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Suppose that $f:A\rightarrow B$ is a function. If $S\subseteq A$, then we define $f(S)$ to be the set $f(S)=\{f(x)\;:\;x\in S\}$.

So for example, if $f:\mathbb{R}\rightarrow \mathbb{R}$ is given by $f(x)=x^2$, then we have $f(\{1,2,3\})=\{1,4,9\}$, we have $f(\{−2,2,3\})=\{4,9\}$, and $f([−2,2])=[0,4]$.

Give an example of a function $f:\mathbb{R} \rightarrow \mathbb{R}$ and two sets $S,T\subseteq \mathbb{R}$ such that $f(S\cap T)\neq f(S)\cap f(T)$.

I need help; I don't have a clue. I have tried multiple examples but cannot seem to find one that works.

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  • $\begingroup$ Hint: Obviously the function $f$ can't be one-to-one. Did you try with $f(x) = x^2$? $\endgroup$
    – Simon S
    Commented Nov 21, 2014 at 0:51
  • $\begingroup$ I did, its the sets Im having trouble with $\endgroup$
    – Joe Neely
    Commented Nov 21, 2014 at 0:52
  • $\begingroup$ You want the ranges $f(S)$ and $f(T)$ to overlap 'more' than $S$ and $T$ do by themselves. How about $S = \{ 0, 1\}$ and $T =$ something else. I really don't want to give it away completely. $\endgroup$
    – Simon S
    Commented Nov 21, 2014 at 0:53
  • $\begingroup$ Thank you, this really helped. $\endgroup$
    – Joe Neely
    Commented Nov 21, 2014 at 0:56

2 Answers 2

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What about $f(x)=x^2$ and $S=\{-1\}$ and $T=\{1\}$?

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Pick a function that isn't injective, like $f(x) = x^2$, and exploit the fact that it isn't injective. Taking $S = \{-3, 1, 2\}$ and $T = \{1, 2, 3\}$ so that $S \cap T = \{1, 2\}$ and $f(S) = f(T) = \{1, 4, 9\}$, notice that: $$ f(S \cap T) = f(\{1, 2\}) = \{1, 4\} \neq \{1, 4, 9\} = \{1, 4, 9\} \cap \{1, 4, 9\} = f(S) \cap f(T) $$

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