Am I abusing set operations to justify an assumption in a simple proof involving the union of connected sets?

Question

I'm working through some analysis textbooks on my own, so I don't want the full answer. I'm only looking for a hint on this problem.

In Rosenlicht's Introduction to Analysis, he proves this proposition:

Let $\{S_i\}$, for $i \in I$, be a collection of connected subsets of a metric space $E$. Suppose there exists $i_0 \in I$ such that for each $i \in I$, we have $S_i \cap S_{i_0} \neq \varnothing$. Then $S = \cup_{i \in I}\ S_i$ is connected.

He starts out by supposing that $S$ is the union of two disjoint open subsets $A$ and $B$ (we showed earlier that this follows from the definition of connectedness), with the goal to show that either $A$ or $B$ is empty. These are his steps:

1. For any $i \in I$, $S_i = \left(A \cap S_i \right) \cup \left(B \cap S_i \right)$ expresses the connected set $S_i$ as the union of two disjoint open subsets. This means that we have either

   a) $A \cap S_i = S_i$ and $B \cap S_i = \varnothing$ or

   b) $A \cap S_i = \varnothing$ and $B \cap S_i = S_i$.

   I follow this because $A$ and $B$ are disjoint, so $A \cap S_i$ and $B \cap S_i$ are clearly disjoint. Also, $A$ and $B$ are open (given) and $S_i$ is open (because it's connected), and the union of two open sets is open. Therefore $A \cap S_i$ and $B \cap S_i$ are open subsets.

2. He then says that "without loss of generality we may assume that $A \cap S_{i_0} = S_{i_0}$."

I don't understand how we can assume this second step. This was my attempt to justify this assumption:

1. Take case a) above (although it doesn't matter which case we take because the only difference is the symbols $A$ and $B$). We have $A \cap S_i = S_i$.

2. From the initial proposition, we have $S_i \cap S_{i_0} \neq \varnothing$. If we assume that $A \cap S_{i_0} \neq S_{i_0}$, we have

\begin{align} A \cap S_{i_0} &\neq S_{i_0} \\ A \cap S_{i_0} \cap S_i & \neq S_{i_0} \cap S_i \ \ \ \text{because $A, S_i$, and $S_{i_0}$ are nonempty} \\ \left(A \cap S_i\right) \cap S_{i_0} & \neq S_{i_0} \cap S_i \\ \Rightarrow S_i \cap S_{i_0} &\neq S_{i_0} \cap S_i \end{align}

which is a contradiction, so $A \cap S_{i_0} = S_{i_0}$ is a valid assumption.

Is this logic correct, or did I abuse set operations to justify this assumption? I believe I'm proving the same proposition as discussed in this question, but I'm fairly certain that I'm confused about something different.

Answer 1

In Step 1 it was shown that $(A\cap S_{i_0})\cup(B\cap S_{i_0})$ is a decomposition of the connected set $S_{i_0}$ into two open disjoint subsets. This means one of them is empty, without loss of generality it's $B\cap S_{i_0}$, then $S_{i_0}=A\cap S_{i_0}$.

Answer 2

No, your logic does not hold. In particular $$A\neq B$$ does not imply $$A\cap C \neq B \cap C$$ since it is possible that, for instance, $C$ is disjoint from $A$ and $B$.

To prove the statement, note that you already have from hypothesis that $A\cup B=S\supseteq S_{i_0}$ and $A\cap B=\emptyset$. It follows that either $A\cap S_{i_0}$ or $B\cap S_{i_0}$ is non-empty, since otherwise they couldn't cover it - but if, as is shown in step one $A\cap S_{i_0}\neq \emptyset \Rightarrow A\supseteq S_{i_0}$, we get the desired assumption if we first assume without loss of generality that $A$ intersects $S_{i_0}$.