 ∅ R 1 𝑜 S
11 6 xpsff1o2 ⊢ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y : Base R × Base S ⟶ 1-1 onto ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y
12 11 a1i ⊢ R ∈ TopSp ∧ S ∈ TopSp → x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y : Base R × Base S ⟶ 1-1 onto ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y
13 f1ocnv ⊢ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y : Base R × Base S ⟶ 1-1 onto ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y → x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y -1 : ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y ⟶ 1-1 onto Base R × Base S
14 f1ofo ⊢ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y -1 : ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y ⟶ 1-1 onto Base R × Base S → x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y -1 : ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y ⟶ onto Base R × Base S
15 12 13 14 3syl ⊢ R ∈ TopSp ∧ S ∈ TopSp → x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y -1 : ran ⁡ x ∈ Base R , y ∈ Base S ⟼ ∅ x 1 𝑜 y ⟶ onto Base R × Base S
16 fvexd ⊢ R ∈ TopSp ∧ S ∈ TopSp → Scalar ⁡ R ∈ V
17 2on ⊢ 2 𝑜 ∈ On
18 17 a1i ⊢ R ∈ TopSp ∧ S ∈ TopSp → 2 𝑜 ∈ On
19 xpscf ⊢ ∅ R 1 𝑜 S : 2 𝑜 ⟶ TopSp ↔ R ∈ TopSp ∧ S ∈ TopSp
20 19 biimpri ⊢ R ∈ TopSp ∧ S ∈ TopSp → ∅ R 1 𝑜 S : 2 𝑜 ⟶ TopSp
21 8 16 18 20 prdstps ⊢ R ∈ TopSp ∧ S ∈ TopSp → Scalar ⁡ R ⨉ 𝑠 ∅ R 1 𝑜 S ∈ TopSp
22 9 10 15 21 imastps ⊢ R ∈ TopSp ∧ S ∈ TopSp → T ∈ TopSp