Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

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

share|improve this question
    
I've tried to edit the first part to be readable LaTex and hope that my changes didn't break anything. In particular, check my edit to the definition of $\bigcup$. Does it match the version of Union Axiom you use? Regarding your question: Usually, $A\cup B$ and $\{A,B\}$ are disjoint (though one can select $A,B$ so that there are common elements of $A\cup B$ and $\{A,B\}$, for example if $B=\{A\}$). –  Hagen von Eitzen Nov 16 '12 at 17:20
    
Kristen, in the future, it is best to modify, edit a post and not repost the same question. –  amWhy Nov 16 '12 at 18:04
    
Actually, I decided to vote to close the earlier post, since this one has activity and such. So I will delete my post identifying this as a duplicate post. –  amWhy Nov 16 '12 at 18:09
    
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). –  amWhy Nov 16 '12 at 19:32
add comment

3 Answers

up vote 2 down vote accepted

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 $$\bigcup_{X\in\{A,B\}}X= A\cup B.$$

share|improve this answer
    
ok that kind of makes sense...do you have any tips for doing problems in set theory. I have a test coming up and I'm having some trouble with these –  Kristen Nov 16 '12 at 18:03
    
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. –  Cameron Buie Nov 16 '12 at 18:17
add comment

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.

share|improve this answer
    
what would my next step be after this one: $$\forall x\colon(X\in\{A,B\}\land x\in X)$$ I main issue with set theory proofs is I always have issues getting started –  Kristen Nov 16 '12 at 17:42
2  
$$x\in \bigcup \{A,B\} \overset{\text{def}}\iff \exists X: X\in\{A,B\} \land x\in X $$ –  Berci Nov 16 '12 at 17:45
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.