5
$\begingroup$

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

$x$ 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?

$\endgroup$
  • 1
    $\begingroup$ 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 $f(S\cap T)$. $\endgroup$ – Namaste Nov 6 '12 at 3:24
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $x\in f(S)\cap f(T)$.

$\endgroup$
2
$\begingroup$

$y \in f(S\cap T)$ 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 $f(S\cap T)\subseteq f(S)\cap f(T)$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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... $\endgroup$ – Namaste Nov 6 '12 at 3:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.