Theorem ist0 22025

Description: The predicate "is a T₀ space". Every pair of distinct points is topologically distinguishable. For the way this definition is usually encountered, see ist0-3 22050. (Contributed by Jeff Hankins, 1-Feb-2010.)

Hypothesis

Ref	Expression
ist0.1	⊢ 𝑋 = ∪ 𝐽

Assertion

Ref	Expression
ist0	⊢ (𝐽 ∈ Kol2 ↔ (𝐽 ∈ Top ∧ ∀𝑥 ∈ 𝑋 ∀𝑦 ∈ 𝑋 (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))

Distinct variable groups:   𝑥,𝑜,𝑦,𝐽   𝑜,𝑋,𝑥,𝑦

Proof of Theorem ist0

Dummy variable 𝑗 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		unieq 4812	. . . 4 ⊢ (𝑗 = 𝐽 → ∪ 𝑗 = ∪ 𝐽)
2		ist0.1	. . . 4 ⊢ 𝑋 = ∪ 𝐽
3	1, 2	eqtr4di 2811	. . 3 ⊢ (𝑗 = 𝐽 → ∪ 𝑗 = 𝑋)
4		raleq 3323	. . . . 5 ⊢ (𝑗 = 𝐽 → (∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) ↔ ∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜)))
5	4	imbi1d 345	. . . 4 ⊢ (𝑗 = 𝐽 → ((∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦) ↔ (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))
6	3, 5	raleqbidv 3319	. . 3 ⊢ (𝑗 = 𝐽 → (∀𝑦 ∈ ∪ 𝑗(∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦) ↔ ∀𝑦 ∈ 𝑋 (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))
7	3, 6	raleqbidv 3319	. 2 ⊢ (𝑗 = 𝐽 → (∀𝑥 ∈ ∪ 𝑗∀𝑦 ∈ ∪ 𝑗(∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦) ↔ ∀𝑥 ∈ 𝑋 ∀𝑦 ∈ 𝑋 (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))
8		df-t0 22018	. 2 ⊢ Kol2 = {𝑗 ∈ Top ∣ ∀𝑥 ∈ ∪ 𝑗∀𝑦 ∈ ∪ 𝑗(∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)}
9	7, 8	elrab2 3607	1 ⊢ (𝐽 ∈ Kol2 ↔ (𝐽 ∈ Top ∧ ∀𝑥 ∈ 𝑋 ∀𝑦 ∈ 𝑋 (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399   = wceq 1538   ∈ wcel 2111  ∀wral 3070  ∪ cuni 4801  Topctop 21598  Kol2ct0 22011

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-10 2142  ax-11 2158  ax-12 2175  ax-ext 2729

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-tru 1541  df-ex 1782  df-nf 1786  df-sb 2070  df-clab 2736  df-cleq 2750  df-clel 2830  df-nfc 2901  df-ral 3075  df-rab 3079  df-v 3411  df-in 3867  df-ss 3877  df-uni 4802  df-t0 22018

This theorem is referenced by:  t0sep  22029  t0top  22034  ist0-2  22049  cnt0  22051  ist0cld  31308