2

If it is the case that $ A \subseteq B,$ then is it always true that the inverse images obey $f^{-1}(A) \subseteq f^{-1}(B),$ where $f$ is a function that is not necessarily injective, and has a codomain containing $B$?

The reason I am asking is due to caution that the inverse image of a set $S$ can be larger than the set $S$.

EDIT My attempt based on the comment by @jjagmath. Let $f:X \to Y$ and $A \subseteq B \subseteq Y$. The definition of inverse image $f^{-1}$ applied to $A$ is $ f^{-1}(A) = \{x \in X: f(x) \in A\}$.

Similarly for $B, f^{-1}(B) = \{x \in X: f(x) \in B\}$.

Since $A \subseteq B$, then $f(x) \in A \implies f(x) \in B$.

Thus, $ f^{-1}(A) \subseteq f^{-1}(B) \; \square$.

amWhy
  • 210,739
Penelope
  • 3,635
  • 7
    Have to tried using the definition of $f^{-1}$? – jjagmath Jan 10 '24 at 02:18
  • @jjagmath let me try, I'll add it to the question as an edit. – Penelope Jan 10 '24 at 02:30
  • 6
    It's a good exercise for starting learner to get used to this kind of routine. To see wheather $f^{-1}(A)\subseteq f^{-1}(B)$ holds, you need to choose an $x\in f^{-1}(A)$, and see how to conclude $x\in f^{-1}(B)$,which by definition of $f^{-1}$ that is how to conclude $f(x)\in B$ from $f(x)\in A$. If you can't work it out, try some examples and find the difficulty, then you can try to construct some counter-examples with the difficulty. That's the routine for this kind of proof. Try yourself and good luck. – Egyptian Jan 10 '24 at 02:37
  • 1
    @Penelope Any work or attempts will be good. – jjagmath Jan 10 '24 at 02:38
  • @jjagmath done. – Penelope Jan 10 '24 at 02:38
  • @MichaelBurr - thanks for the correction, I've updated. – Penelope Jan 10 '24 at 02:50
  • This post is a megaduplicate. See for instance https://math.stackexchange.com/questions/3645591 found [using Approach0](https://approach0.xyz/search/?q=AND site%3Amath.stackexchange.com%2C OR content%3A%24f%5E%7B-1%7D(A) %5Csubseteq f%5E%7B-1%7D(B)%24&p=1) – Anne Bauval Jan 12 '24 at 07:02

1 Answers1

3

Here is one possible approach:

\begin{align*} x\in f^{-1}(A) & \Rightarrow f(x) \in A\subseteq B\\\\ & \Rightarrow f(x) \in B\\\\ & \Rightarrow x\in f^{-1}(B) \end{align*}

Thus the proposed claim is true.

Hopefully this helps!

Átila Correia
  • 18,937