2

The idea of using n-tuples to represent functions is introduced in Terence Tao's Analysis I ex 3.5.2:

Suppose we define an ordered $n$-tuple to be a surjective function $x > : \{i \in N : 1 \leq i \leq n\} \to X$ whose codomain 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 $(x_i)_{1 \leq i \leq n}$. Using this definition, verify that we have $(x_i)_{1 \leq i \leq n} = (y_i)_{1 \leq i \leq n}$ if and only if $x_i = y_i$ for all $1 \leq i \leq n$.

Question: Why does the function need to be surjective?

My Thoughts

Initially I thought this was to ensure the range of the function was fully accounted for by the n-tuple. However, Tao is quite precise in using range and codomain precisely, so I don't think this is an avenue worth pursuing.

Then I thought about the traditional textbook approach of considering the definition of surjective - that every element of the codomain has an element in the domain which is mapped to it. But that line of thinking doesn't seem to help here.

Exploring this site, I can see that surjectivity is only needed for the later part of the exercise:

Also, show that if $(X_i)_{1 \leq i \leq n}$ are an ordered n-tuple of sets, then the Cartesian product, as defined in Definition 3.5.6, is indeed a set. (Hint: use Exercise 3.4.7 and the axiom of specification.)

I can't even begin to attempt this (I'm a beginner) - so I suspect understanding the need for surjectivity may help?

Penelope
  • 3,147

1 Answers1

1

Without having the book at hand, I am slightly speculating. However, as stated, this construction defines $n$-tuples as functions. So the equality $(x_i) = (y_i)$ is an equality of functions.

More often than not, equality of two functions requires three ingredients: (1) equality of domains, (2) equality of codomains, and (3) equality of the values of the two functions at each input from their common domain.

Without surjectivity, the $\Leftarrow$-direction of the claim would no longer be true, as $x_i = y_i$ for all $i$ explicitly only contains the third ingredient of equality of values. Implicitly, the first ingredient of common domain is implied by the shared subindex notation $1 \leq i \leq n$, yet equality of codomains for the two $n$-tuples would remain open.

In contemporary set theory, it is also common to simply define a function to stand for what has traditionally been called its graph. One can deduce from a graph what the domain and range look like, where range is strictly taken to mean the set of actual values of the function. Equality of functions would reduce to equality of these graphs as sets. With this definition, surjectivity would be superfluous to mention when finite sequences (the $n$-tuples from OP) are concerned. Furthermore, note that graph-wise the question Is that function surjective? would no longer be sensible. Rather one should ask Is that function surjective with respect some other set $Y$?, where $Y$ is reasonably some superset of the range. To drive the point home, what would have to be the relationship between the range of a graph-wise function and some superset Y for surjectivity wrt Y to hold?

Forget about this last paragraph until you learn set theory for its own sake.

Linear Christmas
  • 1,285
  • This point is touched on in answers to Exercise 3.5.2 in Analysis by Tao (Set Theory) and Proof verification for exercise 3.5.2 in Tao's Analysis I: Prove that the generalized definition of a Cartesian product is a set, but I think this answer is useful because it's a direct and immediate answer to a direct question about why we need surjectivity. – David K Mar 25 '24 at 03:44
  • Regarding the 3 ingredients for a function: domain, codomain and mapping the same inout to the same output ... (1) as you say the domain is the same thanks to the same indices $1 \leq i \leq n$. (3) the mapping is the same because we're told $x(i) = y(i)$ .. but regarding (2) isn't the range, and hence codomain, the same because every input $i$ is mapped to the same output $x(i) = y(i)$ and there are no other elements of the domain besides those created by $x(i)=y(y)$ ? – Penelope Mar 25 '24 at 13:31
  • To think about this further, if the function was not surjective, there would be an element of the codomain which was not mapped to by an element of the domain. However, by the way the function is defined, there is no such element, because every element of the codomain is a result of being an output from $x(i)$. What am I missing? – Penelope Mar 25 '24 at 13:33
  • @Penelope Instead of using the notation ${(x_i)}{1\leq i \leq n}$ and ${(y_i)}{1\leq i \leq n}$, use ${(x_i)}{1\leq i \leq n}^{X}$ and ${(y_i)}{1\leq i \leq n}^{Y}$. Without surjectivity, there is nothing in the $\Leftarrow$-direction to rule out $X \neq Y$. We can't even say that either $X \subseteq Y$ or $Y \subseteq X$, let alone both. – Linear Christmas Mar 25 '24 at 13:51
  • @Penelope If it helps, I define range to be precisely the set of values that a function obtains. Then yes, the ranges of the tuples are the same. And I let codomains be any superset of the range. And if the equality of functions has equality of codomains, not ranges, then codomain equality is not implied by range equality unless surjectivity holds. Terminology varies, however. What I call range, other might call image. What I call codomain, others might refer to as range. – Linear Christmas Mar 25 '24 at 13:57
  • I've printed this out and have been re-reading for 2 hours and I still don't understand. Perhaps I can start by understanding which statement the ⇐ -direction refers to as failing if subjectivity is not asserted? – Penelope Mar 25 '24 at 17:25
  • 1
    @Penelope For every $n$-tuple ${(x_i)}{1\leq i \leq n}^{X}$ and ${(y_i)}{1\leq i \leq n}^{Y}$, the equivalence $$ {(x_i)}{1\leq i \leq n}^{X} = {(y_i)}{1\leq i \leq n}^{Y} \Leftrightarrow \forall k \in {1, \dotsc, n } ,( x_k = y_k ) .$$ The $\Leftarrow$-direction requires surjectivity as part of the definition of $n$-tuple – Linear Christmas Mar 25 '24 at 19:52
  • We're getting to the root cause of my error... so let me repeat what I think you're saying. The LHS has two functions $x$ and $y$ are equal. The RHS of the $\iff$ says that the two functions give the same output for the same input. Regarding the LHS, functions are equal if the domain, codomain and mapping is the same. The RHS only appears to say the mapping is the same, but actually the domain is too by definition $1 \leq k \leq n$. The only thing left is the codomain. The RHS says the range of $x$ and $y$ are the same. Here we need subjectivity to say the range is also the domain? Correct? – Penelope Mar 25 '24 at 20:58
  • 1
    @Penelope If you define range as image of domain (i.e., set of values that function actually obtains), then yes. I assume you meant to say "range is also codomain" in your penultimate sentence. – Linear Christmas Mar 25 '24 at 22:00
  • Yes, I am using range to mean the image, the actual values. Thanks Linear for helping me zoom into the issue at the root of my misunderstanding. I will sleep on it as I still feel intuitively that two facts (1) common domain, and (2) same mapping, imply the range is also the codomain without the need to additionally assert it by definition ... but I am closer to resolving this than I was in the last 48 hours! Thanks again! – Penelope Mar 25 '24 at 22:09
  • 1
    @Penelope You're welcome! Think about the functions $f\colon \mathbb{R} \to \mathbb{R}$ and $g\colon \mathbb{R} \to [0, +\infty)$ where, for all $x \in \mathbb{R}$, $f(x) = g(x) := x^2$. Using the three-step definition of equality for functions, $f \neq g$ since codomains don't match: $\mathbb{R} \neq [0, +\infty)$. You cannot deduce from "$g(x) = f(x)$ everywhere" what were the codomains of $f$ or $g$. Unless, of course, we demanded that the functions were surjective. But here $f$ was not surjective. – Linear Christmas Mar 25 '24 at 22:40
  • I have asked a more specific question about what I think is the root cause of my misunderstanding here https://math.stackexchange.com/questions/4887890/basic-question-about-function-equality-range-vs-codomain – Penelope Mar 26 '24 at 13:52
  • 1
    @Penelope In linked question, you admit that "codomain may be any superset of range". Now there are two options. Either you accept the definition of equality of functions, or you don't. If you accept the definition, the ranges of functions being equal can't, in general, imply equality of codomains. The functions $f$, $g$ above show this to be the case. Both of their ranges are $[0, +\infty)$, but their codomains are still different. If you don't accept the definition, that's a whole other matter. Then it becomes a metaquestion about "why equality of codomains, not ranges" is in the definition. – Linear Christmas Mar 26 '24 at 14:16
  • 1
    @Penelope And such metaquestions are ultimately a (somewhat subjective) matter of utility. Note that there is not much use in replacing "equality of codomain" with "equality of ranges" precisely because it would be superfluous, redundant. The conditions (1) and (3) would, as you argue, already imply equality of ranges. So there is little point in demanding this separately. Therefore, there are two staightforward options. Either keep the "equality of codomains" as is, or drop it altogether, and demand only (1) and (3). Both of these variants of definition can be found in literature. – Linear Christmas Mar 26 '24 at 14:21
  • 1
    @Penelope If you choose to drop the condition (2) altogether, you essentially fall to a situation equivalent to the one I described in the third paragraph of my answer. Keep in mind: you can only use one definition at a time, not jump back and forth, unless the definitions are equivalent. Here the definitions are not equivalent, as, again, the example for $f$, $g$ demonstrated. If you drop equality of codomains, then $f=g$. – Linear Christmas Mar 26 '24 at 14:26
  • Thanks for your continued patience - if the root of this confusion for me is the meta-question, then that is progress for me. – Penelope Mar 26 '24 at 16:18
  • Interestingly, Terrence Tao seems to have changed his definition of a function between book editions. In the 4th edition of Analysis I he says the "codomains" have to be the same ... but in the 3rd edition he uses "range" ... it seems this is a topic that generates. a lot of discussion and mind-changing! – Penelope Mar 26 '24 at 21:34
  • @Penelope I doubt Tao changed the definition itself, only the word. The word "range" is used in different meanings. I mentioned this possibility at the end of my second comment here: "Terminology varies, however ..." – Linear Christmas Mar 26 '24 at 22:19