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So this is a question out of Real Analysis book that I'm working through, and I'm quite stuck.

Give a counterexample to one of the following four formulas for images and inverse images of sets (the other three are true)

  1. $f(X_1 \cup X_2) = f(X_1) \cup f(X_2$),
  2. $f^{-1}(Y_1 \cup Y_2) = f^{-1}(Y_1) \cup f^{-1}(Y_2)$,
  3. $f(X_1 \cap X_2) = f(X_1) \cap f(X_2)$,
  4. $f^{-1}(Y_1 \cap Y_2) = f^{-1}(Y_1) \cap f^{-1}(Y_2)$

I went through Overview of basic results about images and preimages and from other resources I can find online, all 4 of these are true... I can't figure out what I am missing.

Thanks for the help

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  • $\begingroup$ The fifth property in your linked post is exactly about 3. check it out. Have you actually read that question completely? $\endgroup$ – Henno Brandsma May 6 '18 at 6:24
  • $\begingroup$ I did, I must have skimmed and missed that. However when going through that answer I fear I need some clarification. If f is constant, then isn't it true that I don't have a bijection any longer? In my text it states that a bijection is needed for an inverse function to exist, and since we're involving the inverses here I imagined we were assuming a bijection. $\endgroup$ – jaibhavaya May 6 '18 at 13:31
  • $\begingroup$ No, $f^{-1}[A] = \{x \in X: f(x) \in A\}$, there is no need for a bijection. The $f^{-1}[A]$ is just a notation for a set. $\endgroup$ – Henno Brandsma May 6 '18 at 13:33
  • $\begingroup$ Well that is precisely where my confusion was then. My book had something like "... then the function is a bijection and has an inverse $f^{-1}(X)$ ". Thank you for the help! Marking the answer below correct because it said essentially the same thing as you did. $\endgroup$ – jaibhavaya May 6 '18 at 13:36
  • $\begingroup$ Wait.. but copying from Wikipedia: "Not all functions have inverse functions. In order for a function f: X → Y to have an inverse it must have the property that for every y in Y there must be one, and only one x in X so that f(x) = y. This property ensures that a function g: Y → X will exist having the necessary relationship with f.". So f cannot be a constant function and still have an inverse. $\endgroup$ – jaibhavaya May 6 '18 at 13:58
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3 is false, if F is a contant map and $X_1, X_2$ disjoint.
Exercise. Prove for 3 that the
left hand side is a subset of the right hand side.

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  • $\begingroup$ See my comments above. $\endgroup$ – jaibhavaya May 6 '18 at 14:01

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