On page 64 (3.4 Images and Inverse Images) of "Analysis I" by Terence Tao, it says:

Note that the set $f(S)$ ($f$ is a function) is well-defined thanks to the axiom of replacement (Axiom 3.6). One can also define $f(S)$ using the axiom of specification (Axiom 3.5) instead of replacement, but we leave this as a challenge to the reader.

right below the definition 3.4.1 (image of sets).

(In case any ambiguity caused by incompleteness of pre-knowledge, here is the definition of function given in the same book:

Definition 3.3.1 (Functions). Let $X$, $Y$ be sets, and let $P(x,y)$ be a property pertaining to an object $x\in X$ and an object $y \in Y$, such that for every $x\in X$, there is exactly one $y \in Y$ for which $P(x,y)$ is true (this is sometimes known as the vertical line test). Then we define the function $f:X\rightarrow Y$ defined by $P$ on the domain $X$ and range $Y$ to be the object which, given any input $x\in X$, assigns an output $f(x) \in Y$, defined to be the unique object $f(x)$ for which $P(x, f(x))$ is true. Thus, for any $x\in X$ and $y \in Y$, $$y=f(x) \Leftrightarrow P(x, y) \textrm{ is true.}$$


It is quite obvious for me to see how to define f(S) by using axiom of replacement. However, it bothers me for a bit too long to "beat off" the challenge left to the reader.

My thinking so far is to somehow construct a property $P(y)\ $that just depends on $y \in Y$ and somehow relates to the domain $S \subseteq X$ (motivation is simply from the format of the set given by axiom of specification $ \lbrace x \in A\mid P(x) \rbrace $). I am not sure but I guess that I did not understand the dependence between the property and the variables very well. That means, I am not sure in which situation the appearance of two variables like ($x$ and $y$) would be allowed to construct a property $P(x)$ or $P(y)$, or even it's not possible to do this. I don't know...

So I wish I could get some useful hint or enlightenment. In fact, I would be more excited about the feeling of reading three lines then "Ah-Ha! That's how it's done", than a complete solution (though I accept it definitely).

By the way, I am a pre-service math teacher right now (not an undergrad student), thus this is not a question for any assignment.

I appreciate for any help on this question.


  • $\begingroup$ To make your question more self-contained (and increase the chance that you get an answer) it would be best to say what $f$ is. Separation is not an adequate substitute for every use of replacement, so the answer to this question will need to take the definition of $f$ into account. $\endgroup$ Commented May 20, 2015 at 5:50
  • $\begingroup$ f is just a function (map). Thanks~! $\endgroup$ Commented May 20, 2015 at 5:52
  • $\begingroup$ Is it a set of ordered pairs, or a (definable) class of ordered pairs? (I suspect the former, but I want to make sure.) $\endgroup$ Commented May 20, 2015 at 6:05
  • $\begingroup$ Sorry, the concept of ordered pair has not been formally introduced yet up to this chapter of the book. And which set are you talking about, f(S) or the {x∈A∣P(x)} ? $\endgroup$ Commented May 20, 2015 at 6:13
  • $\begingroup$ I meant to ask whether $f$ itself is a set of ordered pairs (one common definition of "function" is a set of ordered pairs $f$ satisfying a certain property.) But if the concept of ordered pair hasn't been formally defined yet in the book, that must not be the definition of "function" that is being used. And with the other definition of "function" I can think of, the exercise would be false (i.e. separation might not suffice instead of replacement.) So I am confused; I'd better leave the answer to someone who has the book. $\endgroup$ Commented May 20, 2015 at 6:20

2 Answers 2


Basically, by definition of function, the function $f\colon X\to Y$ is defined by a property $P$ such that $P(x,y)$ is true when $y=f(x)$.

We have $f(S):=\{f(x):x\in S\}$ defined by axiom of replacement. So, this is a short form to represent $\{y:y=f(x)\text{ for some } x\in S\}$ what is the same that $\{y:P(x,y)\text{ for some } x\in S\}$.

To use the axiom of specification, we need a reference set, we say $A$, to define a set like $\{x\in A:P(x)\}$ for some property $P$.

In this case, we can use the set $Y$ of $f\colon X\to Y$ as reference set. Thus, defining $Q(y):=P(x,y)\text{ for some } x\in S$, we can state $f(S):=\{y\in Y:Q(y)\}$, what use the axiom of specification.

Edit. The property $Q$ is only used for ilustrative purpose. So, the final definition of image is $$f(S):=\{y\in Y:P(x,y)\text{ for some } x\in S\}.$$ Note that the property $P(x,y)$ is already defined by the definition of function for $f\colon X\to Y$.

  • $\begingroup$ Thank you very much, Cristhian! The solution is plausible and makes sense. The only thing I am still not sure is your definition of Q(y):=P(x,y) for some x∈S is okay or not. In general, I am not sure when and how can we redefine a property so that one of the variable (in this case it is x) can be "legitimately eliminated" (in this case, we do this by defining Q(y) to be a property that actually depends on x and y. i.e. P(x,y)). Could you also explain this? But anyway, I really appreciate for your help. : ) $\endgroup$ Commented May 20, 2015 at 18:57
  • $\begingroup$ Oh! Sure. I used $Q$ to ilustrate the idea that we can put the expression "$P(x,y)\text{ for some } x\in S$" where we see $Q(x)$. By other hand, the variable $x$ is not eliminated in $Q$; it is in the scope of the existencial quantifier $\text{ for some }$. In briefly, I used $Q$ beacuse it's "easy" to see that we use the axiom of specification in a expression like $\{x\in X: P(x)\text{ is true}\}$. $\endgroup$ Commented May 20, 2015 at 20:56

I’m just learning this so take it with a grain of salt, but here’s how I’d construct it.

First, a small picture of what we’re dealing with. We’re trying to define the inner set on the right.

A mapping of two sets with a function

To use the Axiom of Specification, we need two things:

  • An original set $A$.
  • A property $P(a)$ that specifies whether some $a$ should be included.

Since we're using the Axiom of Specification, we want to start with a wider set than $f(S)$ and find a way to “narrow it down”. By definition, $f(S)$ is a subset of $Y$. So let’s pick $Y$ as our original set.

Next, we need a property that would “filter out” any values in $Y$ that don’t belong in $f(S)$. We can write a condition for $y$ that says “include if there exists $x$ in $S$ such that $f(x) = y$”. Intuitively, this corresponds to the dots in the inner right circle—because are reachable from the red lines.

Now that we picked the original set and the condition, we can use Specification: $$f(S):=\{y\in Y: P(y) \text{ is true}\}.$$ where $P(y) = \exists x\in S: f(x) = y$.

Hopefully this is right!

  • $\begingroup$ I guess this says the same thing as the other answer, but I found it a bit clearer to not “unroll” the definition of the function itself. $\endgroup$ Commented Nov 1, 2021 at 0:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .