Exercise 3.3.1 (Equality of functions)

Proof.

• (Reflexive) We have to prove that $f=f,$ which by Definition 3.3.7 is equivalent to

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:f(x)\stackrel{Y}{\text{=}}f(x).$$

  But the above property follows directly from the reflexive property of equality $\stackrel{Y}{\text{=}}$ on $Y$.

• (Symmetric) Suppose that $f=g$. We now have to show that $g=f,$ which by Definition 3.3.7 means that

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:g(x)\stackrel{Y}{\text{=}}f(x).$$

  Since $f=g$, we have $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:f(x)\stackrel{Y}{\text{=}}g(x)$ by definition. Since $\stackrel{Y}{\text{=}}$ is symmetric, we have $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:g(x)\stackrel{Y}{\text{=}}f(x)$. By Definition 3.3.7, this means $g=f$, as desired.

• (Transitivity) Suppose that $f=g$ and $g=h.$ We now have to demonstrate that $f=h.$ Since $f=g,$ we have $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:f(x)\stackrel{Y}{\text{=}}g(x).$ Since $g=h,$ we have $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:g(x)\stackrel{Y}{\text{=}}h(x).$ Since $\stackrel{Y}{\text{=}}$ is transitive, we have $\forall <!--\; FUNCTION\; APPLICATION\; -->x\in X:f(x)\stackrel{Y}{\text{=}}h(x).$ By Definition 3.3.7 this means that $f=h,$ as desired.