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I have the following exercise:

If $f:X\rightarrow Y$ is a function and $A\subset X$ is a set. It is true that $f\cap(A\times Y)$ is a function?

At first I thought this was the restriction function, but I checked the definition of the restriction of a function and it does not match this one. I thought that since f is a function, and $f \subset X\times Y$ then that was enough to see that $f\cap(A\times Y)$ is a function, but I think it is not enough, could you please help me. Thank you.

Haus
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$f\cap(A\times Y)$ is indeed a function:

  • Its elements are elements of $X \times Y$.
  • Each element is paired with just one element of $Y$.

However it is not a function from $X$ to $Y$. It is a function from $A$ to $Y$.

mathcounterexamples.net
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  • Thanks, when you mention "Each element is paired with only one element of ", how can you prove that? I think that would be enough to see what a function is, right? – Haus Nov 20 '20 at 12:31
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    Can you specify exactly what $f\cap(A\times Y)$ is defined as? It just looks like a set to me. – Derek Luna Nov 20 '20 at 12:31
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    I think I understand what the idea is supposed to be but man that's confusing notation. – Derek Luna Nov 20 '20 at 12:33
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    @DerekLuna Yes, it is a set. In set theory, functions are encoded as sets. –  Nov 20 '20 at 12:33
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    That's why I included the word "just". I know functions are sets. But not all sets are functions. – Derek Luna Nov 20 '20 at 12:34
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    @DerekLuna And the subsequent three lines explain why the intersection of the set $f$ with the set $A\times Y$ is a function in the set-theoretic sense. –  Nov 20 '20 at 12:35
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    If you have an intersection of a function and a set, it is not exactly clear it's a function just because you list the two requirements for it to be one as bullet points. The confusing part was the notation itself. What about this is so hard for you to understand? – Derek Luna Nov 20 '20 at 12:38
  • @Haus "Each element is paired with only one element of " because this is the case for $f$ itself as it is supposed to be a function. By making the intersection, all ordered pairs having the first element in $A$ will remain. – mathcounterexamples.net Nov 20 '20 at 15:24
  • See also What is the set-theoretic definition of a function?. – mathcounterexamples.net Nov 20 '20 at 15:37