Theorem basgen 23016

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 22996	. . . 4 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
2	1	3adant3 1132	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3		tgtop 23001	. . . 4 ⊢ (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
4	3	3ad2ant1 1133	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
5	2, 4	sseqtrd 4049	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6		simp3 1138	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
7	5, 6	eqssd 4026	1 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Syntax hints:   → wi 4   ∧ w3a 1087   = wceq 1537   ∈ wcel 2108   ⊆ wss 3976  ‘cfv 6573  topGenctg 17497  Topctop 22920

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2158  ax-12 2178  ax-ext 2711  ax-sep 5317  ax-nul 5324  ax-pow 5383  ax-pr 5447  ax-un 7770

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 847  df-3an 1089  df-tru 1540  df-fal 1550  df-ex 1778  df-nf 1782  df-sb 2065  df-mo 2543  df-eu 2572  df-clab 2718  df-cleq 2732  df-clel 2819  df-nfc 2895  df-ral 3068  df-rex 3077  df-rab 3444  df-v 3490  df-dif 3979  df-un 3981  df-in 3983  df-ss 3993  df-nul 4353  df-if 4549  df-pw 4624  df-sn 4649  df-pr 4651  df-op 4655  df-uni 4932  df-br 5167  df-opab 5229  df-mpt 5250  df-id 5593  df-xp 5706  df-rel 5707  df-cnv 5708  df-co 5709  df-dm 5710  df-iota 6525  df-fun 6575  df-fv 6581  df-topgen 17503  df-top 22921

This theorem is referenced by:  basgen2  23017