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$f(S \cap T) \subseteq f(S) \cap f(T)$

$x$ lies in ($S \cap T$), which means the domain has fewer elements than the domain of $S$ and $T$, since $x$ must be in $S$ and $T$. All $f(x)$ values of $x$, which resides in ($S \cap T$) is also a member of $f(S) \cap f(T)$, because $f(S)$ encompasses all $x$ in $S$ even those in ($S \cap T$) and the same can be said about $f(T)$.

Can you give me the solution?

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    Yes, assuming $S \cap T$ is not empty (if empty, then it's trivially a subset of the right hand side), $x \in S \cap T$, which means $x \in S$ and $x \in T$ Good. Since $f$ is a function, $f$ is a mapping of such each such x to $f(S \cap T)$. Being in $S \cap T$ is not the same as being in $f(S\cap T)$.2012-11-06
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    **Not** duplicate of question referred to above.2012-11-06
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    @AndréNicolas: I messed up on that one. You are absolutely right. how can I clean my close vote?2012-11-06
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    @Thomas: Don't know how to remove a close vote. Probably doesn't matter, there will presumably not be further close votes. Unless someone finds an earlier question on MSE that is essentially the same (there are).2012-11-06
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    Re-open request: http://meta.math.stackexchange.com/a/6519/1543 I don't think the "corrected" duplicate target actually answers the question here.2012-11-06
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    See also http://math.stackexchange.com/questions/939730/2014-09-21

3 Answers 3