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

$x$ 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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    $\begingroup$ 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 $f(S\cap T)$. $\endgroup$ – amWhy Nov 6 '12 at 3:24
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    $\begingroup$ Not duplicate of question referred to above. $\endgroup$ – André Nicolas Nov 6 '12 at 3:31
  • $\begingroup$ @AndréNicolas: I messed up on that one. You are absolutely right. how can I clean my close vote? $\endgroup$ – Thomas Nov 6 '12 at 3:36
  • $\begingroup$ @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). $\endgroup$ – André Nicolas Nov 6 '12 at 3:40
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    $\begingroup$ Re-open request: meta.math.stackexchange.com/a/6519/1543 I don't think the "corrected" duplicate target actually answers the question here. $\endgroup$ – Willie Wong Nov 6 '12 at 11:38

For this one you take an element $x\in f(S\cap T)$. You want to prove that this element is also in $f(S)\cap f(T)$. That is, you want to prove that $x\in f(S)$ and also $x\in f(T)$. Now since $x\in f(S\cap T)$ you know that there is some $y\in S\cap T$ such that $x = f(y)$. Now $y\in S\cap T$ so in particular $y\in S$, so that means $x = f(y) \in f(S)$. The same argument works to show that $x\in f(T)$.

Hence in all $x$ is both an element of $f(S)$ and $f(T)$, so $x\in f(S)\cap f(T)$.


$y \in f(S\cap T)$ means $y=f(x)$ for some $x\in S\cap T$.

That means $y=f(x)$ for some $x$ for which $x\in S$ and $x\in T$.

That implies $y=f(x)$ for some $x\in S$, so $y\in f(S)$, and for the same reason, $y\in f(T)$.

Since $y \in f(S)$ and $y\in f(T)$, we have $y\in f(S)\cap f(T)$.

So we have proved that if $y\in f(S\cap T)$ then $y\in f(S)\cap f(T)$.

That is true of every value of $y$.

Thus every member of $f(S\cap T)$ is a member of $f(S)\cap f(T)$.

That's what it means to say $f(S\cap T)\subseteq f(S)\cap f(T)$.


Let $f(x) \in f(S \cap T)$. This means that $x \in S \cap T$, so $x \in S$ and $x \in T$. You fill in this rest.

  • $\begingroup$ Brian, that's pretty much as far as the OP got (i.e., he's got a handle on that...It's the rest of his post that's confused, or fuzzy... $\endgroup$ – amWhy Nov 6 '12 at 3:28

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