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Prove that if f is injective, then $f(A \cap B) = f(A)\cap f(B)$

My answer:

i) $f(A \cap B) \subset f(A) \cap f( B )$

Take an $x \in A \cap B$.

$x \in A \cap B \implies x \in A \land x \in B$

$x \in A \implies f(x) \in f(A) $

$x \in B \implies f(x) \in f(B) $

So, $x \in A \cap B \implies f(x) \in f(A) \land f(x) \in f(B) $

But, $x \in A \cap B \implies f(x) \in f(A \cap B)$

So, $f(x) \in f(A \cap B) \implies f(x) \in f(A) \land f(x) \in f(B)$

Therefore, $f(A \cap B ) \subset f(A) \cap f(B)$

ii) $f(A) \cap f( B ) \subset f(A \cap B) $

f is injective, so $f(x) = f(y) \implies x=y$

Take an $x \in A, y \in B : f(x) = f(y)$

$x \in A \implies f(x) \in f(A) $

$y \in B \implies f(y) \in f(B) $

But $x = y \implies x \in B \implies f(x) \in f(B) $

Since $f(x) \in f(A) \land f(x) \in f(B) \implies f(x) \in f(A \cap B)$

So $f(A) \cap f( B ) \subset f(A \cap B) $

Is that ok?

Silva
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    I guess you can make your first containment proof more rigorous if that's something you like. Let $y \in f(A \cap B)$. By definition, there exists $x \in A \cap B$ so that $f(x) = y$. Since $x \in A$, you have $y \in f(A)$. Similarly, $x \in B$ implies $y \in f(B)$. – MathNewbie May 23 '15 at 20:55
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    Your part ii) is not correct. In order to prove that $f(A) \cap f( B ) \subset f(A \cap B)$, you must start by letting $x$ be a particular but arbitrary element of $f(A) \cap f( B )$, and then conclude (using the fact that $f$ is injective as necessary) that $x$ is an element of $f(A \cap B)$. You didn’t start part ii) by choosing a particular but arbitrary element of $f(A) \cap f( B )$, so it’s not clear what your statements in part ii have shown. @Walter has given a correct proof of part ii. (He used the letter $y$ for the particular but arbitrary element of $f(A) \cap f( B )$.) – Steve Kass May 23 '15 at 21:18

1 Answers1

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(ii)

if $y \in f(A) \cap f(B)$ then $y \in f(A) \land y \in f(B)$. Hence $\exists x \in A$ such that $f(x) = y$ and $\exists x' \in B$ such that $f(x') = y$. Then, because f is injective, $x = x' \in A \cap B$ so $y = f(x) = f(x') \in f(A \cap B)$. Hence $f(A) \cap f(B) \subset f(A \cap B)$.

I think this is a more direct proof than the one you posed.

Stanley
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