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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 4536 | . . . 4 ⊢ (𝑌 ∈ 𝐽 → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ 𝑍)) | |
2 | sseq2 3947 | . . . 4 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ⊆ 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) | |
3 | 1, 2 | sylan9bbr 511 | . . 3 ⊢ ((𝑍 = (∪ 𝐽 ∖ 𝑋) ∧ 𝑌 ∈ 𝐽) → (𝑌 ∈ 𝒫 𝑍 ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) |
4 | 3 | pm5.32da 579 | . 2 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → ((𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋)))) |
5 | elin 3903 | . 2 ⊢ (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ∈ 𝒫 𝑍)) | |
6 | elssuni 4871 | . . . 4 ⊢ (𝑌 ∈ 𝐽 → 𝑌 ⊆ ∪ 𝐽) | |
7 | incom 4135 | . . . . . 6 ⊢ (𝑋 ∩ 𝑌) = (𝑌 ∩ 𝑋) | |
8 | 7 | eqeq1i 2743 | . . . . 5 ⊢ ((𝑋 ∩ 𝑌) = ∅ ↔ (𝑌 ∩ 𝑋) = ∅) |
9 | reldisj 4385 | . . . . 5 ⊢ (𝑌 ⊆ ∪ 𝐽 → ((𝑌 ∩ 𝑋) = ∅ ↔ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) | |
10 | 8, 9 | syl5bb 283 | . . . 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 575 | . 2 ⊢ ((𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅) ↔ (𝑌 ∈ 𝐽 ∧ 𝑌 ⊆ (∪ 𝐽 ∖ 𝑋))) |
13 | 4, 5, 12 | 3bitr4g 314 | 1 ⊢ (𝑍 = (∪ 𝐽 ∖ 𝑋) → (𝑌 ∈ (𝐽 ∩ 𝒫 𝑍) ↔ (𝑌 ∈ 𝐽 ∧ (𝑋 ∩ 𝑌) = ∅))) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ↔ wb 205 ∧ wa 396 = wceq 1539 ∈ wcel 2106 ∖ cdif 3884 ∩ cin 3886 ⊆ wss 3887 ∅c0 4256 𝒫 cpw 4533 ∪ cuni 4839 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 ax-5 1913 ax-6 1971 ax-7 2011 ax-8 2108 ax-9 2116 ax-ext 2709 |
This theorem depends on definitions: df-bi 206 df-an 397 df-tru 1542 df-fal 1552 df-ex 1783 df-sb 2068 df-clab 2716 df-cleq 2730 df-clel 2816 df-ral 3069 df-rab 3073 df-v 3434 df-dif 3890 df-in 3894 df-ss 3904 df-nul 4257 df-pw 4535 df-uni 4840 |
This theorem is referenced by: clddisj 46197 |
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