0

If and A and B are separated and C is a connected subset of $A\cup B$, then either $C\subset A$ or $C\subset B$.

Proof: Suppose by contradiction that it's $\underline{not}$ the case that either $C\subset A$ or $C\subset B$, then by demorgan's law $C\not\subset A$ and $C\not\subset B$. If A and B are separated meaning that $\overline{A}\cap B=\varnothing=A\cap\overline{B}$ and $C\subset A\cup B$, but then $C\subset A$ or $C\subset B$. Hence, by contradiction $C\subset A$ or $C\subset B$.

I feel like its not enough. Thank you.

cemsicles
  • 442
  • 2
  • 12

2 Answers2

1

"but then $C \subset A$ or $C \subset B$" assumes what you are trying to prove.

If neither $C \subset A$ nor $C \subset B$, then $C = (C \cap A) \cup (C \cap B)$ with both $C \cap A \neq \varnothing$ and $C \cap B \neq \varnothing$. But the separation of $A$ and $B$ separates these two pieces of $C$. Therefore, $C$ is not connected.

Eric Towers
  • 67,037
1

But $A \cap B$ empty; $C \subset A \cup B$; do NOT necessarily imply that one of $C \subset A$ or $C \subset B$ has to hold. So your line of reasoning isn't correct there.

Suppose $C$ intersects both $A$ and $B$. Then $C$ connected implies that $C \cap A$ is connected to $C \cap B$. This contradicts $A$ and $B$ being separated.

Mike
  • 20,434
  • Connected does not necessarily mean path connected. – John Douma Apr 23 '18 at 23:13
  • Yes indeed. Editting – Mike Apr 23 '18 at 23:15
  • The idea is that connectedness defines an equivalence class; a set is connected if every member is in the same such equivalence class, and $A$ and $B$ are separated if no member of $A$ is in the same such equivalence class as any member of $B$. – Mike Apr 23 '18 at 23:24