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I have a simple question which is very trivial for the other people, I guess. However, I never fully understand the argument completely. Take any $X\neq\emptyset$ and let $\mathbf{S}$ be an arbitrary class of subsets of $X$. If $\mathbf{S}$ is empty than class of finite intersections of sets is a single element class $\left\{X\right\}$. Moreover, all unions of sets in this class is the two-element class $\left\{ \emptyset,X \right\}$.

For the first one argument is the following, any element $x\in X$ will be in the finite intersection of sets because there is no such set. I can see the logic but this seems very counter-intuitive. To me, if there is no set in the class intersection should be $\emptyset$. For the second one i.e. union one, I can't see the logic. Since only $X$ is in this class how come all unions yield $\emptyset$?. Thanks very much!

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  • $\begingroup$ If $\mathbf S$ is really arbitrary (hence has no connection with $X$) then the class of finite intersections is not $\{X\}$. So what is the relation between $\mathbf S$ and $X$? $\endgroup$ – drhab May 25 '14 at 13:58
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A good thing to keep in mind is the similarity with the multiplication and sum. The intersection of nothing is the whole space, as the product of no real numbers is $1$: $$ \prod_{x\in\emptyset} x = 1,\qquad \bigcap_{S\in\emptyset} S = X $$ Why? Fir the second, cearly $\bigcap_{S\in\emptyset} S \subseteq X$. The other direction goes as follows: fix any $x\in X$. The goal is to show that $x\in \bigcap_{S\in\emptyset} S$. This is equivalent to saying that for all $S\in\emptyset$ one has $x\in S$... which is vacuously true, since there is no $S\in\emptyset$

(if you are more confortable with a proof by contradiction for the last part, suppose $x\notin \bigcap_{S\in\emptyset} S$: then by definition this means there exists $S\in\emptyset$ for which $x\notin S$, which can't be true since there is no $S$ to begin with. So $x\in\bigcap_{S\in\emptyset} S$).

As for the union: all finite unions possible from a set of $k$ sets include unions of $0$ sets, unions of $1$ set, unions of $2$ sets,..., unions of $k$ sets. In our case, $k=1$ (you only have $X$ in $\{X\}$, so all possible unions are $\emptyset$ (union of $0$ elements: $\bigcup_{S\in\emptyset}S=\emptyset$) and $X$ (union of one element).

(Here, think about the sum: $\bigcup_{S\in\emptyset}S=\emptyset$, as $\sum_{x\in\emptyset}S=0$.)

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