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Theorem basgen 23011
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 22991 . . . 4 ((𝐽 ∈ Top ∧ 𝐵𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
213adant3 1131 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3 tgtop 22996 . . . 4 (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
433ad2ant1 1132 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
52, 4sseqtrd 4036 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6 simp3 1137 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
75, 6eqssd 4013 1 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)
Colors of variables: wff setvar class
Syntax hints:  wi 4  w3a 1086   = wceq 1537  wcel 2106  wss 3963  cfv 6563  topGenctg 17484  Topctop 22915
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706  ax-sep 5302  ax-nul 5312  ax-pow 5371  ax-pr 5438  ax-un 7754
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1540  df-fal 1550  df-ex 1777  df-nf 1781  df-sb 2063  df-mo 2538  df-eu 2567  df-clab 2713  df-cleq 2727  df-clel 2814  df-nfc 2890  df-ral 3060  df-rex 3069  df-rab 3434  df-v 3480  df-dif 3966  df-un 3968  df-in 3970  df-ss 3980  df-nul 4340  df-if 4532  df-pw 4607  df-sn 4632  df-pr 4634  df-op 4638  df-uni 4913  df-br 5149  df-opab 5211  df-mpt 5232  df-id 5583  df-xp 5695  df-rel 5696  df-cnv 5697  df-co 5698  df-dm 5699  df-iota 6516  df-fun 6565  df-fv 6571  df-topgen 17490  df-top 22916
This theorem is referenced by:  basgen2  23012
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