Theorem basgen 21170

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 21150	. . . 4 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
2	1	3adant3 1166	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3		tgtop 21155	. . . 4 ⊢ (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
4	3	3ad2ant1 1167	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
5	2, 4	sseqtrd 3866	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6		simp3 1172	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
7	5, 6	eqssd 3844	1 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Syntax hints:   → wi 4   ∧ w3a 1111   = wceq 1656   ∈ wcel 2164   ⊆ wss 3798  ‘cfv 6127  topGenctg 16458  Topctop 21075

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1894  ax-4 1908  ax-5 2009  ax-6 2075  ax-7 2112  ax-8 2166  ax-9 2173  ax-10 2192  ax-11 2207  ax-12 2220  ax-13 2389  ax-ext 2803  ax-sep 5007  ax-nul 5015  ax-pow 5067  ax-pr 5129  ax-un 7214

This theorem depends on definitions:  df-bi 199  df-an 387  df-or 879  df-3an 1113  df-tru 1660  df-ex 1879  df-nf 1883  df-sb 2068  df-mo 2605  df-eu 2640  df-clab 2812  df-cleq 2818  df-clel 2821  df-nfc 2958  df-ral 3122  df-rex 3123  df-rab 3126  df-v 3416  df-sbc 3663  df-dif 3801  df-un 3803  df-in 3805  df-ss 3812  df-nul 4147  df-if 4309  df-pw 4382  df-sn 4400  df-pr 4402  df-op 4406  df-uni 4661  df-br 4876  df-opab 4938  df-mpt 4955  df-id 5252  df-xp 5352  df-rel 5353  df-cnv 5354  df-co 5355  df-dm 5356  df-iota 6090  df-fun 6129  df-fv 6135  df-topgen 16464  df-top 21076

This theorem is referenced by:  basgen2  21171