Intuitively, when the collection $\mathscr C$ grows larger, its union grows larger and its intersection grows smaller. Going the other way, when the collection $\mathscr C$ grows smaller, its union grows smaller and its intersection grows larger. Taken to the extreme, when the collection $\mathscr C$ is as small as possible (empty), its union is as small as possible (empty) and its intersection is as large as possible (the whole space).
More technically, a point $x\in X$ is in $\bigcup \mathscr C$ if there is a member $C$ of the collection $\mathscr C$ with $x\in C$. When $\mathscr C$ is empty, this can't happen, so no point qualifies to be in $\bigcup \mathscr C$.
Similarly, a point $x\in X$ is in $\bigcap \mathscr C$ if $x\in C$ for every member $C$ of the collection $\mathscr C$. When $\mathscr C$ is empty, this is vacuously true (you can't demonstrate a member of $\mathscr C$ that fails to contain $x$).
So, the union of a family "starts out empty" and grows as you add sets to the family (more points qualify to belong), and the intersection of a family "starts out universal" and shrinks as you add sets to the family (fewer points qualify to belong).
Edit: I can't comment any more or the comments will be moved to chat; so I will "cheat" and comment here (sorry for the breach of protocol). No, it doesn't go beyond $X$ itself because we specified that $\mathscr C$ was a collection of "subsets of $X$" to begin with. I know that's a little vague, but $X$ is the universal set in this context.
The proof is somewhat dependent on the definitions you use.
The simple (as you basically have only one possible definition) part is the intersection of bounded sets. You either define bounded as having a radius such that every member of the set has absolute value (or norm) less or equal to the radius, or more general that the distance of any two points in the set have distance $d(x,y)$ less or equal to the radius (I use the latter), I don't require the radius to be the smallest bound here (if one bound exists there exists an infimum of them which is also a bound by the axiom of largest lower bound):
Let $B_j$ be bounded sets which means that there exists a radius $r_j$ such that $d(x,y)\le r_j$ for all $x,y\in B_j$. Now consider the intersection $B = \bigcap B_j$, then we have that $d(x,y)\le r_o$ for any $x,y\in B_o$, but since $B \subseteq B_o$ we have for any $x, y\in B$ that $x,y\in B_o$ and therefore $d(x,y)\le r_o$, so $r_o$ is a radiuos of $B$.
To prove that intersection of closed sets is closed we have to do that differently depending on the definition:
- Closed being that the complement is open: Use deMorgan theorem and use that $\overline{F_j}$ is open, therefore $\bigcap F_j = \overline{\bigcup \overline {F_j}}$, but the union of open sets is open so the intersection is the complement of an open set.
- Closed being containing all it's limit points. Suppose we have $a$ being a limit point of $F =\bigcap F_j$, then there exists a sequence $a_k\in F$ such that $\lim_{k\to\infty}a_k = a$, but since $F_j\subseteq F$ we have that $a_k$ is a sequence in $F_j$ and therefore $a\in F_j$ as $F_j$ is closed. And since $a\in F_j\subseteq F$ we have $a\in F$. So $F$ contains all its limit ponts.
Also for compactness it's a matter of definition:
- Compact being closed and bounded: The intersection of closed is closed, and intersection of bounded is bounded. Therefore intersection of compact is compact.
- Compact being that open cover has a finite subcover: This is a lot trickier (and may be out of your scope), I will need to use more assumptions here. Suppose we have a open cover of $F \subset \cup \Omega_j$. We extend this to a cover of any $F_o$ by observing that for each $\phi\in F_o\setminus F$ we have an open set $\Omega_\Phi$ that doesn't intersect $F$ (assuming the space is regular Hausdorff - for example the set of real numbers). Now we have an open cover of $F_o$ which have a finite subcover, and excluding the ammended sets to cover the whole $F_o$ will make it a cover of $F$.
Best Answer
The empty set is bounded: $\forall x \in \emptyset : (x \le 1) \land (x \ge -1)$ is certainly a (vacuusly) true statement.
The union of two such sets need not be bounded: Take $A = (-\infty, 1]$, $B= [-1,\infty)$, then $A \cup B= \mathbb{R}$ but $A \cap B = [-1,1]$.
To see the statement: let $a_0$ be an upperbound for $A$ and $b_0$ a lower bound for $B$. Let $x \in A \cap B$ be arbitrary. Then $x \in A$ so $x \le a_0$ an $x\in B$ too, so $b_0 \le x$. So $A \cap B \subseteq [a_0, b_0]$ hence $A \cap B$ is bounded.