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The question is as follows:

Let f be a mapping of the space $\mathcal{X}$ to the space $\mathcal{Y}$ . Let $\{A_t: t \subset T\}$ be a collection of subsets of $\mathcal{X}$, where $T$ is an arbitrary index set. If for any $A \subset \mathcal{X}$, $f$ is defined as $f(A)=\{f(x):x\in A\}$, then $$(a) \;\;\;\;\;\; f(\bigcup_{t\in T}A_t)=\bigcup_{t\in T}f(A_t)$$ $$(b) \;\;\;\;\;\; f(\bigcap_{t\in T}A_t)\subset \bigcap_{t \in T}f(A_t)$$ $$(c) \;\;\;\;\;\; \mathrm{Give\;an\;example\;where\;f(\bigcap_{t\in T}A_t) \neq \bigcap_{t\in T}f(A_t)}$$

Here is my attempted work:

Attempt

For $(a)$ $$f(\bigcup_{t\in T}A_t)=\{f(x):x\in \bigcup_{t\in T}A_t\}=\bigcup_{t\in T}\{f(x):x\in A_t\}=\bigcup_{t\in T}f(A_t)$$

For $(b)$, $\bigcap_{t\in T}f(A_t)$ can have quite a bit of overlapping values (e.g if we create sets for which a symmetric function is over). Hence, it will certainly contain values for which $f(\bigcap_{t\in T}A_t)$ may not, and so it must be the case that $\bigcap_{t\in T}f(A_t)$ contains $f(\bigcap_{t\in T}A_t)$.

For $(c)$, consider $f(x)=x^2$, where we sets $A_1=(-2,0), A_2=(0,2)$. Observe that $$f(A_1)=(0,4),\;f(A_2)=(0,4)$$ $$\implies f(A_1)\cap f(A_2)=(0,4)$$ But $A_1,A_2$ are clearly disjoint, so it follows that $$f(A_1 \cap A_2)=f(\emptyset)=\emptyset$$ $$\implies f(A_1 \cap A_2)\neq f(A_1)\cap f(A_2)$$

I was honestly not sure how to prove $(a),(b)$ (especially $(b)$). Do my arguments seem valid or shall I go back to the cutting board. They didn't seem all that difficult to formulate at first glance but as I proceeded, the overthinking began to ruin me. -_-

If anyone could guide me down the right road, I'd be much obliged.

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  • $\begingroup$ For (A) I think you are on the right track. You need to show that if an element is chosen arbitrarily from the left hand side of the equation it is also in the right hand side of the equation. You then must show that if you start with an arbitrary element in the right hand side of the equation it is also in the left hand side. I think what you have is a good start, but maybe be more explicit so that you can see it more clearly. $\endgroup$ – Zeta10 Feb 17 '15 at 7:22
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I would recommend not attempting to prove equality of sets directly. To prove sets $A$ and $B$ are equal you should prove that $A\subseteq B$ and $B\subseteq A$. To prove $A\subseteq B$, let $x\in A$ and show that this necessarily means $x\in B$. I will prove $f(\bigcup_{t\in\mathcal T}A_t)\subseteq\bigcup_{t\in\mathcal T}f(A_t)$ and let you have a go at the other inclusions you must prove.

Let $x\in f(\bigcup_{t\in\mathcal T}A_t)$. Then there exists $y\in\bigcup_{t\in\mathcal T}A_t$ such that $f(y)=x$. In particular, $y\in A_{t_0}$ for some $t_0\in\mathcal T$. This implies $x=f(y)\in f(A_{t_0})$. Now if $A\subseteq B$, it follows that $f(A)\subseteq f(B)$ (you might like to prove this fact). Hence $x\in f(A_{t_0})\subseteq f(\bigcup_{t\in\mathcal T}A_t)$, completing the proof.

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  • $\begingroup$ I felt this would be a better approach to (a), but wasn't sure. Thank you for your answer! Do you find (b) to be sufficient or does that also need some fixing? $\endgroup$ – user146925 Feb 19 '15 at 22:31

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