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Theorem basgen 22995
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 ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Proof of Theorem basgen
StepHypRef Expression
1 tgss 22975 . . . 4 ((𝐽 ∈ Top ∧ 𝐵𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
213adant3 1133 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3 tgtop 22980 . . . 4 (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
433ad2ant1 1134 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
52, 4sseqtrd 4020 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6 simp3 1139 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
75, 6eqssd 4001 1 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)
Colors of variables: wff setvar class
Syntax hints:  wi 4  w3a 1087   = wceq 1540  wcel 2108  wss 3951  cfv 6561  topGenctg 17482  Topctop 22899
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-11 2157  ax-12 2177  ax-ext 2708  ax-sep 5296  ax-nul 5306  ax-pow 5365  ax-pr 5432  ax-un 7755
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-3an 1089  df-tru 1543  df-fal 1553  df-ex 1780  df-nf 1784  df-sb 2065  df-mo 2540  df-eu 2569  df-clab 2715  df-cleq 2729  df-clel 2816  df-nfc 2892  df-ral 3062  df-rex 3071  df-rab 3437  df-v 3482  df-dif 3954  df-un 3956  df-in 3958  df-ss 3968  df-nul 4334  df-if 4526  df-pw 4602  df-sn 4627  df-pr 4629  df-op 4633  df-uni 4908  df-br 5144  df-opab 5206  df-mpt 5226  df-id 5578  df-xp 5691  df-rel 5692  df-cnv 5693  df-co 5694  df-dm 5695  df-iota 6514  df-fun 6563  df-fv 6569  df-topgen 17488  df-top 22900
This theorem is referenced by:  basgen2  22996
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