Consider the following theorem from proofwiki:
Let $S_i, T_i$ be sets for $i \in \mathbb{N}$.
Then:
$$\displaystyle \forall n \in \mathbb{N}_{>0}: \bigcup_{i \mathop = 1}^n S_i \Delta \bigcup_{i \mathop = 1}^n T_i \subseteq \bigcup_{i \mathop = 1}^n \left({S_i \Delta T_i}\right),$$
where $S \Delta T = (T\backslash S) \cup (S \backslash T)$ denotes the symmetric difference between $S$ and $T$.
The statement of the theorem makes it look as though there would be a problem with the following conjecture:
Let $S_\alpha, T_\alpha$ be sets for $\alpha \in I$.
Then:
$$\bigcup_{\alpha \in I} S_\alpha \Delta \bigcup_{\alpha \in I} T_\alpha \subseteq \bigcup_{\alpha \in I} \left({S_\alpha \Delta T_\alpha}\right).$$
The proof given cites the following theorem:
Let $I$ be an indexing set. Let $S_\alpha, T_\alpha$ be sets, for all
$\alpha \in I$. Then: $$\displaystyle \left({\bigcup_{\alpha \mathop
\in I} S_\alpha}\right) \setminus \left({\bigcup_{\alpha \mathop \in
I} T_\alpha}\right) \subseteq \bigcup_{\alpha \mathop \in I}
\left({S_\alpha \setminus T_\alpha}\right)$$
And then concludes
\begin{align}
&\displaystyle \bigcup_{i \mathop = 1}^n S_i \Delta \bigcup_{i \mathop = 1}^n T_i \\
=&\displaystyle \left({\bigcup_{i \mathop = 1}^n S_i \setminus \bigcup_{i \mathop = 1}^n T_i}\right) \cup \left({\bigcup_{i \mathop = 1}^n T_i \setminus \bigcup_{i \mathop = 1}^n S_i}\right) \\
\subseteq& \displaystyle \bigcup_{i \mathop = 1}^n \left({S_i \setminus T_i}\right) \cup \bigcup_{i \mathop = 1}^n \left({T_i \setminus S_i}\right) \\
=& \displaystyle \bigcup_{i \mathop = 1}^n \left({\left({S_i \setminus T_i}\right) \cup \left({T_i \setminus S_i}\right)}\right) \\
=& \displaystyle \bigcup_{i \mathop = 1}^n \left({S_i \Delta T_i}\right).
\end{align}
I think the same reasoning would hold, if we replaced the finite unions by $\bigcup_{\alpha \in I}$.
Am i missing something?
Best Answer
I think the theorem should generalize. It may help to interpret the sets in natural language: $$ \left( \bigcup_{\alpha \in I} S_\alpha \right) \triangle \left( \bigcup_{\alpha \in I} T_\alpha \right)$$ is the set of elements that are in at least one $S_\alpha$ and no $T_\alpha$ whatsoever, or vice versa, whereas $$\bigcup_{\alpha \in I} ( S_\alpha \triangle T_\alpha)$$ is the set of elements that are in at least one $S_\alpha$ and not the corresponding $T_\alpha$, or vice versa. It should be clear that the first criterion is more restrictive than the second.