Tao Analysis I - 3.4.4

Exercise 3.4.4

Let $f:X \to Y$ be a function from one set $X$ to another set $Y$, and let $U$, $V$ be subsets of $Y$. Show that $f^{-1}(U \cup V)= f^{-1}(U) \cup f^{-1}(V)$, that $f^{-1}(U \cap V)= f^{-1}(U) \cap f^{-1}(V)$, and that $f^{-1}(U \setminus V)= f^{-1}(U) \setminus f^{-1}(V)$.

Before we attempt the exercise, let's remind ourselves of Definition 3.4.5 for inverse images. If $U$ is a subset of $Y$, we define the set $f^{-1}(U)$ to be the set

$$f^{-1}(U) := \{x \in X: f(x) \in U\}$$

That is, $f^{-1}$ consists of all the elements of $X$ which map into $U$:

$$f(x) \in U \iff x \in f^{-1}(U)$$

We call $f^{-1}(U)$ the inverse image of $U$.

Show $f^{-1}(U \cup V)= f^{-1}(U) \cup f^{-1}(V)$

Let's apply Definition 3.4.5 to $f^{-1}(U \cup V)$,

$$f^{-1}(U \cup V) := \{x \in X: f(x) \in U \cup V\}$$

This is the set of $x$ such that either $f(x) \in U$ or $f(x) \in V$, or both, using the definition of union.

Let's consider both cases:

• $f(x) \in U$. Definition 3.4.5 tells us $f(x) \in U \iff x \in f^{-1}(U)$
• $f(x) \in V$. Definition 3.4.5 tells us $f(x) \in U \iff x \in f^{-1}(V)$

Since either or both of these are true, we have either or both of the following are true:

• $x \in f^{-1}(U)$
• $x \in f^{-1}(V)$

This is equivalent to $x \in f^{-1}(U) \cup f^{-1}(V)$.

Thus, we have shown $f^{-1}(U \cup V)= f^{-1}(U) \cup f^{-1}(V)$. $\square$

Note that we didn't have to show inclusion in both directions because the definitions we used are bidirectional $\iff$.

Show $f^{-1}(U \cap V)= f^{-1}(U) \cap f^{-1}(V)$

Let's apply Definition 3.4.5 to $f^{-1}(U \cap V)$,

$$f^{-1}(U \cap V) := \{x \in X: f(x) \in U \cap V\}$$

This is the set of $x$ such that $f(x) \in U$ and $f(x) \in V$, using the definition of intersection.

Let's consider the two cases, both of which must be true:

• $f(x) \in U$. Definition 3.4.5 tells us $f(x) \in U \iff x \in f^{-1}(U)$
• $f(x) \in V$. Definition 3.4.5 tells us $f(x) \in U \iff x \in f^{-1}(V)$

Since both of these are true, then both of the following are true:

• $x \in f^{-1}(U)$
• $x \in f^{-1}(V)$

This is equivalent to $x \in f^{-1}(U) \cap f^{-1}(V)$.

Thus, we have shown $f^{-1}(U \cap V)= f^{-1}(U) \cap f^{-1}(V)$. $\square$

Note that we didn't have to show inclusion in both directions because the definitions we used are bidirectional $\iff$.

Show $f^{-1}(U \setminus V)= f^{-1}(U) \setminus f^{-1}(V)$

Let's apply Definition 3.4.5 to $f^{-1}(U \setminus V)$,

$$f^{-1}(U \setminus V) := \{x \in X: f(x) \in U \setminus V\}$$

This is the set of $x$ such that $f(x) \in U$ and $f(x) \notin V$. 

Let's consider the two cases, both of which must be true:

• $f(x) \in U$. Definition 3.4.5 tells us $f(x) \in U \iff x \in f^{-1}(U)$
• $f(x) \notin V$. Definition 3.4.5 tells us $f(x) \in U \iff x \notin f^{-1}(V)$

Since both of these are true, then both of the following are true:

• $x \in f^{-1}(U)$
• $x \notin f^{-1}(V)$

This is equivalent to $x \in f^{-1}(U) \setminus f^{-1}(V)$.

Thus, we have shown $f^{-1}(U \setminus V)= f^{-1}(U) \setminus f^{-1}(V)$. $\square$

Note that we didn't have to show inclusion in both directions because the definitions we used are bidirectional $\iff$.