Proof Wrong: Where f(A∩B) ≠ f(A) ∩ f(B)

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SUMMARY

The discussion centers on the mathematical proof that f(A∩B) is not equal to f(A) ∩ f(B). Participants emphasize that to establish this inequality, one must demonstrate that f(A) ∩ f(B) is not a subset of f(A∩B). A counterexample is suggested, involving a function f and distinct elements p and q from sets A and B, respectively, where f(p) equals f(q). This highlights the necessity of proving both inclusions to validate the inequality.

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woundedtiger4
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I know that f(A∩B) is not equal to f(A) ∩ f(B) but i don't know that where am i wrong in the following proof...:( can someone please give me an intuitive example?
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In order to prove "[itex]X= Y[/itex]" you must prove both "[itex]X\subseteq Y[/itex]" and [itex]Y\subseteq X[/itex].

You want to prove that [itex]f(A\cap B)[/itex] is not equal to [itex]f(A)\cap f(B)[/itex] and your give proof shows only that [itex]f(A\cap B)\subseteq f(A)\cap f(B)[/itex].

So look for a counter example in which [itex]f(A)\cap f(B)[/itex] is NOT a subset of [itex]f(A\cap B)[/itex].

That is, find a function f and values p and q, p in A, q in B such that f(p)= f(q).
 
HallsofIvy said:
In order to prove "[itex]X= Y[/itex]" you must prove both "[itex]X\subseteq Y[/itex]" and [itex]Y\subseteq X[/itex].

You want to prove that [itex]f(A\cap B)[/itex] is not equal to [itex]f(A)\cap f(B)[/itex] and your give proof shows only that [itex]f(A\cap B)\subseteq f(A)\cap f(B)[/itex].

So look for a counter example in which [itex]f(A)\cap f(B)[/itex] is NOT a subset of [itex]f(A\cap B)[/itex].

That is, find a function f and values p and q, p in A, q in B such that f(p)= f(q).
Thank you sir.
I understand your point & I can guess the best example is one to many which is not true, & i have tried to show it in proof.
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