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I got the following very basic question:

Why is an image - in contrast to a preimage - defined by using the existential quantifier?

Let $f\colon A \to B$ be a mapping and $M$ a subset of $A$ and $N$ a subset of $B$. Then, the image of $M$ can be defined as:

  1. $f(M)=\{y\in B: \exists x \in M:(x)=y\}$

and the preimage of N as:

  1. $f^{-1}(N)=\{x\in A:() \in N\}$.

Best and thank you very much,

Niki

FD_bfa
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Nikiiiii
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    It would be helpful if you gave the definition that you're working with. The image of a function $f\colon X\to Y$ is usually not defined with an existential quantifier, it's defined as the set ${f(x)\mid x\in X}$. – Randy Marsh Mar 18 '23 at 00:25
  • sorry, you are right - I have updated my post. In the proofs i have seen, the above definitions are assumed. e.g., https://math.stackexchange.com/questions/359693/overview-of-basic-results-about-images-and-preimages – Nikiiiii Mar 18 '23 at 01:07

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If $f$ were simply a relation instead of a function, then both definitions would need an existential quantifier. However, the definition of a function is that given $x\in A$, there is one and only one corresponding element of $B$, and that element is given the name $f(x)$. So there's no need for an existential quantifier in the definition of $f^{-1}(N)$, since there's already a notation for "that other object that must exist for $x$ to be in $f^{-1}(N)$".

Greg Martin
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