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Theorem indistps2ALT 23182
Description: The indiscrete topology on a set 𝐴 expressed as a topological space, using direct component assignments. Here we show how to derive the direct component assignment version indistps2 23180 from the structural version indistps 23179. (Contributed by NM, 24-Oct-2012.) (New usage is discouraged.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
indistps2ALT.a (Base‘𝐾) = 𝐴
indistps2ALT.j (TopOpen‘𝐾) = {∅, 𝐴}
Assertion
Ref Expression
indistps2ALT 𝐾 ∈ TopSp

Proof of Theorem indistps2ALT
StepHypRef Expression
1 indistps2ALT.a . . . 4 (Base‘𝐾) = 𝐴
2 fvex 6894 . . . 4 (Base‘𝐾) ∈ V
31, 2eqeltrri 2859 . . 3 𝐴 ∈ V
4 indistopon 23169 . . 3 (𝐴 ∈ V → {∅, 𝐴} ∈ (TopOn‘𝐴))
53, 4ax-mp 5 . 2 {∅, 𝐴} ∈ (TopOn‘𝐴)
61eqcomi 2771 . . 3 𝐴 = (Base‘𝐾)
7 indistps2ALT.j . . . 4 (TopOpen‘𝐾) = {∅, 𝐴}
87eqcomi 2771 . . 3 {∅, 𝐴} = (TopOpen‘𝐾)
96, 8istps 23102 . 2 (𝐾 ∈ TopSp ↔ {∅, 𝐴} ∈ (TopOn‘𝐴))
105, 9mpbir 234 1 𝐾 ∈ TopSp
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   = wceq 1569  wcel 2142  Vcvv 3454  c0 4285  {cpr 4590  cfv 6536  Basecbs 17275  TopOpenctopn 17480  TopOnctopon 23078  TopSpctps 23100
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734  ax-sep 5256  ax-nul 5268  ax-pow 5335  ax-pr 5403  ax-un 7734
This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-nf 1813  df-sb 2096  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3416  df-v 3456  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-pw 4563  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-opab 5173  df-mpt 5192  df-id 5555  df-xp 5666  df-rel 5667  df-cnv 5668  df-co 5669  df-dm 5670  df-iota 6492  df-fun 6538  df-fv 6544  df-top 23062  df-topon 23079  df-topsp 23101
This theorem is used by: (None)
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