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Since $A \cup B = \left\{ x \,\vert\, x \in A \lor x \in B \right\}$ doesn't it mean that this holds in general $\left\{ f(x) \,\vert\, x \in X \right\} = \bigcup_{x \in X}\left\{ f(x) \right\}$ ($f$ is some function $f : X \rightarrow Y$)?

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    $\begingroup$ Well, you always have $A = \bigcup_{a \in A} \{a\}$ whether or not functions are involved. This is pretty much the same thing. $\endgroup$ – Randall Oct 8 '18 at 14:07
  • $\begingroup$ This question arose when I was looking at "Incomplete Database" theory and two definitions where given: $Q(D) := \left\{Q(D) \,\vert\, D \in I \right\};$ $possible(Q, I) := \bigcup_{D \in I} Q(D);$ So the question actually is: $Q(I)$ and $possible(Q, I)$ are equal, aren't they? $\endgroup$ – Mega-X Oct 8 '18 at 14:16

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