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Theorem basgen 12279
 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 12262 . . . 4 ((𝐽 ∈ Top ∧ 𝐵𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
213adant3 1002 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3 tgtop 12267 . . . 4 (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
433ad2ant1 1003 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
52, 4sseqtrd 3136 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6 simp3 984 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
75, 6eqssd 3115 1 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ w3a 963   = wceq 1332   ∈ wcel 1481   ⊆ wss 3072  ‘cfv 5127  topGenctg 12165  Topctop 12194 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-13 1492  ax-14 1493  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122  ax-sep 4050  ax-pow 4102  ax-pr 4135  ax-un 4359 This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1737  df-eu 2003  df-mo 2004  df-clab 2127  df-cleq 2133  df-clel 2136  df-nfc 2271  df-ral 2422  df-rex 2423  df-v 2689  df-sbc 2911  df-un 3076  df-in 3078  df-ss 3085  df-pw 3513  df-sn 3534  df-pr 3535  df-op 3537  df-uni 3741  df-br 3934  df-opab 3994  df-mpt 3995  df-id 4219  df-xp 4549  df-rel 4550  df-cnv 4551  df-co 4552  df-dm 4553  df-iota 5092  df-fun 5129  df-fv 5135  df-topgen 12171  df-top 12195 This theorem is referenced by:  basgen2  12280
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