Tao Analysis I - 3.3.1

Exercise 3.3.1

Show that the definition of equality in Definition 3.3.8 is reflexive, symmetric, and transitive. Also verify the substitution property: if $f, \tilde{f}: X \mapsto Y$ and $g, \tilde{g} : Y \mapsto Z$ are functions such that $f = \tilde{f}$ and $g = \tilde{g}$, then $g \circ f = \tilde{g} \circ \tilde{f}$.

Let's remind ourselves of Definition 3.3.8.

Definition 3.3.8 (Equality of functions). Two functions $f : X \mapsto Y$, $g : X' \mapsto Y'$ are said to be equal if their domains and codomains agree (i.e., $X = X'$ and $Y = Y'$ ), and furthermore that $f(x) = g(x)$ for all $x \in X$. If $f(x)$ and $g(x)$ agree for some values of $x$ in the domain, but not others, then we do not consider $f$ and $g$ to be equal. If two functions $f$, $g$ have different domains, or different ranges, we also do not consider them to be equal.

Let's remind ourselves of what reflexive, symmetric, transitive mean:

• https://staff.cdms.westernsydney.edu.au/cgi-bin/cgiwrap/zhuhan/dmath/dm_readall.cgi?page=18&part=2

Reflexive

Let's show the Definition 3.3.8 for equality of functions is reflexive, 

$$f=f$$

Referring to the definition: two functions, $f$ and $g$, are said to be equal if their domains and codomains agree, and furthermore that $f(x) = g(x)$ for all $x$ in the domain of $f$. 

Here, the domains and codomains are the same, if $f$ is also $g$. Furthermore, $f(x) = g(x)$ for all $x$ in the domain of $f$, because $f$ is $g$. 

Therefore, Definition 3.3.8 for equality of functions is reflexive. $\square$

Symmetric

Let's show the Definition 3.3.8 for equality of functions is symmetric,

$$(f=g)\implies (g=f)$$

Referring to the definition: two functions, $f$ and $g$, are said to be equal if their domains and codomains agree, and furthermore that $f(x) = g(x)$ for all $x$ in the domain of $f$. 

So, if $f=g$ then we know their domains and codomains agree. If $g=f$ then their domains and codomains also agree. 

Furthermore, if $f=g$ then we know $f(x) = g(x)$ for all $x$ in the domain of $f$. If $g=f$ then it should be the case that $g(x) = f(x)$ for all $x$ in the domain of $g$. Since the domains of $f$ and $g$ agree, then this is indeed the case.

Thus we have shown $(f=g)\implies (g=f)$, that is, the Definition 3.3.8 for equality of functions is symmetric. $\square$.

Transitive

Let's show the Definition 3.3.8 for equality of functions is transitive,

$$(f=g) \land (g=h) \implies (f=h)$$

Referring to the definition: two functions, $f$ and $g$, are said to be equal if their domains and codomains agree, and furthermore that $f(x) = g(x)$ for all $x$ in the domain of $f$. 

So, if $f=g$ then we know their domains and codomains agree. If $g=h$ then their domains and codomains also agree. Thus the domains and codomains of all $f$, $g$ and $h$ agree.

Furthermore, if $f=g$ then we know $f(x) = g(x)$ for all $x$ in the domain of $f$. Also, if $g=h$ then we know $g(x) = h(x)$ for all $x$ in the domain of $g$. Since the domains of all $f$, $g$ and $h$ agree, we can say $f(x)=g(x)=h(x)$ for all $x$ in their domains, which agree.

From this we conclude the domains and codomains of $f$ and $h$ agree. Furthermore, that $f(x) = h(x)$ for all $x$ in the domain of $f$. 

Thus, we have shown the Definition 3.3.8 for equality of functions is transitive. $\square$.

Substitution Property

We need to verify the substitution property: if $f, \tilde{f}: X \mapsto Y$ and $g, \tilde{g} : Y \mapsto Z$ are functions such that $f = \tilde{f}$ and $g = \tilde{g}$, then $g \circ f = \tilde{g} \circ \tilde{f}$.

Let's start by considering $g \circ f$. The domain of $f$ is $X$, and the codomain is $Y$. The domain of $g$ is $Y$, and the codomain is $Z$. Therefore, the domain of $g \circ f$ is $X$ and. the codomain is $Z$.

The same argument tells us the domain of $\tilde{g} \circ \tilde{f}$ is $X$ and. the codomain is $Z$.

So the domains and codomains of $g \circ f$ and $\tilde{g} \circ \tilde{f}$ agree.

Now let's consider the mappings. Since $f = \tilde{f}$, we know that $f(x) = \tilde{f}(x)$ for all $x$ in the domain of $f$. Similarly, since $g = \tilde{g}$, we know that $g(x) = \tilde{g}(x)$ for all $x$ in the domain of $g$.

Let's consider $g \circ f = g \circ \tilde{f}$. We can say 

$$g(f(x)) = g(\tilde{f}(x))$$

because $f(x) = \tilde{f}(x)$ for all $x$ in the domain of $f$, which agrees with the domain of $\tilde{f}$.

Applying the same argument, 

$$g(f(x)) = g(\tilde{f}(x)) = \tilde{g}(\tilde{f}(x))$$

because $g(x) = \tilde{g}(x)$ for all $x$ in the domain of $g$, which agrees with the domain of $\tilde{g}$.

So, $g \circ f$ and $\tilde{g} \circ \tilde{f}$ have the same domain and codomain, and map $x$ in the same way.

Thus we have shown $g \circ f = \tilde{g} \circ \tilde{f}$. $\square$