Difficulty understanding why a set is saturated with respect to a map The following is the beginning of Example 7 on page 143 of Munkres' Topology.

Example 7. The product of two quotient maps need not be a quotient map.
Let $X = \mathbb{R}$ and let $X^*$ be the quotient space obtained from $X$ by identifying the subset $\mathbb{Z}_+$ to a point $b$; let $p : X \to X^*$ be the quotient map.
  Let $\mathbb{Q}$ be the subspace of $\mathbb{R}$ consisting of the rational numbers; let $i : \mathbb{Q} \to \mathbb{Q}$ be the idenitity map.
  We show that $$p \times i : X \times \mathbb{Q} \to X^* \times \mathbb{Q}$$ is not a quotient map.
For each $n$, let $c_n = \frac{\sqrt{2}}{n}$, and consider the straight line in $\mathbb{R}^2$ with slopes $1$ and $-1$, respectively, through the point $n \times c_n$.
  Let $U_n$ consist of all points of $X \times \mathbb{Q}$ that lie above both of these lines or beneath both of them, and also between the vertical lines $x = n - \frac{1}{4}$ and $x = n + \frac{1}{4}$.
  Then $U_n$ is open in $X \times \mathbb{Q}$; it contains the set $\{ n \} \times \mathbb{Q}$ because $c_n$ is not rational.
Let $U$ be the union of the sets $U_n$; then $U$ is open in $X \times \mathbb{Q}$.
  It is saturated to $p \times i$ because it contains the entire set $\mathbb{Z}_+ \times \{ q \}$ for each $q \in \mathbb{Q}$.
....

I can't understand why $U$ is saturated with respect
to $p × i$. Can anyone who has studied Munkres' Topology illustrate the matter?
 A: We have $X=\Bbb R$, and $X^*$ is the result of identifying $\Bbb Z^+\subseteq\Bbb R$ to a point; $p:X\to X^*$ is the quotient map. We also have the identity map $i:\Bbb Q\to\Bbb Q$, and the example is to show that $$p\times i:X\times\Bbb Q\to X^*\times\Bbb Q$$ is not a quotient map.
For each $n$ Munkres sets $c_n=\frac{\sqrt2}n$. Let $\ell^+$ and $\ell^-$ be the lines through $n\times c_n$ of slope $1$ and $-1$, respectively. Then $U_n$ consists of all points of $X\times\Bbb Q$ that lie either above both $\ell^+$ and $\ell^-$ or below both of these lines, and between the lines $x=n-\frac14$ and $x=n+\frac14$. This set is open in $X\times\Bbb Q$, and it contains the set $\{n\}\times\Bbb Q$ because $c_n\notin\Bbb Q$.
Thus $U=\bigcup_{n\in\Bbb Z^+}U_n$ is open in $X\times\Bbb Q$, and the claim is that it is saturated with respect to $p\times i$. The reason is given in the same sentence:

It is saturated with respect to $p\times i$ because it contains the entire set $\Bbb Z_+\times\{q\}$ for each $q\in\Bbb Q$.

I’ll try to expand on this. $U$ is saturated with respect to $p\times i$ if $$U=(p\times i)^{-1}\big[(p\times i)[U]\big]\;.$$ Another, often more useful, way to look at this is that $U$ is saturated if and only if it contains every fibre of $p\times i$ that it intersects. (The fibres of a map $f:Y\to Z$ are the set $f^{-1}[\{z\}]$ for $z\in Z$.) What are the fibres of $p\times i$? They’re the sets of the form $(p\times i)^{-1}[\{x\times q\}]$ such that $x\in X^*$ and $q\in\Bbb Q$. In other words, they’re the sets $\{x\times q\}$ such that $x\in\Bbb R\setminus\Bbb Z^+$ and $q\in\Bbb Q$, and the set $\Bbb Z^+\times\{q\}$ such that $q\in\Bbb Q$. All of these except the sets $\Bbb Z^+\times\{q\}$ for $q\in\Bbb Q$ are singletons: if $U$ intersects one of them, it automatically contains it. Thus, we need only worry about the sets of the form $\Bbb Z^+\times\{q\}$ with $q\in\Bbb Q$. 
Fix $q\in\Bbb Q$. For each $n\in\Bbb Z^+$ we know that $n\times q\in U_n\subseteq U$, so $\Bbb Z^+\times\{q\}\subseteq U$. Thus, $U$ contains every one of the fibres $\Bbb Z^+\times\{q\}$ of $p\times i$. And as I noted above, every other fibre of $p\times i$ is a singleton and hence either disjoint from or completely contained in $U$. $U$ therefore contains every fiber of $p\times i$ that it meets, so 
$$U=(p\times i)^{-1}\big[(p\times i)[U]\big]\;,$$
and $U$ is therefore saturated with respect to $p\times i$.
(I’ve addressed only your explicit question; if you have questions about the rest of the proof, let me know.)
Added: It occurs to me that it might be helpful to someone to have a picture of $X^*$:

The red point is the image under $p$ of $\Bbb Z^+$. The tail to the left is the image of $(\leftarrow,1)$. The innermost circle is the image of the interval $(1,2)$; the next circle out is the image of $(2,3)$; and in general the $n$-th circle out is the image of the interval $(n,n+1)$.
