0

(Exercise 3.5.2 in Analysis by Tao) Suppose we define an ordered $n$-tuple to be a surjective function $x: \{ i \in \mathbb{N}: 1 \le i \le n\} \to X$ whose range is some arbitrary set $X$ (so different ordered $n$- tuples are allowed to have different ranges); we then write $x_i$ for $x(i)$, and also write $x$ as $(x_i)_{1\le i\ \le n}$. Using this definition, verify that we have $(x_i)_{1\le i\ \le n} = (y_i)_{1\le i\ \le n}$ if and only if $x_i = y_i$ for all $1 \le i \le n$.

This question seems not difficult, but I am struggling with it. I especially don't know how to use the surjective condition given. Any help would be appreciated.

shk910
  • 3,599
  • The "surjective" thing is some mistake, an ordered $n$-tuple is simply a function $[1,n] \to X$ and that's it. Given two $x,y : [1,n] \to X$, of course these are equal if and only if $x(i) = y(i)$ for all $i \in [1,n]$ (also, it is not desirable to ask whether such tuples with different ranges are equal or not). – Sasha Feb 23 '20 at 22:58
  • I'm actually on the exact same question. I think the surjective condition is to ensure that the ranges are equal, which is necessary for Tao's definition of function equality. For example, you could have $\forall i \ 1\leq i \leq n \ x(i) = y(i)$ BUT that would not necessarily imply that $(x_i){1\leq i \leq n} = (y_i){1\leq i \leq n}$ because the ranges could be different...and therefore, by Tao's definition provided earlier in the book, the functions would not be equivalent. – S.C. Mar 03 '20 at 02:57

1 Answers1

1

Before reading this answer, please consider trying to solve this on your own. I will not provide a complete proof here, as I will deliberately skip the details (which you will take care of :)).

I will write $x$ instead of $(x_i)_{1\leq i\leq n}$ and $y$ instead of $(y_i)_{1\leq i\leq n}$, just as the exercise indicates, for sake of brevity.

Suppose that $x=y$. Then $x_i = y_i$ $\forall i$ follows immediately from the equality of functions. As a matter of fact, it is possible to prove the conclusion by only assuming that $x$ and $y$ have the same domain and codomain (surjectivity will be needed in order to proceed in this manner).

Conversely, let $x_i = y_i$ $\forall i$, and let $Y$ be the codomain of $y$. Use the definition of surjectivity to show that $X=Y$. Furthermore, $x$ and $y$ have the same domain by the definition provided in the excercise.

Note that surjectivity is in fact needed. The claim in Sasha's comment fails to hold in many instances: if $x(i)=y(i)$ $\forall 1 \leq i \leq n$, it might be the case that $x$ and $y$ possess different codomains.

DucassiiiN
  • 11