Theorem ist0 23258

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 23283. (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 4894	. . . 4 ⊢ (𝑗 = 𝐽 → ∪ 𝑗 = ∪ 𝐽)
2		ist0.1	. . . 4 ⊢ 𝑋 = ∪ 𝐽
3	1, 2	eqtr4di 2788	. . 3 ⊢ (𝑗 = 𝐽 → ∪ 𝑗 = 𝑋)
4		raleq 3302	. . . . 5 ⊢ (𝑗 = 𝐽 → (∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) ↔ ∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜)))
5	4	imbi1d 341	. . . 4 ⊢ (𝑗 = 𝐽 → ((∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦) ↔ (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))
6	3, 5	raleqbidv 3325	. . 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 3325	. 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 23251	. 2 ⊢ Kol2 = {𝑗 ∈ Top ∣ ∀𝑥 ∈ ∪ 𝑗∀𝑦 ∈ ∪ 𝑗(∀𝑜 ∈ 𝑗 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)}
9	7, 8	elrab2 3674	1 ⊢ (𝐽 ∈ Kol2 ↔ (𝐽 ∈ Top ∧ ∀𝑥 ∈ 𝑋 ∀𝑦 ∈ 𝑋 (∀𝑜 ∈ 𝐽 (𝑥 ∈ 𝑜 ↔ 𝑦 ∈ 𝑜) → 𝑥 = 𝑦)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1540   ∈ wcel 2108  ∀wral 3051  ∪ cuni 4883  Topctop 22831  Kol2ct0 23244

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-ext 2707

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-ex 1780  df-sb 2065  df-clab 2714  df-cleq 2727  df-clel 2809  df-ral 3052  df-rex 3061  df-rab 3416  df-v 3461  df-ss 3943  df-uni 4884  df-t0 23251

This theorem is referenced by:  t0sep  23262  t0top  23267  ist0-2  23282  cnt0  23284  ist0cld  33864