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This question may seem kinda silly but in constructing a well organized proof about the associativity of function compositions I need to clear my confusion.

Here's the definition of composition of relations.See definition of this article from Wikipedia.

I want to use this in case of function composition. If two functions $ f,g $ are defined by , $ f:X→Y , g:Y→Z $ then their composition $ g \circ f : X→ Z $ is defined by , $ g \circ f =\{(x,z) \in X×Z : \exists y \in Y : (x,y) \in f \ \text{and} \ (y,z) \in g\} $.

Now imagine the function $ f: \mathbb{R} → \mathbb{R} $ is defined by the formula $ f(x) = \frac{x^{2}-9}{x-3} $ We can take a pair $(x,y)=(3,6) \in \mathbb{R} × \mathbb{R} $ but $\notin f $ . Can't this similar happen when we say $ (x,y) \in f $ in the definition of $ g \circ f $ ? So, how can we use these sets $ X,Y,Z $ flawlessly in the definition when for some of the members of these sets actually don't follow $ (x,y) \in f \ \text{and} \ (y,z) \in g $ . Or,does the pair $ (x,y) $ actually mean only those values for which the function $ f $ is defined ?

peek-a-boo
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    If $(x,y)\notin f$, then you simply ignore that pair. Move on to a different choice. As you have noticed, for any $y\in\Bbb{R}$, the pair $(3,y)\in\Bbb{R}^2$ does not belong to $f$, i.e $(3,y)\notin f$ and therefore $(3,y)\notin g\circ f$. THis is simply because of how $f$ is defined (its domain is $\Bbb{R}\setminus{3}$, not $\Bbb{R}$). – peek-a-boo Sep 04 '23 at 20:32
  • What is $f(3)$? – Paul Frost Sep 04 '23 at 21:55

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Based on the formulation of your question, I guess that you have not properly assimilated the meaning of the definition of $g \circ f$.

The definition

$ g \circ f =\{(x,z) \in X×Z : \exists y \in Y : (x,y) \in f \ \text{and} \ (y,z) \in g\} $

means, in plain English, that $g \circ f$ is the set of those $(x, z)$ for which there exists $y$ such that both $(x, y) ∈ f$ and $(y, z) ∈ g$ hold.

The notation $(x, y) ∈ f$ in this definition is not a statement that $(x, y) ∈ f$ (saying "it always holds"), it's a condition on $x$ and $y$ that must hold (for some $y$) for $(x, z)$ to belong to $g ◦ f$ (saying "if this holds then..."). Thus, there is no logic flaw.

Jean Abou Samra
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