Union and Intersection, Problem with {a} In some book (Deiser, Mengenlehre, 3. Aufl.) about set theory the author defines the general union and intersection:
$\cap$M = {x|for all a $\in$ M also x $\in$ a}
$\cup$M = {x|there exists some a $\in$ M with x $\in$ a}.
He then brings the example: $\cap${a} = $\cup${a} = a. If you just apply the definition it does not make sense to me. Do you see that by just applying the definition? What'd make sense is this: $\cap${{a}} = $\cup${{a}} = a. Am I right or do I make a mistake?
 A: Let's take a few examples.
Let's say $a=\{1,2,3\}$ and $b=\{4,5,6\}$. Then:
$\cup\{a,b\}=\{1,2,3,4,5,6\}$ (everything in either $a$ or $b$).
Now, let's look at $\cup\{a\}$. This is:
$\cup\{a\}=\{1,2,3\}$ (everything in $a$).
So, as you can see, $\cup\{a\}=a$ because they both have the same elements.
A: Yes, it follows from the definition.
$\bigcap\{a\}$ is by definition the set of all $x$ such that $x\in y$ for every member $y$ of the set $\{a\}$. The only member of the set $\{a\}$ is $a$, so by definition $x\in\bigcap\{a\}$ if and only if $x\in a$. The sets $\bigcap\{a\}$ and $a$ therefore have the same members and hence are the same set.
Similarly, $\bigcup\{a\}$ is by definition the set of all $x$ such that $x$ belongs to at least one member of $\{a\}$. The only member of $\{a\}$ is $a$, so if $x$ belongs to $\bigcup\{a\}$, then $x$ must belong to $a$. Conversely, if $x\in a$, then certainly $x$ belongs to at least one member of $\{a\}$, so $x\in\bigcup\{a\}$. Thus, $\bigcup\{a\}\subseteq a$ and $a\subseteq\bigcup\{a\}$, so $\bigcup\{a\}=a$.
A: To prove $\bigcup\{a\}=a$ we need to prove (i) $\bigcup\{a\}\subseteq a$; and  (ii) $a\subseteq \bigcup\{a\}$.
For (i) let $x\in\bigcup\{a\}$. Then for some $p\in\{a\}$ we have $x\in p$. But there is only one such $p$, namely $a$. So $x\in a$ and we are done.
For (ii) let $x\in a$. Then $x\in a\in \{a\}$, so by definition $x\in\bigcup\{a\}$.
[I have assumed that we have proved that $\{a\}$ (is the definition in your book $\{a\}=\{a,a\}$, where $\{a,b\}$ is already defined?) has precisely one element $a$. ]
