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Let $A$ be a set. $y \in f(A)$ iff $f(x) = y$ for some $x \in A$.

Suppose $A = \{2, 3\}$ and $f(x) = x^2$. Then $f(A) = \{4, 9\}. f(-2) \in f(A)$, but $-2 \notin A$. Is it contradicting the definition or is it just a case where the image of $f \neq$ codomain of $f$? Do we use the definition above to compute $f(A)$?

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PornStar
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1 Answers1

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If $A$ is a subset of the domain of $f$, then one helpful way to think about $f(A)$ might perhaps be $$ f(A) = \{f(x) : x \in A\}. $$ You can check that this agrees with the definition you gave above.

As for your example, assuming $-2$ is in the domain of $f$, it is true that $f(-2) \in f(A) = \{f(2),f(3)\} = \{4,9\}$ since $f(-2) = 4$, but this causes no contradiction! Indeed, by your definition $4 \in f(A)$ if and only if there is some $x \in A$ such that $f(x) = 4$; since $2 \in A$ and $f(2) = 4$, then we're a-ok. The important thing to note is that your definition does NOT imply that if $f(x) \in f(A)$ then $x \in A$, rather it implies that if $f(x) \in f(A)$ then there is some $\tilde{x} \in A$ such that $f(\tilde{x}) = f(x)$.

Tom
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  • $f(x) \in f(A) \to \exists \tilde x \in A$ such that $f(\tilde x) =f(x)$. So, $4 \in f(A) \to -2 \in A$ such that $f(-2) = f(2)$? – PornStar Jun 12 '14 at 19:04
  • @PornStar $f(x) \in f(A) \implies \exists\tilde{x} \in A$ such that $f(\tilde{x}) = f(x)$. So, $f(-2) \in f(A) \implies \exists \tilde{x} \in A$ such that $f(\tilde{x}) = f(-2)$. – Tom Jun 12 '14 at 19:09
  • This is great! Do we word the converse just the same with the arrow in the reverse direction or is there a better way to word the converse so that there are no ambiguities and other problems like the one in the OP of this thread? – PornStar Jun 12 '14 at 19:20
  • @PornStar Truthfully: $f(x) \in f(A) \iff \exists \tilde{x} \in A$ such that $f(\tilde{x}) = f(x)$. You can then deduce the "other direction" that if $\tilde{x} \in A$ and $f(\tilde{x}) = f(x)$ for some $x$ in the domian of $f$, then $f(x) \in f(A)$. – Tom Jun 12 '14 at 19:23
  • This is awesomely helpful. Thank you, Tom! – PornStar Jun 12 '14 at 19:27