# Pedantic question on function notation and the meaning of domain

Suppose we have a function $f: A\to B$. Then we know, without specifying what $f$ is, that $f$ may or may not map to every element $b\in B$. If $f$ does map to every element $b\in B$ then it's surjective.

However, does the notation $f:A\to B$ imply that every element $a\in A$ is mapped to some unique $b\in B$?

Furthermore, is there a difference between domain and natural domain? If there is, how can we tell whether $A$ is a/the natural domain from the notation $f: A\to B$?

Suppose we consider the function defined this way: $f: \mathbb{R}\to \mathbb{R}, f=\frac{1}{x}$ Clearly, $f$ is not defined at $x=0$. Would it therefore be incorrect to write $f: \mathbb{R}\to \mathbb{R}, f=\frac{1}{x}$ instead of $f: \mathbb{R}\backslash \{0\}\to \mathbb{R}, f=\frac{1}{x}$?

The question is a bit clumsy. You are welcome to ask for clarification if this isn't clear enough.

• Yes, for every $a$ in $A$ there is a unique $b$ in $B$ such that $f(a)=b$ --- that's what a function is. The notion of "natural domain" only comes up in specific contexts, not in the generality you discuss. E.g., the square root function on the reals has "natural domain" the non-negative reals, since it can be defined on that domain but not on any larger one. May 1, 2015 at 9:35
• To your last paragraph, yes writing $\mathbb R \to \mathbb R$ would leave part of $f$ undefined. May 1, 2015 at 19:19

Suppose we have a function $f: A\to B$. Then we know, without specifying what $f$ is, that $f$ may or may not map to every element $b\in B$. If $f$ does map to every element $b\in B$ then it's surjective.

That is correct.

However, does the notation $f:A\to B$ imply that every element $a\in A$ is mapped to some unique $b\in B$?

Depends on what you mean by unique. Each $a \in A$ maps to exactly 1 $b \in B$. But each $b \in B$ may be mapped from any number of $a \in A$, or not be mapped from any at all.

Furthermore, is there a difference between domain and natural domain?

Natural domain is the largest set of values for which a function is defined, which, ironically, isn't a well defined concept.

Sometimes natural domain will be used to mean "all real numbers for which the function is defined" or "all complex numbers for which a function is defined".

If there is, how can we tell whether $A$ is a/the natural domain from the notation $f: A\to B$?

It depends on which definition of "natural domain" you are using. If you can determine that the natural domain is equal to $A$, then the natural domain is equal to $A$. If you can determine that it is not equal, then it is not equal.

For example, under the definition of natural domain being "all real numbers for which a function is defined", you then

$$f(x) = x^{-1} ~:~ \mathbb R^{+} \to \mathbb R$$

Here $A=\mathbb R^{+}$ is not the natural domain, since the function could have been defined for $R^{-}$ as well. The natural domain is not $\mathbb R$, since the function isn't defined for $0$.

Suppose we're given a function $f: A\to B$. This means that every $a\in A$ maps to a unique element $b\in B$. This means there's only one $b\in B$ it maps to. We still allow that multiple elements of $A$ map to the same element of $B$. We call $A$ the domain of $f$, because all possible $x$ for which $f(x)$ is defined (i.e. has meaning within our context) lie in $A$.

Now the word context is very important when you want to say what the natural domain is, because the term natural domain is very loosely defined. It is the set of values that you would normally consider the largest possible domain. This is where context comes in, because you can imagine that "largest" isn't really saying anything. I'll give an example to demonstrate.

Suppose we have $f=\frac1x$. Then most of the time you would say that the largest possible domain we can have for $f$ is $\mathbb R\backslash\{0\}$. But I'm shure that you agree with me that $\mathbb C\backslash\{0\}$ is larger.

So which one would you consider the natural domain. The answer is bassically in the words. You must decide which one is (more) natural. And that in turn depends on the context you're in.

Note that we can't say anyting about the natural domain without actually knowing what $f$ is. If we only have $f: A\to B$, then the term natural domain is quite meaningless.

As you might have noticed, there's no avoiding a little hand-waving. The best thing I think you can do is to try asking yourself this:

"If I do not specify any domain for my function, what would people assume the implied domain is."

I would say that gives you your natural domain.

N.B. Overall I would just avoid the term natural domain and just clearly specify a certain domain.

In general, the notation $f:A \to B$ implies that every element $a \in A$ is mapped to some element $b∈B$, which is not necessarily unique.

However, the definition of a function means that this element is unique in terms of a, ie. for any $a\in A$ there exists only one $b \in B$ such that $f(a)=b$. If you are requiring uniqueness in terms of b as well, ie. for any $b \in B$ there can exist only one $a \in A$ such that $f(a)=b$, then the mapping is said to be injective. To write this formally, you could state:

\begin{align} f(x)=f(x') \Rightarrow x = x' \end{align}

This does not necessarily mean that every element of $B$ does have a corresponing element from $A$; It is very easily to see that the domain {1,2} can map injectively to {1,2,3} with each number in $A$ mapping to itself in $B$. This would meet the criteria for being an injective function, but would not be surjective.

Regarding domain vs. natural domain, I believe that the natural domain is the largest domain for which the mapping is defined. Every other domain must necessarily be a subset of the natural domain.

In your example, $A$ would be the natural domain as $f$ maps every single element in $A$ to a corresponding element in $B$, and any other domain will undoubtedly be a subset of it - the function is not specified to map elements outside of $A$ because we don't know what $A$ actually is.

As a counter-example, if we knew that $f: \mathbb R \to \mathbb R$, and were then given $f: A\to B$ with $A \subset \mathbb R$, then A would not be the natural domain any longer, as we would be aware that there exists a larger domain for which the function operates, ie. $\mathbb R$.

• Careful about the largest domain part. For example $x\mapsto x^{-1}$ can have many "largest domains" wich are disjoint. Just pick any multiplicative group such as $\mathbb Z_p^{\times}$ for $p$ prime and the function can't be extended sensibly to a larger domain, still you could argue the most natural domain of the inverse would be $\mathbb K \setminus \{0\}$ for some reasonably natural field $\mathbb K$. May 6, 2015 at 8:00