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Let $f:\;\mathbb R\longmapsto\;\mathbb R$, $A,B\subset\mathbb R$. Suppose we have $f(A)\subseteq B$, then prove

(1) $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$;

(2) $f^{-1}(A \cup B)=f^{-1}(A) \cup f^{-1}(B)$;

(3) $f^{-1}(A \cap B)=f^{-1}(A) \cap f^{-1}(B)$.

By 357725 , 291777 and 228711 , one can prove the above 3 formulas as follows

(1) $$ \begin{aligned} f^{-1}(\overline{A}) &:=\{x \in \operatorname{dom}(f): f(x) \in \overline A \} \\ &=\{x \in \operatorname{dom}(f): f(x) \notin A \} \\ &=: \overline{f^{-1}(A)}. \end{aligned} $$ (2) $$ \begin{aligned} f^{-1}(A \cup B) &:=\{x \in \operatorname{dom}(f): f(x) \in A \cup B\} \\ &=\{x \in \operatorname{dom}(f): f(x) \in A \text { or } f(x) \in B\} \\ &=\{x \in \operatorname{dom}(f): f(x) \in A\} \cup\{x \in \operatorname{dom}(f): f(x) \in B\} \\ &=: f^{-1}(A) \cup f^{-1}(B) . \end{aligned} $$ (3) $$ \begin{aligned} f^{-1}(A \cap B) &:=\{x \in \operatorname{dom}(f): f(x) \in A \cap B\} \\ &=\{x \in \operatorname{dom}(f): f(x) \in A \text { and } f(x) \in B\} \\ &=\{x \in \operatorname{dom}(f): f(x) \in A\} \cap\{x \in \operatorname{dom}(f): f(x) \in B\} \\ &=: f^{-1}(A) \cap f^{-1}(B). \end{aligned} $$ It seems that I prove the processes without using the condition Suppose we have $f(A)\subseteq B$.

Is this condition unnecessary? Or am I missing something? A counterexample?

Martin Sleziak
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John Stone
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1 Answers1

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Indeed, $f(A) \subseteq B$ is unnecessary for the exercises as they are currently presented.

Consider the following set of exercises:

(a) $A \subseteq B \implies f^{-1}(A) \subseteq f^{-1}(B)$

(b) $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$;

(c) $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$.

Such a set of exercises may be found in the beginning chapter of Topology by Munkres, and indeed — there is no requirement that $f(A) \subseteq B$. Your proofs are definitely valid. The only requirement is that $A,B$ are subsets of the range of $f$, which you said they are.

The point of the exercises is to prove that the preimage of a function preserves inclusions, unions, intersections, and set differences.

polite proofs
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