Let $f \colon A \to B$ be a function and let $p$ and $q$ be subsets of $A$. How can we prove by counterexample that $f(p) \cap f(q)$ is not a subset of $f(p \cap q)$?
Could anyone show me how to solve this problem?
Please?
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Note that you mean "is not necessarily a subset" instead of "is not a subset". There are choices for $f, A, B, p, q$ where $f(p) \cap f(q)$ is a subset of $f(p \cap q)$, but it doesn't always hold. Furthermore, you should somehow indicate what you have tried yourself. This will help people give a response at an appropriate level. Finally a hint: pick $p$ and $q$ disjoint. – Magdiragdag Apr 30 '14 at 20:15
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Here's one possibility. Suppose $f:A\to B$ is a function which is not injective. You could take $f(x)=x^2$, for instance. In any case, if there are two elements $a$ and $a'$ for which $f(a)=f(a')$ but $a\neq a'$ then $f(\{a\}\cap\{a'\})=f(\varnothing)=\varnothing$ but $f(\{a\})\cap f(\{a'\})=\{f(a)\}\neq\varnothing$.
user134824
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Take $A=\{p,q\}$ with $p\ne q$ and $B=\{b\}$.
There is only one function $f:A\rightarrow B$ and it send $u$ and $v$ both to $b$.
Taking $P=\{p\}$ and $Q=\{q\}$ we find that $f(P\cap Q)=f(\emptyset)=\emptyset$.
$f(P)\cap f(Q)=\{b\}$ and is not a subset $f(P\cap Q)=\emptyset$.
Personally I would call this the most basic counterexample.
drhab
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Thank you so much. Then how can I prove if f is a one to one function then f(P) intersect f(Q) is a subset of f (P intersect Q)? By using contradiction method? Lets say f is not one to one function and f(p) intersect f(q) is subset of f(p intersect q). – chris Apr 30 '14 at 20:27
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@chris Let $f$ be one to one an let $x\in f(P)\cap f(Q)$. Then $x=f(u)$ for some $u\in P$ and $x=f(v)$ for some $v\in Q$. Then $f(u)=x=f(v)$ and consequently $u=v$ since $f$ is one to one. Then $u=v\in P\cap Q$ and $x=f(u)=f(v)\in f(P\cap Q)$. This proves that $f(P)\cap f(Q)\subset f(P\cap Q)$ if $f$ is one to one. – H.B. May 01 '14 at 12:14