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I am trying to prove that Proving $f\left(\bigcup\limits_{i \in I} A_i\right) = \bigcup\limits_{i \in I} f(A_i)$. Here is my attempt.

Given $y \in Y$, we have: \begin{align*} y \in f\left(\bigcup\limits_{i \in I} A_i\right) & \iff \exists a \in \bigcup\limits_{i \in I} A_i, \; y = f(a) \\ & \iff \exists i \in I, \; a \in A_i, \; y = f(a) \\ & \iff \exists i \in I, \; y \in f(A_i) \\ & \iff y \in \bigcup\limits_{i \in I} f(A_i). \end{align*}

How does this look?

J.-E. Pin
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Brad G.
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2 Answers2

2

This solution is completely correct.

Matheus Andrade
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    This is more suitable as a comment. – Anurag A Jul 21 '21 at 03:16
  • @AnuragA I'd agree if this was a more advanced question but I don't really think there's any meaningful improvement to be made on OP's proof and thought it better the question wasn't left unanswered (and I don't think anyone else is really going to answer anything else... but I'd be happy to be wrong). – Matheus Andrade Jul 21 '21 at 03:21
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    @AnuragA It is better to have it as an answer, so the asker has something to mark as accepted and show the matter is closed. – Mike Earnest Jul 21 '21 at 03:21
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    +1 This answered the question. – Umberto P. Jul 21 '21 at 03:36
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Let me write down little additional specification based on, that $(\exists x \in X)R(x)\Leftrightarrow (\exists x)(x \in X\land R(x))$:

$$\begin{align*} &\exists a \in \bigcup\limits_{i \in I} A_i, \; y = f(a) \\ & \iff (\exists a)\Big(a \in \bigcup\limits_{i \in I} A_i \land y = f(a)\Big)\\ &\iff (\exists a)\Big((\exists i)(i \in I \land a \in A_i) \land y = f(a)\Big)\\ &\iff (\exists a)(\exists i)\Big(i \in I \land a \in A_i \land y = f(a)\Big)\\ \end{align*}$$ Now you can reorder/move/group existential quantifiers and logical and and get 3'd line.

zkutch
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