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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?

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  • $\begingroup$ I think it should. Every step except the one from line 2 to line 3 (invoking the general theorem) is just manipulating the definition of symmetric difference. $\endgroup$ Oct 31 '18 at 20:19
  • $\begingroup$ Often times these sort of statements are given as an exercise of inductive proofs. Things like DeMorgan laws are far easier to prove in general than they are to prove by induction. But since people don't stop to think whether or not it is the case, and they want easy induction homework, we keep seeing people trying to prove that for all $n\in\Bbb N$, $X\setminus\bigcup_{k<n} X_k=\bigcap_{k<n}(X\setminus X_k)$, or stuff like that. This is another good example. $\endgroup$
    – Asaf Karagila
    Nov 1 '18 at 17:59
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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.

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  • $\begingroup$ Thank you, i edited the wiki accordingly. $\endgroup$
    – cdwe
    Nov 1 '18 at 16:30

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