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Theorem indistps2ALT 23293
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 23291 from the structural version indistps 23290. (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 6886 . . . 4 (Base‘𝐾) ∈ V
31, 2eqeltrri 2857 . . 3 𝐴 ∈ V
4 indistopon 23280 . . 3 (𝐴 ∈ V → {∅, 𝐴} ∈ (TopOn‘𝐴))
53, 4ax-mp 5 . 2 {∅, 𝐴} ∈ (TopOn‘𝐴)
61eqcomi 2769 . . 3 𝐴 = (Base‘𝐾)
7 indistps2ALT.j . . . 4 (TopOpen‘𝐾) = {∅, 𝐴}
87eqcomi 2769 . . 3 {∅, 𝐴} = (TopOpen‘𝐾)
96, 8istps 23213 . 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 1570  wcel 2145  Vcvv 3450  c0 4278  {cpr 4585  cfv 6527  Basecbs 17348  TopOpenctopn 17553  TopOnctopon 23189  TopSpctps 23211
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732  ax-sep 5248  ax-nul 5259  ax-pow 5326  ax-pr 5390  ax-un 7734
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-pw 4558  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5103  df-opab 5167  df-mpt 5186  df-id 5542  df-xp 5653  df-rel 5654  df-cnv 5655  df-co 5656  df-dm 5657  df-iota 6483  df-fun 6529  df-fv 6535  df-top 23173  df-topon 23190  df-topsp 23212
This theorem is used by: (None)
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