Let $f:X \rightarrow Y$ be a function. For any subset $B$ of $Y$, define $f^{-1}(B) = \{x \in X | f(x) \in B\}$ as the preimage of $B$. Prove that for any collection of subsets $\{... Let $f:X \rightarrow Y$ be a function. For any subset $B$ of $Y$, define $f^{-1}(B) = \{x \in X | f(x) \in B\}$ as the preimage of $B$. Prove that for any collection of subsets $\{B_i\}_{i \in I}$ of $Y$, $f^{-1}(\bigcup_{i \in I} B_i) = \bigcup_{i \in I} f^{-1}(B_i)$.
Understand the Problem
The problem asks us to prove that the preimage of the union of a collection of sets is equal to the union of the preimages of those sets. We need to show that an element belongs to the left side of the equation if and only if it belongs to the right side.
Answer
$f^{-1}\left(\bigcup_{i \in I} A_i\right) = \bigcup_{i \in I} f^{-1}(A_i)$
Answer for screen readers
$f^{-1}\left(\bigcup_{i \in I} A_i\right) = \bigcup_{i \in I} f^{-1}(A_i)$
Steps to Solve
- Define the sets and function
Let $X$ and $Y$ be sets, and let $f: X \to Y$ be a function. Let ${A_i}{i \in I}$ be a collection of subsets of $Y$, where $I$ is an index set. We want to prove that $$ f^{-1}\left(\bigcup{i \in I} A_i\right) = \bigcup_{i \in I} f^{-1}(A_i) $$
- Show that $f^{-1}\left(\bigcup_{i \in I} A_i\right) \subseteq \bigcup_{i \in I} f^{-1}(A_i)$
Suppose $x \in f^{-1}\left(\bigcup_{i \in I} A_i\right)$. This means that $f(x) \in \bigcup_{i \in I} A_i$. By definition of the union, there exists some $i_0 \in I$ such that $f(x) \in A_{i_0}$. Then, by definition of the preimage, $x \in f^{-1}(A_{i_0})$. Since $x \in f^{-1}(A_{i_0})$ and $i_0 \in I$, it follows that $x \in \bigcup_{i \in I} f^{-1}(A_i)$. Therefore, $f^{-1}\left(\bigcup_{i \in I} A_i\right) \subseteq \bigcup_{i \in I} f^{-1}(A_i)$.
- Show that $\bigcup_{i \in I} f^{-1}(A_i) \subseteq f^{-1}\left(\bigcup_{i \in I} A_i\right)$
Suppose $x \in \bigcup_{i \in I} f^{-1}(A_i)$. This means that there exists some $i_0 \in I$ such that $x \in f^{-1}(A_{i_0})$. By definition of the preimage, $f(x) \in A_{i_0}$. Since $f(x) \in A_{i_0}$ and $i_0 \in I$, it follows that $f(x) \in \bigcup_{i \in I} A_i$. Then, by definition of the preimage, $x \in f^{-1}\left(\bigcup_{i \in I} A_i\right)$. Therefore, $\bigcup_{i \in I} f^{-1}(A_i) \subseteq f^{-1}\left(\bigcup_{i \in I} A_i\right)$.
- Conclude the proof
Since $f^{-1}\left(\bigcup_{i \in I} A_i\right) \subseteq \bigcup_{i \in I} f^{-1}(A_i)$ and $\bigcup_{i \in I} f^{-1}(A_i) \subseteq f^{-1}\left(\bigcup_{i \in I} A_i\right)$, we can conclude that $$ f^{-1}\left(\bigcup_{i \in I} A_i\right) = \bigcup_{i \in I} f^{-1}(A_i) $$
$f^{-1}\left(\bigcup_{i \in I} A_i\right) = \bigcup_{i \in I} f^{-1}(A_i)$
More Information
This result shows that the preimage operation preserves unions of sets. This is a fundamental property in set theory and is often used in more advanced mathematical proofs.
Tips
A common mistake is to confuse the preimage $f^{-1}(A)$ with the inverse function $f^{-1}$, which only exists if $f$ is bijective. The preimage is defined for any function, regardless of whether it's invertible. Also, not understanding how to prove the equality of two sets, which requires showing that each one is a subset of the other.
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