I understand that a function is a relation with certain restrictions; a set, at the end, but how is $f(x)$ defined, the action of applying in a way $f$ to $x$? Can it only be defined using "if and only if" condition? like in

$$f(x)=y\iff (x,y)\in f$$

Isn't there an algebraic form for function application that define them without using logical statements? I feel like if you need to define a function, you need to involve a certain logical operator.

  • $\begingroup$ What does $(x,y) \in f$ mean in your expression? $\endgroup$
    – Yuriy S
    Oct 25, 2018 at 16:20
  • $\begingroup$ I'm not sure if you are asking about how to define a function or how to define the concept of function. $\endgroup$
    – ajotatxe
    Oct 25, 2018 at 16:21
  • $\begingroup$ I edited the question $\endgroup$
    – Garmekain
    Oct 25, 2018 at 16:24

2 Answers 2


One way is a function $f: X \to Y$ is a set

$$f \in \{A \subseteq X \times Y: \forall x \in X, (x,y) \in A \land (x,z) \in A \implies y=z\}$$

So $f$ is a particular set of ordered pairs such that each element in the domain maps to one and only one element in the range.


" algebraic form for function application "

I don't know what that means. And surely trig functions can't be defined "algebraically".

without using logical statements?

What's wrong with "logical statements"? We certainly can't define anything with illogical statements.

And all definitions are like this. "A triangle is isosceles $\iff$ the triangle has two equal sides"


Intuitively we have an idea and concept of "function" in that if we have two sets $X$ and $Y$ we can imagine a mapping in which every element $x$ of $X$ will get assigned to an element $y$ of $Y$. And the "function" is the mapping as a whole.

Okay, that's seems utterly clear to me but not at all rigorous. How do we formally define "mapping" and have the idea of mapping expressed as an object?

Since a mapping $X$ to $Y$ is actually a collection of individual maps of $x_\alpha$ to $y_\alpha$ then a function will be expressible as set $\{x_\alpha \mapsto y_\alpha\}$ where the notation $x_\alpha \mapsto y_\alpha$ means a specific assigning $x_\alpha\mapsto y_\alpha$.

But what is a "assigning". Well, that's nothing more or less than saying "here we take this $x_\alpha$ and tie it to this $y_\alpha$" which nothing more or less than making an ordered pair $(x_\alpha,y_\alpha)$ where we say the first term is paired to the second pair.

So we define a function $f:X\to Y$ (or $f$ for short) as a subset of $X\times Y$ where $X\times Y = \{(x,y)$ ordered pairs where $x \in X$ and $y \in Y\}$. However as we are assigning every specific element of $X$ to one specific element of $Y$ the set $f \subset X\times Y$ must have the condition that for every $x \in X$ there is exactly one and only one $(x,y) \in f$.

So that is the definition of a function. The vague concept of a "mapping" is now a concrete "object"; a set of ordered pairs.

Now your question about $f(x) = y \iff (x,y) \in f$ is just one of notation. We define the notation of $f(x) = y$ to mean $x \mapsto y$ is some "mapping" so that $(x,y) \in f \subset X\times Y$ for a specific subset $f$ which ... follows the rule that for every $x \in Y $ there is exactly one $(x,y) \in f$.

And the $\iff$ should not be thought of as being some thing that only exists in boolean logic world but instead is simply a matter of how we declare definitions. For example "$X$ is a pelican $\iff$ $X$ is an aquatic bird that ... is whatever it is that makes a pelican a pelican" is the definition of pelican.

... Oh... maybe you are wondering how we can do something like taking a "rule" such as $f(x) = x^3 + 7x^2 + 3$ and have it somehow be a set of ordered pairs.

Well, $f(x) = x^3 + 7x^2 + 3$ is simply shorthand notation for $f = \{(x,y) \in \mathbb R\times \mathbb R| y = x^3 + 7x^2 + 3\}$.

Of maybe I should say $f = \{(x,y) \in \mathbb R\times \mathbb R| y = x^3 + 7x^2 + 3\}$ is simply the concrete objectification of the abstract mapping conceptual rule of $f(x) = x^3 + 7x^2 +3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.