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I just wonder if my following solution is true.

Let $X,Y$ be sets, let $f:X\to Y$ be a function, let $\{Y_i\}_{i\in I}$ be a family of subsets of $Y$. (Note: I use equalities instead of mutual containment)

$$\begin{align}f^{-1}\left[\bigcup_{i\in I} Y_i\right] &= \{x \in X: \mbox{there exists an}\quad i \in I\mbox{ such that } y \in Y_i,f(x)=y\} \\&=\bigcup_{i \in I} f^{-1}\left[Y_i\right] \end{align}$$

I initially do not know how to get from the left to right, but when I put both sets in set notation, they turn out to be the same, hence the one line proof. Something go ultimately wrong?

VIVID
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Daniel
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  • Do the $X_i$ serve any purpose here? Whether what is written is enough is a matter of comfort, I guess. – Dylan Moreland Jul 23 '12 at 21:15
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    There’s nothing wrong with your proof that $$f^{-1}\left[\bigcup_{i\in I}Y_i\right]=\bigcup_{i\in I}f^{-1}[Y_i];,$$ except that the $\infty$ above the $\bigcup$ is incorrect. What do the sets $X_i$ have to do with it? – Brian M. Scott Jul 23 '12 at 21:15
  • sorry about the $X_i$, I wrote out the whole question, and the question I posted is just part of the problems. and yes you guys are right about the infinity, that is typo – Daniel Jul 23 '12 at 21:17

1 Answers1

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The statement is true, and your argument is essentially right, but I would say that you are skipping steps to achieve that identification. (Also, you should not have those $\infty$'s on top of the union symbol). I would instead add:

$$\begin{align*} f^{-1}\left[\bigcup_{i\in I}Y_i\right] &= \left\{ x\in X\;\left|\; f(x)\in\bigcup_{i\in I}Y_i\right\}\right.\\ &=\Biggl\{x\in X\;\Biggm|\; \exists i\in I\text{ such that }f(x)\in Y_i\Biggr\}\\ &= \bigcup_{i\in I}\{x\in X\mid f(x)\in Y_i\}\\ &= \bigcup_{i\in I}f^{-1}[Y_i]. \end{align*}$$ The first equality is by definition of inverse image; the second by definition of the union; the third is by definition of union; and the fourth by definition of inverse image.

Arturo Magidin
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  • yes you are right about skipping steps, since I am not sure what step I skipped too – Daniel Jul 23 '12 at 21:22
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    @jsk: In my opinion, you skipped what my first and third steps were. That is, you jumped from "f(x) must be in the union" to "f(x) must be in one of the sets", and from "f(x)$ must be in one of the sets" to "it's the union of these sets". By making them explicit, the process is a bit clearer. – Arturo Magidin Jul 23 '12 at 21:23
  • yes you are absolutely right – Daniel Jul 23 '12 at 21:25
  • So this is true for arbitrary union. I also know this relation holds for finite intersection. Can we see such relation in arbitrary intersection? – Mariana Nov 21 '21 at 01:32
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    @Mariana: The inverse image of an arbitrary intersection equals the intersection of the inverse images. It is as easy to verify as the above was nine+ years ago. – Arturo Magidin Nov 21 '21 at 02:01
  • @ArturoMagidin Exactly. I verified. Thanks – Mariana Nov 21 '21 at 02:17