Define $f : X \to Y$. Prove $f[ \bigcup_{i}A_i] = \bigcup_{i} f[A_i]$ for a family $\{A_i\}$ of subsets of $X$.

Question

Define $f : X \to Y$. Prove $f[ \bigcup_{i}A_i] = \bigcup_{i} f[A_i]$ for a family $\{A_i\}$ of subsets of $X$.

My Attempted Proof.

We first show $f\left[ \bigcup_{i}A_i\right] \subset \bigcup_{i} f\left[A_i \right]$. Pick $x \in \bigcup_{i}A_i$, we show $f(x) \in \bigcup_{i} f\left[A_i \right]$. Since $x \in \bigcup_{i}A_i$, we have $x \in A_i$ for some $i$, hence we have $f(x) \in f[A_i]$ for this $i$, and thus $f(x) \in \bigcup_{i} f[A_i]$. Therefore it follows that $f\left[ \bigcup_{i}A_i\right] \subset \bigcup_{i} f\left[A_i \right]$.

Conversely we show $f\left[ \bigcup_{i}A_i\right] \supset \bigcup_{i} f\left[A_i \right]$. Fix $i$, take $x \in A_i$, then $f(x) \in f[A_i]$ and $f(x) \in \bigcup_{i} f[A_i]$. Since $x \in A_i$ for some $i$, we have $x \in \bigcup_{i} A_i$ certainly and hence $f(x) \in f\left[\bigcup_{i} A_i\right]$, and thus it follows that $f\left[ \bigcup_{i}A_i\right] \supset \bigcup_{i} f\left[A_i \right]$. $\ \ \ \square$

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Is my proof correct? If so any comments on my proof writing and how to make my arguments clearer to the reader are greatly appreciated.

Answer 1

Your proof is basically correct, but there is one part where your wording isn't really right.

Conversely we show $f\left[ \bigcup_{i}A_i\right] \supset \bigcup_{i} f\left[A_i \right]$. Fix $i$, take $x \in A_i$, then $f(x) \in f[A_i]$ and $f(x) \in \bigcup_{i} f[A_i]$.

The second sentence is written backwards. You want to prove that $f\left[ \bigcup_{i}A_i\right] \supset \bigcup_{i} f\left[A_i \right]$, so you want to prove that if $y\in \bigcup_{i} f\left[A_i \right]$ then $y\in f\left[ \bigcup_{i}A_i\right]$. So your initial assumption should not be that you have $x\in A_i$, but that you have $y\in \bigcup_{i} f\left[A_i \right]$. You can then use the definition of union and image to conclude that there exists $i$ and $x\in A_i$ such that $y=f(x)$. For instance, you might write:

Let $y\in \bigcup_{i} f\left[A_i \right]$. Then there exists $i$ such that $y\in f\left[A_i \right]$, which means there exists $x\in A_i$ for that $i$ such that $y=f(x)$.