Theorem basgen 22904

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 22884	. . . 4 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
2	1	3adant3 1132	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ (topGen‘𝐽))
3		tgtop 22889	. . . 4 ⊢ (𝐽 ∈ Top → (topGen‘𝐽) = 𝐽)
4	3	3ad2ant1 1133	. . 3 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐽) = 𝐽)
5	2, 4	sseqtrd 3967	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) ⊆ 𝐽)
6		simp3 1138	. 2 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → 𝐽 ⊆ (topGen‘𝐵))
7	5, 6	eqssd 3948	1 ⊢ ((𝐽 ∈ Top ∧ 𝐵 ⊆ 𝐽 ∧ 𝐽 ⊆ (topGen‘𝐵)) → (topGen‘𝐵) = 𝐽)

Syntax hints:   → wi 4   ∧ w3a 1086   = wceq 1541   ∈ wcel 2113   ⊆ wss 3898  ‘cfv 6486  topGenctg 17343  Topctop 22809

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2115  ax-9 2123  ax-10 2146  ax-11 2162  ax-12 2182  ax-ext 2705  ax-sep 5236  ax-nul 5246  ax-pow 5305  ax-pr 5372  ax-un 7674

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1544  df-fal 1554  df-ex 1781  df-nf 1785  df-sb 2068  df-mo 2537  df-eu 2566  df-clab 2712  df-cleq 2725  df-clel 2808  df-nfc 2882  df-ne 2930  df-ral 3049  df-rex 3058  df-rab 3397  df-v 3439  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4283  df-if 4475  df-pw 4551  df-sn 4576  df-pr 4578  df-op 4582  df-uni 4859  df-br 5094  df-opab 5156  df-mpt 5175  df-id 5514  df-xp 5625  df-rel 5626  df-cnv 5627  df-co 5628  df-dm 5629  df-iota 6442  df-fun 6488  df-fv 6494  df-topgen 17349  df-top 22810

This theorem is referenced by:  basgen2  22905