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Theorem basgen 21012
 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 20992 . . . 4 ((𝐽 ∈ Top ∧ 𝐵𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
213adant3 1125 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3 tgtop 20997 . . . 4 (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
433ad2ant1 1126 . . 3 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
52, 4sseqtrd 3788 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6 simp3 1131 . 2 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
75, 6eqssd 3767 1 ((𝐽 ∈ Top ∧ 𝐵𝐽𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∧ w3a 1070   = wceq 1630   ∈ wcel 2144   ⊆ wss 3721  ‘cfv 6031  topGenctg 16305  Topctop 20917 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1869  ax-4 1884  ax-5 1990  ax-6 2056  ax-7 2092  ax-8 2146  ax-9 2153  ax-10 2173  ax-11 2189  ax-12 2202  ax-13 2407  ax-ext 2750  ax-sep 4912  ax-nul 4920  ax-pow 4971  ax-pr 5034  ax-un 7095 This theorem depends on definitions:  df-bi 197  df-an 383  df-or 827  df-3an 1072  df-tru 1633  df-ex 1852  df-nf 1857  df-sb 2049  df-eu 2621  df-mo 2622  df-clab 2757  df-cleq 2763  df-clel 2766  df-nfc 2901  df-ne 2943  df-ral 3065  df-rex 3066  df-rab 3069  df-v 3351  df-sbc 3586  df-dif 3724  df-un 3726  df-in 3728  df-ss 3735  df-nul 4062  df-if 4224  df-pw 4297  df-sn 4315  df-pr 4317  df-op 4321  df-uni 4573  df-br 4785  df-opab 4845  df-mpt 4862  df-id 5157  df-xp 5255  df-rel 5256  df-cnv 5257  df-co 5258  df-dm 5259  df-iota 5994  df-fun 6033  df-fv 6039  df-topgen 16311  df-top 20918 This theorem is referenced by:  basgen2  21013
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