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Mirrors > Home > MPE Home > Th. List > Mathboxes > opndisj | Structured version Visualization version GIF version |
Description: Two ways of saying that two open sets are disjoint, if 𝐽 is a topology and 𝑋 is an open set. (Contributed by Zhi Wang, 6-Sep-2024.) |
Ref | Expression |
---|---|
opndisj | ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅))) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | elpwg 4488 | . . . 4 ⊢ (𝑌 ∈ 𝐽 → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ 𝑍)) | |
2 | sseq2 3901 | . . . 4 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ⊆ 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) | |
3 | 1, 2 | sylan9bbr 514 | . . 3 ⊢ ((𝑍 = (∪ 𝐽 ∖ 𝑋) ∧ 𝑌 ∈ 𝐽) → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) |
4 | 3 | pm5.32da 582 | . 2 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → ((𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))) |
5 | elin 3857 | . 2 ⊢ (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍)) | |
6 | elssuni 4825 | . . . 4 ⊢ (𝑌 ∈ 𝐽 → 𝑌 ⊆ ∪ 𝐽) | |
7 | incom 4089 | . . . . . 6 ⊢ (𝑋 ∩ 𝑌) = (𝑌 ∩ 𝑋) | |
8 | 7 | eqeq1i 2743 | . . . . 5 ⊢ ((𝑋 ∩ 𝑌) = ∅ ↔ (𝑌 ∩ 𝑋) = ∅) |
9 | reldisj 4338 | . . . . 5 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑌 ∩ 𝑋) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) | |
10 | 8, 9 | syl5bb 286 | . . . 4 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑋 ∩ 𝑌) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) |
11 | 6, 10 | syl 17 | . . 3 ⊢ (𝑌 ∈ 𝐽 → ((𝑋 ∩ 𝑌) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) |
12 | 11 | pm5.32i 578 | . 2 ⊢ ((𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) |
13 | 4, 5, 12 | 3bitr4g 317 | 1 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 209 ∧ wa 399 = wceq 1542 ∈ wcel 2113 ∖ cdif 3838 ∩ cin 3840 ⊆ wss 3841 ∅c0 4209 𝒫 cpw 4485 ∪ cuni 4793 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1802 ax-4 1816 ax-5 1916 ax-6 1974 ax-7 2019 ax-8 2115 ax-9 2123 ax-ext 2710 |
This theorem depends on definitions: df-bi 210 df-an 400 df-tru 1545 df-fal 1555 df-ex 1787 df-sb 2074 df-clab 2717 df-cleq 2730 df-clel 2811 df-ral 3058 df-rab 3062 df-v 3399 df-dif 3844 df-in 3848 df-ss 3858 df-nul 4210 df-pw 4487 df-uni 4794 |
This theorem is referenced by: clddisj 45703 |
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