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

Suppose there is a $x$ that is in $S$, but not in $T$, then there is a value $y$ such that $f(x) = x$, that is in $f(S)$, but not in $f(S \cap T)$. Suppose there is a $x$ that is in $T$, but not in $S$, then there is a value $y$ such that $f(x) = x$ that is in $f(T)$, but is not in $f(S \cap T)$.

*correction made

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  • $\begingroup$ No. If $x$ is in $S$ but not $T$, and $y$ is in $T$ but not in $S$, it's possible there are $z$ and $w$ in $S\cap T$ such that $f(z)=f(y)$ and $f(w)=f(x)$. $\endgroup$
    – Gerry Myerson
    Nov 4, 2012 at 2:37
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    $\begingroup$ You might want to begin your proof with this: Let $y \in f(S \cap T)$ so $y = f(x)$ for some $x \in $ ... $\endgroup$
    – wj32
    Nov 4, 2012 at 2:46
  • $\begingroup$ ahh im too confused now $\endgroup$
    – Gladstone Asder
    Nov 4, 2012 at 2:49
  • $\begingroup$ Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. $\endgroup$
    – Martin Sleziak
    Nov 4, 2012 at 6:50

3 Answers 3

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No. In general, to show that $A\subset B$, it is not sufficient to show that there is $b\in B$ so that $b \notin B$. Take $A=\{1,2\}$ and $B=\{2,3\}$. Then $3 \notin A$ but $B\not \subset A$. You need to show that for every $a\in A$, $a\in B$ to show that $A\subset B$. I hope that helps.

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What if there's an element that is in the $S \cap T$ that also maps to $y$? In other words, what if $f$ isn't injective? I honestly stopped reading there.

Usually when showing that a set is a subset of another, you want to take a direct approach. I'm not saying that a contradiction wouldn't work, and I believe that in more complicated situations it would, but for simple properties like this, you should try a direct approach first:

I'll be informal (or at least not use symbols), so you can seen if you understand the proof and write it yourself.

You pick an element in the set on the left, and you see that its preimage sits in $S\cap T$. This just follows from the definition of the preimage. So the preimage sits in both $S$ and $T$. Now apply $f$ to the elements in the preimage, what do you get?

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  • $\begingroup$ i just had to add such that f(x) = x, right? $\endgroup$
    – Gladstone Asder
    Nov 4, 2012 at 2:58
  • $\begingroup$ No, that doesn't make sense to me. You need to start with an element in $f(S \cap T)$, not with an element in $S$ or $T$. Try to see if you can prove it with a straightforward approach (i.e. direct proof and not a contradiction). Once you understand it, play around with the proof and see if you can prove it with a contradiction. I honestly haven't tried doing it using a contradiction, but if I were to do so, I don't think it'd work the way you started it. $\endgroup$
    – eliya gwetta
    Nov 4, 2012 at 3:02
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How about $S = \{1,\ 2,\ 3\}$ and $T = \{2,\ 3,\ 4\}$ with $f$ defined as $$f(1) = 1,\ f(2) = 1,\ f(3) = 2,\ f(4) = 2$$ Then $1\notin T$ but every value in $f(S) = \{1,\ 2\}$ is in $f(S\cap T) = \{1,\ 2\}$. Similarly with $4\notin S$.

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