Section Functions and Sets
We conclude this section with a connection between subsets and functions. A bit of notation first. If \(f\) is a function from a set \(X\) to a set \(Y\text{,}\) and if \(A\) is a subset of \(X\) and \(B\) is a subset of \(Y\text{,}\) we define \(f(A)\) and \(f^{-1}(B)\) as
\begin{equation*}
f(A) = \{f(a) \mid a \in C\}\text{,}
\end{equation*}
and
\begin{equation*}
f^{-1}(B) = \{a \in A \mid f(a) \in B\}\text{.}
\end{equation*}
We call \(f(A)\) the image of the set \(A\) under \(f\) and \(f^{-1}(B)\) is the preimage of the set \(B\) under \(f\text{.}\) Note that \(f^{-1}(B)\) is defined for any function, not just invertible functions. So it is important to recognize that the use of the notation \(f^{-1}(B)\) does not imply that \(f\) is invertible.
When we work with continuous functions in later sections, we will need to understand how a function behaves with respect to subsets. One result is in the following lemma.
Lemma 2.11.
Let \(f : X \to Y\) be a function and let \(\{A_{\alpha}\}\) be a collection of subsets of \(X\) for \(\alpha\) in some indexing set \(I\text{,}\) and \(\{B_{\beta}\}\) be a collection of subsets of \(Y\) for \(\beta\) in some indexing set \(J\text{.}\) Then
\(f\left(\bigcup_{\alpha \in I} A_{\alpha}\right) = \bigcup_{\alpha \in I} f(A_{\alpha})\) and
\(f^{-1}\left(\bigcup_{\beta \in J} B_{\beta}\right) = \bigcup_{\beta \in J} f^{-1}(B_{\beta})\text{.}\)
Proof.
Let \(f : X \to Y\) be a function and let \(\{A_{\alpha}\}\) be a collection of subsets of \(X\) for \(\alpha\) in some indexing set \(I\text{.}\) To prove part 1, we demonstrate the containment in both directions.
Let \(b \in f\left(\bigcup_{\alpha \in I} A_{\alpha}\right)\text{.}\) Then \(b = f(a)\) for some \(a \in \bigcup_{\alpha \in I} A_{\alpha}\text{.}\) It follows that \(a \in A_{\rho}\) for some \(\rho \in I\text{.}\) Thus, \(b \in f(A_{\rho}) \subseteq \bigcup_{\alpha \in I} f(A_{\alpha})\text{.}\) We conclude that \(f\left(\bigcup_{\alpha \in I} A_{\alpha}\right) \subseteq \bigcup_{\alpha \in I} f(A_{\alpha})\text{.}\)
Now let \(b \in \bigcup_{\alpha \in I} f(A_{\alpha})\text{.}\) Then \(b \in f(A_{\rho})\) for some \(\rho \in I\text{.}\) Since \(A_{\rho} \subseteq \bigcup_{\alpha \in I} A_{\alpha}\text{,}\) it follows that \(b \in f\left(\bigcup_{\alpha \in I} A_{\alpha}\right)\text{.}\) Thus, \(\bigcup_{\alpha \in I} f(A_{\alpha}) \subseteq f\left(\bigcup_{\alpha \in I} A_{\alpha}\right)\text{.}\) The two containments prove part 1.
For part 2, we again demonstrate the containments in both directions. Let \(a \in f^{-1}\left(\bigcup_{\beta \in J} B_{\beta}\right)\text{.}\) Then \(f(a) \in \bigcup_{\beta \in J} B_{\beta}\text{.}\) So there exists \(\mu \in J\) such that \(f(a) \in B_{\mu}\text{.}\) This implies that \(a \in f^{-1}(B_{\mu}) \subseteq \bigcup_{\beta \in J} f^{-1}(B_{\beta})\text{.}\) We conclude that \(f^{-1}\left(\bigcup_{\beta \in J} B_{\beta}\right) \subseteq \bigcup_{\beta \in J} f^{-1}(B_{\beta})\text{.}\)
For the reverse containment, let \(a \in \bigcup_{\beta \in J} f^{-1}(B_{\beta})\text{.}\) Then \(a \in f^{-1}(B_{\mu})\) for some \(\mu \in J\text{.}\) Thus, \(f(a) \in B_{\mu} \subseteq \bigcup_{\beta \in J} B_{\beta}\text{.}\) So \(a \in f^{-1}\left(\bigcup_{\beta \in J} B_{\beta}\right)\text{.}\) Thus, \(\bigcup_{\beta \in J} f^{-1}(B_{\beta}) \subseteq f^{-1}\left(\bigcup_{\beta \in J} B_{\beta}\right)\text{.}\) The two containments verify part 2.
At this point it is reasonable to ask if
Lemma 2.11 would still hold if we replace unions with intersections. We leave that question for
Exercise 7.
Another result is contained in the next activity.
Activity 2.8.
Let \(X\text{,}\) \(Y\text{,}\) and \(Z\) be sets, and let \(f: X \to Y\) and \(g: Y \to Z\) be functions. Let \(C\) be a subset of \(Z\text{.}\) There is a relationship between \((g \circ f)^{-1}(C)\) and \(f^{-1}(g^{-1}(C))\text{.}\) Find and prove this relationship.
