Let $A$ and $B$ be sets and let $f:A\rightarrow B$ be a function. Let $I$ and $K$ be nonempty sets, let $\big\{U_i\big\}_{i\in I}$ be a family of subsets of $A$ indexed by $I$, and let $\big\{V_k\big\}_{k\in K}$ be a family of subsets of $B$.

Lemma. $f\big ( \bigcup_{i \in I} \big) = \bigcup_{i \in I} f \big( U_i \big)$.

Prove: $f^{-1} \big( \bigcup_{k\in K} V_k \big) = \bigcup_{k \in K} f^{-1} \big( V_k \big)$

The following is an incorrect proof, and I'm supposed to locate the flaw(s) embedded in the proof. I believe there are two mistakes in the proof, so I would like verification of the the problems I acknowledged after presenting the "proof," as well as any additional mistakes that I may have missed. Thank you in advance for taking the time to look at this post.


Applying $f$ $\,$ to $\,$ $f^{-1} \big( \bigcup_{k\in K} V_k \big)$, we obtain

$$f\big(f^{-1} \big( \bigcup_{k\in K} V_k \big) \big) = \bigcup_{k\in K} V_k \tag{1} $$

Applying $f$ to $\bigcup_{k \in K} f^{-1} \big( V_k \big)$, and using the lemma, we incur

$f\big(\bigcup_{k \in K} f^{-1} \big( V_k \big) \big) = \bigcup_{k \in K} f\big (f^{-1}\big) \big( V_k \big) = \bigcup_{k\in K} V_k \tag{2}$

Because applying $f$ to both sides of the equation $f^{-1} \big( \bigcup_{k\in K} V_k \big) = \bigcup_{k \in K} f^{-1} \big( V_k \big)$ yields the same result, we deduce that the equation is true, hence proving the reuslt. $\blacksquare$

First, the deduction that $f$ and $f^{-1}$ "cancel" each other is not necessarily the case, for we do not know if $f$ is bijective, only that it is a function.

Lastly, the second part of the proof which renders equation $(2)$ has an incorrect use of the lemma. The lemma only works for the set image $f$, not the function $f$, so we cannot interchange the $f$ from the "outside" to the "inside" of the union function.

Are my observations incorrect? If so, would you mind showing my how? Further, are there any other mistakes in the "proof"?


Note that "$f$ cancels $f^{-1}$" is used twice: Once in $(1)$, and once in the second "$=$" of $(2)$.

Your complaint about the set-image does not apply, though - we are talking about $f$ applied to sets here throughout. The application of the lemma is the only correct step in te "proof", I suppose - just let $U_i=f(V_i)$. - There is however a notational error at that point: The $f(f^{-1})(V_k)$ is supposed to be $f(f^{-1}(V_k))$ or $(f\circ f^{-1})(V_k)$

The final conclusion of the kind $f(A)=f(B)\implies A=B$ once again illegally assume that $f$ is injective.

  • $\begingroup$ Okay, I see how my second objection is no good. It seems the only problem with the argument is that $f$ is (illegally) assumed to have an inverse. $\endgroup$
    – J. Dunivin
    Nov 4 '15 at 21:38
  • $\begingroup$ Equation (1) in the proof is invalid.For example if $\phi \ne S\subset \cup_kV_k=B\backslash \{f(x):x\in A\}$ the LHS is $\phi$, then the RHS is $S$. $\endgroup$ Nov 4 '15 at 23:54
  • $\begingroup$ A simple direct proof is $ \forall x : x\in f^{-1}\cup_kV_k\iff f(x)\in \cup_kV_k\iff \exists k (f(x)\in V_k)\iff \exists k (x\in f^{-1}V_k)\iff x\in \cup_kf^{-1}V_k$. $\endgroup$ Nov 5 '15 at 0:04

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