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I'm working on some set theory problems and I've run across some issues. I need to prove:

(Sorry if this looks messy but I dont know exactly how to type this out. It's a union of a collection of sets, by the way.) $\bigcup_{X\in\{A,B\}} X=A\cup B.$

So I start off using the definition of $\bigcup$ and I get:

$\forall x\colon(\exists X\colon X\in\{A,B\}\land x\in X)$

So my question is...can I go ahead and assume that $X$ is an element of $A \cup B$ since it is an element of $\{A,B\}$?

And then my next step would look like:

$(\forall X)(X \in A \cup B \Rightarrow x \in X)$

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    Kristen, if you have found an answer that has helped you, you can indicate so by "accepting" it (by clicking on the grayed-out arrow to the left of the answer).2012-11-16

3 Answers 3

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On the one hand, suppose that $x\in\bigcup_{X\in\{A,B\}}X.$ Then there is some $X\in\{A,B\}$ such that $x\in X$ (by definition). Since $X\in\{A,B\}$, then $X=A$ or $X=B$, so $x\in A$ or $x\in B$, and in any case $x\in A\cup B:=\{y:y\in A\text{ or }y\in B\}$. Therefore, $\bigcup_{X\in\{A,B\}}X\subseteq A\cup B.$

On the other hand, suppose that $x\in A\cup B$. By definition, $x\in A$ or $x\in B$, so there is some $X\in\{A,B\}$ such that $x\in X$. Hence, $x\in\bigcup_{X\in\{A,B\}}X,$ and therefore, $\bigcup_{X\in\{A,B\}}X\supseteq A\cup B.$

By extensionality, it follows that $\bigcup_{X\in\{A,B\}}X= A\cup B.$

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    There are a *lot* of tips and tricks (too many to list). The best way to come by them is to do a lot of proofs and read a lot of proofs. One recommendation that I *can* give you (that will serve you well) is that it's very important to get a sense of what definitions **mean**, not just how to express them in the symbolic language.2012-11-16
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You were exchanging the $\in$ and $\subseteq$ notions. The set $\{A, B\}$ has exactly two elements (unless $A=B$), so either $X=A$ or $X=B$.

So, we can conlcude, that $X$ is a subset of $A\cup B$, not an element.

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    $x\in \bigcup \{A,B\} \overset{\text{def}}\iff \exists X: X\in\{A,B\} \land x\in X $2012-11-16
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Here is how I would solve this, using the rules of predicate logic. Using a slightly different notation, let's see which elements $\;x\;$ are in the left hand side set: \begin{align} & x \in \langle \cup X : X \in \{A,B\} : X \rangle \\ \equiv & \qquad \text{"definition of $\;\cup\;$-quantification"} \\ & \langle \exists X : X \in \{A,B\} : x \in X \rangle \\ \equiv & \qquad \text{"definition of $\;\{\ldots,\ldots\}\;$"} \\ & \langle \exists X : X = A \lor X = B : x \in X \rangle \\ \equiv & \qquad \text{"logic: split range of quantification"} \\ & \langle \exists X : X = A : x \in X \rangle \;\lor\; \langle \exists X : X = B : x \in X \rangle \\ \equiv & \qquad \text{"logic: one-point rule, twice"} \\ & x \in A \;\lor\; x \in B \\ \equiv & \qquad \text{"definition of $\;\cup\;$"} \\ & x \in A \cup B \\ \end{align} By set extensionality, this proves the original statement.