Tao Analysis I - 3.5.3

Exercise 3.5.3

Show that the definitions of equality for ordered pair and ordered $n$-tuple are consistent with the reflexivity, symmetry, and transitivity axioms, in the sense that if these axioms are assumed to hold for the individual components $x$, $y$ of an ordered pair $(x, y)$, then they hold for the ordered pair itself.

The approach is the same for both ordered pair and $n-tuple$ but for clarity we'll do the special easier case of an ordered pair first, then do the more general $n$-tuple.

Ordered Pair

Two ordered pairs $(x,y)$ and $(x',y')$ are equal if and only if $x=x'$ and $y=y'$, by Definition 3.5.1.

Let's consider reflexivity, $A=B \implies B=A$. So let's assume $(x,y)=(x',y')$. This means $x=x'$ and $y=y'$. Since reflexivity holds for $x, y, x', y'$, we can say $x'=x$ and $y'=y$, which is equivalent to $(x',y')=(x,y)$. Thus we have shown reflexivity, $(x,y)=(x',y') \implies (x',y')=(x,y)$. $\square$

Now let's consider symmetry, $(x,y)=(x,y)$. This is true because we know the components obey the equivalence property $x=x$ and $y=y$, which means the ordered pair is equal to itself by definition 3.5.1. $\square$

Finally, consider transitivity, $A=B \land B=C \implies A=C$. Let's start with $(x,y)=(x',y')$ and $(x',y')=(x'',y'')$. Using definition 3.5.1, we know that $x=x', y=y'$ and also $x'=x'', y'=y''$. Since these individual components obey the equivalence property of transitivity, we can say $x=x'', y=y''$. By definition 3.5.1 this tells us $(x,y)=(x'',y'')$. We have shown ordered pairs obey transitivity, $(x,y)=(x',y')$ and $(x',y')=(x'',y'')$ implies $(x,y)=(x'',y'')$. $\square$

n-Tuples

Definition 3.5.6 generalises equality of ordered pairs to n-tuples:

Two ordered $n$-tuples $(x_i)_{1 \leq i \leq n}$ and $(y_i)_{1 \leq i \leq n}$ are said to be equal iff $x_i = y_i$ for all $1 \leq i \leq n$.

Let's consider reflexivity, $A=B \implies B=A$. Given two equal n-tuples, $(x_i)_{1 \leq i \leq n} = (y_i)_{1 \leq i \leq n}$, we know by the above definition 3.5.1 that $x_i = y_i$ for all $1 \leq i \leq n$. Since these individual components $x_i$ and $y_i$ obey reflexivity, we can say $y_i = x_i$ for all $1 \leq i \leq n$. This is equivalent to $(y_i)_{1 \leq i \leq n} = (x_i)_{1 \leq i \leq n}$, thus we have shown reflexivity. $\square$

Let's now consider symmetry, $(x_i)_{1 \leq i \leq n}=(x_i)_{1 \leq i \leq n}$. For this to be true we must have $x_i = x_i$ for all $1 \leq i \leq n$. This is indeed the case as the individual components $x_i$ obey symmetry. This we have shown symmetry. $\square$

Finally, consider transitivity, $A=B \land B=C \implies A=C$. Let's start with $(x_i)_{1 \leq i \leq n} = (y_i)_{1 \leq i \leq n}$, and $(y_i)_{1 \leq i \leq n} = (z_i)_{1 \leq i \leq n}$. This tells us $x_i = y_i$ for all $1 \leq i \leq n$, and $y_i = z_i$ for all $1 \leq i \leq n$. Since these individual elements obey transitivity, we have $x_i = z_i$ for all $1 \leq i \leq n$. This is equivalent to $(x_i)_{1 \leq i \leq n} = (z_i)_{1 \leq i \leq n}$. Thus, we have shown transitivity. $\square$.