Theorem basgen 12031

Description: Given a topology 𝐽, show that a subset 𝐵 satisfying the third antecedent is a basis for it. Lemma 2.3 of [Munkres] p. 81 using abbreviations. (Contributed by NM, 22-Jul-2006.) (Revised by Mario Carneiro, 2-Sep-2015.)

Assertion

Ref	Expression
basgen	⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Proof of Theorem basgen

Step	Hyp	Ref	Expression
1		tgss 12014	. . . 4 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
2	1	3adant3 969	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3		tgtop 12019	. . . 4 ⊢ (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
4	3	3ad2ant1 970	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
5	2, 4	sseqtrd 3085	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6		simp3 951	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
7	5, 6	eqssd 3064	1 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Syntax hints:   → wi 4   ∧ w3a 930   = wceq 1299   ∈ wcel 1448   ⊆ wss 3021  ‘cfv 5059  topGenctg 11917  Topctop 11946

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 671  ax-5 1391  ax-7 1392  ax-gen 1393  ax-ie1 1437  ax-ie2 1438  ax-8 1450  ax-10 1451  ax-11 1452  ax-i12 1453  ax-bndl 1454  ax-4 1455  ax-13 1459  ax-14 1460  ax-17 1474  ax-i9 1478  ax-ial 1482  ax-i5r 1483  ax-ext 2082  ax-sep 3986  ax-pow 4038  ax-pr 4069  ax-un 4293

This theorem depends on definitions:  df-bi 116  df-3an 932  df-tru 1302  df-nf 1405  df-sb 1704  df-eu 1963  df-mo 1964  df-clab 2087  df-cleq 2093  df-clel 2096  df-nfc 2229  df-ral 2380  df-rex 2381  df-v 2643  df-sbc 2863  df-un 3025  df-in 3027  df-ss 3034  df-pw 3459  df-sn 3480  df-pr 3481  df-op 3483  df-uni 3684  df-br 3876  df-opab 3930  df-mpt 3931  df-id 4153  df-xp 4483  df-rel 4484  df-cnv 4485  df-co 4486  df-dm 4487  df-iota 5024  df-fun 5061  df-fv 5067  df-topgen 11923  df-top 11947

This theorem is referenced by:  basgen2  12032