Show that $\bigcup_i f(A_i) = f(\bigcup_i A_i)$, where $A_i$ are subsets of $X$ and $f: X\to Y$. It seems intuitively obvious but yet I cannot prove it....
Answer
This is easily proved by ‘element-chasing’: assume that $x$ is an element of the lefthand side and chase through the definition of union to show that $x$ is an element of the righthand side, then do the opposite. I’ll do one direction and let you try the other.
Suppose that $x\in\bigcup_{i\in I}f[A_i]$. By the definition of union this means that there is at least one $i_0\in I$ such that $x\in f[A_i]$. Since $x\in f[A_{i_0}]$, there is some $y\in A_{i_0}$ such that $x=f(y)$. That’s about as much as we can get directly from the hypothesis, so let’s see where we’re trying to go: we want to show that $x\in f\left[\bigcup_{i\in I}A_i\right]$.
Okay, we know that $x=f(y)$; is $y\in\bigcup_{i\in I}A_i$? If it is, then certainly $x=f(y)\in f\left[\bigcup_{i\in I}A_i\right]$, and we’re home free. And the answer is yes: $y\in A_{i_0}\subseteq\bigcup_{i\in I}A_i$, so $y\in\bigcup_{i\in I}A_i$, and therefore $x\in f\left[\bigcup_{i\in I}A_i\right]$. Since $x$ was an arbitrary element of $\bigcup_{i\in I}f[A_i]$, we’ve shown that $$\bigcup_{i\in I}f[A_i]\subseteq f\left[\bigcup_{i\in I}A_i\right]\;.$$
Now you try the same approach to prove that $$f\left[\bigcup_{i\in I}A_i\right]\subseteq\bigcup_{i\in I}f[A_i]\;;$$ once you’ve done that, you can conclude that $$\bigcup_{i\in I}f[A_i]=f\left[\bigcup_{i\in I}A_i\right]\;.$$
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