# Thread: Math Notation: How to explicitly specify the domain of a function symbolically?

1. How can I specify the domain of a function symbolically?

Assume I have a very simple linear function.

$f(x)=2x$

If I want to say that the domain of this function is all the real numbers, how would I write this with symbols? Do I need to use set-builder notation? What would the set-builder notation for something like this look like? What other notations are available?

I suppose I could just write "for all real x" followed by the function, like in course text books and on tests. But I want to learn the formal, symbolic way of writing this with less words. Because I know I will have to learn it sooner or later in my studies. I would prefer sooner. My teachers are holding back on this, as if I am not prepared for these fancy symbols yet or some nonsense like that. Or maybe they just don't know it themselves...

If there is no mention of what the domain of the function is, is it always assumed to be the real number set then? Implicit domain definition, eh? What about explicit then?

One way I can think of is writing something like this.

$\displaystyle f(x)=2x|x\in\Bbb{R}$

Is this unconventional?

Another way I could think of is something like this.

$D=\left\{\Bbb{R}\right\}$

I use the upper case D to mean "domain". This is now a set of all real numbers, referenced by "D". But how do I connect the domain to the function rule? What is the notation for this?

I am thinking about doing something like this.

$D=\left\{x|x\in\Bbb{R}\right\}$

Is this legit? How else would I connect the domain set to the function rule?

$f:X\to Y$

Is this what I am looking for in fact? I have seen notations like this on the web, on Wikipedia, among other places.

How is this interpreted? My understanding is that X and Y are two sets, and this is read as "X to Y", relating set X to set Y. Does the colon carry any special meaning? The lower case "f" is the name of the function, of course. So upper case X and Y letters are used for the related sets, and lower case x and f(x) for the discrete values within these sets?

So my set D would correspond to the set X in this example, correct? It's the domain set?

What about the Y or the f(x)? How do I describe this with symbols?

If there are no constraints on Y, then the function is defined for all discrete x values of set X?

$Y=\left\{f(x)\right\}$

Something like that?... maybe?

What are your thoughts? Am I on the right track or am I completely lost? Perhaps if someone were to write a few examples for me I would better understand this notation. It doesn't seem too hard, but my teachers won't hear of it. So I have to rely on the web as my teacher.

And what's the deal with x values equal to zero?

Here is a very simple example.

$g(x)=1/x$

I understand we are not supposed to divide by zero, because that would mean the end of the universe. But when writing down the domain of such function, my teachers always just state the following.

$x\neq 0$

Is this really the "domain" of the function? Because I thought the "domain" of a function was supposed to tell us what x values the function is defined for, not what x values it is not defined for. Isn't this wrong?

The domain is supposed to include "what is", not "what is not". Or am I wrong? The x equal to zero is a point of exclusion, not inclusion.

Who cares what x values the function isn't defined for? If I know what x values it is defined for then I can easily figure out what x values it is not defined for. Especially when there is only a single point at which the function is not defined.

So this is really a short hand version of something more formal like this.

$D=\left\{x|x\in\Bbb{R}, x\neq 0\right\}$

Or simply like this.

$D=\left\{\Bbb{R}, x\neq 0\right\}$

But this can also be described alternatively, like this.

$D=\left\{x|x\in\Bbb{R}, x\lt 0 \quad \text{and} \quad x\gt 0\right\}$

Or simply like this.

$D=\left\{x\lt 0 \quad \text{and} \quad x\gt 0\right\}$

So my question is this! Are these exclusion points supposed to be defined as part of the domain of a function?  Reply With Quote

2.

3. Hi samir! Let me try to give some answers... Originally Posted by samir One way I can think of is writing something like this.

$\displaystyle f(x)=2x|x\in\Bbb{R}$

Is this unconventional?
Indeed. This is unconventional.

Properly we can say that we have the function $f: \mathbb R \to \mathbb R$ given by $f(x)=2x$.
Or alternatively the function $f: \mathbb R \to \mathbb R$ given by $x \mapsto 2x$.
In this notation the first $\mathbb R$ is the domain, and $f$ is required to be defined for every element of the domain. Quote:
Another way I could think of is something like this.

$D=\left\{\Bbb{R}\right\}$
$\mathbb R$ is already a set, so $\left\{\Bbb{R}\right\}$ is a set that contains a set, which is not what we want here.

The symbol $D$ has no meaning unless we define it, which we'd typically do with words.
So we might say, that if we define $D$ to be the domain of $f$, that we have $D=\mathbb R$. Quote:
I am thinking about doing something like this.

$D=\left\{x|x\in\Bbb{R}\right\}$

Is this legit? How else would I connect the domain set to the function rule?
This is legit set builder notation.
It's the same as $D=\mathbb R$ though. Quote:
$f:X\to Y$

Is this what I am looking for in fact? I have seen notations like this on the web, on Wikipedia, among other places.

How is this interpreted? My understanding is that X and Y are two sets, and this is read as "X to Y", relating set X to set Y. Does the colon carry any special meaning? The lower case "f" is the name of the function, of course. So upper case X and Y letters are used for the related sets, and lower case x and f(x) for the discrete values within these sets?
Yes. This is how we say symbolically that $f$ is a function with domain $X$ and codomain $Y$. Quote:
If there are no constraints on Y, then the function is defined for all discrete x values of set X?

$Y=\left\{f(x)\right\}$

Something like that?... maybe?
A function $f$ is required to be defined for all values $x \in X$.
However, it's not required that every value $y$ in $Y$ has a corresponding $x$.
The image of $f$ is the set of all reachable values and is written as $f(X)$.
So we have that $f(X) = \left\{f(x)\right\}$, which is a subset of (or equal to) the codomain $Y$. Quote:
And what's the deal with x values equal to zero?

Here is a very simple example.

$g(x)=1/x$
The domain has to exclude $0$.

Notations for this are:
$$\mathbb R^* = \mathbb R^\times = \mathbb R_{x\ne 0} = \left\{x\in\Bbb{R} \mid x\neq 0\right\} = \left\{x \mid x\in\Bbb{R}, x\neq 0\right\} = ...$$  Reply With Quote

4. The "standard" way to indicate the domain is something like:

$f: D \to B$

Sometimes mathematicians get fussy about "how big you make $B$", it should, at the very least, contain the range of $f$.

So, for your first example, we write:

$f: \Bbb R \to \Bbb R$ given by $f(x) = 2x$, or:

$f: \Bbb R \to \Bbb R$ given by $f: x \mapsto 2x$.

If we just want to exclude a single point from our domain set, like $x_0$ ($x_0 = 0$ in your example $\dfrac{1}{x}$), we can write:

$f: D - \{x_0\} \to B$, so, for example:

$f: \Bbb R - \{0\} \to \Bbb R$ given by $f(x) = \dfrac{1}{x}$.

If the definition of the domain set $D$ is more complicated, we may have to settle for:

$f: D \to B$, where $D = \{x \in A: \text{some properties hold for }x\}$, hopefully the set $A$ that $D$ is a subset of is "well-understood" (it is quite possible for the chain of subsets to be quite long, if the definition of $D$ is really complicated, we might have $D \subseteq A \subseteq X \subseteq S$, etc.).

As you can see, a function $f$ has several "parts".

1. A domain set (the source of values $f$ acts upon).

2. A co-domain set (the type of thing we get "when we're done", or the "target set", or the set the values lie IN). There are reasons, often, for making this set larger than it needs to be: the range of values may be difficult to describe, explicitly, but it may be easier to just say "whatever they are, they're all real numbers". An important distinction: while $f$ takes values IN its co-domain, such values need not be the entire co-domain (in fact, there is a special word for functions where the range = the co-domain, they are called ONTO).

3. A "rule" for pairing a value $f(x)$ (in the co-domain) to any given $x$ in the domain. This is often done using the "=" sign, like so:

$f(x) =$ some expression in $x$.

For example, the function $f: \Bbb R \to \Bbb R$ given by the rule $f(x) = x^2$ is actually the set of all pairs of real numbers (a subset of $\Bbb R \times \Bbb R$, which we often represent by a PLANE, or in desperate times, a sheet of paper) of the form:

$(x,x^2)$, so it makes perfect sense, for example, to say:

$(-2,4) \in f$.

There is actually something a bit subtle going on with part 3: a function, by its very nature, can only take a specific $x$ to one and only one place in the co-domain. This doesn't mean that two DIFFERENT $x$'s cannot have the same "target" (for example, -3 and 3 both map to 9 under a squaring function), but one $x$ in the domain cannot wind up in "two different targets". If some rule allows this, we call it "ill-defined", and conversely, we require a function NOT be "ill-defined", that it must be "well-defined". An example of an "ill-defined" rule is:

$f(x) = \pm x$.

Such an $f$ is not a function, but merely a RELATION.

So when we say:

$f: A \to B$, given by $f: x \to f(x)$, we mean a lot of things all at once:

1. $x \in A$
2. $f(x) \in B$

For EVERY $x \in A$, $f(x)$ is "defined", uniquely. No ambiguity. We often say: $x$ DETERMINES $f(x)$. (Given $x$, and the rule $f$, everyone should always come up with the same element $f(x)$ of $B$, every time).

It is possible to describe some functions in ways that "look completely different". Here is a "toy example".

Let $D = \{1,2,3\}$, and let $B = \{1,2,3,4,5,6,7,8,9\}$.

We can define $f: D \to B$ by:

(1)

$f(x) = x^2$, or by:

(2)

$f:$
$1 \mapsto 1$
$2 \mapsto 4$
$3 \mapsto 9$

or

(3)

$f = \{(1,1),(2,4),(3,9)\} \subset D \times f(D) \subset D \times B$

It is also perfectly permissible to say such things as:

"Let $f$ be the function with domain $D$, where $D$ is the set...(followed by a set-builder, or otherwise unambiguous description of $D$), which takes values in the set $B$...(followed by a suitable description of $B$)" which sends the variable $x$ to $f(x) =$ (followed by some expression in $x$...not "all" expressions result in a function, but most of the ones you encounter will). Certain sets are so often used, they usually need no description:

$\Bbb N$, the set of natural numbers.
$\Bbb Z$, the set of integers.
$2\Bbb Z$, the set of even integers.
$\Bbb Q$, the set of rational numbers.
$\Bbb R$, the set of real numbers.
$\Bbb C$, the set of complex numbers.

$f: X \to Y$, the (usual) reading of this is:

"$f$ such that (this is what the colon means) $f$ maps $X$ into $Y$.

Functions and mappings are often used interchangeably (this changes when you get to abstract algebra, but until then I wouldn't worry about it). Just as maps (the real ones, like in an atlas) don't keep "all the information" of the "real world", functions often "condense" the domain into an "image" (this is sort of like what file compression does, it squeezes out "the repetitious bits"), which keeps SOME, but not necessarily ALL of the original information about the domain.

For example, the squaring function defined from and to the reals loses "sign information" about the domain element. We can take a square root of a square, but we don't always get back where we started from (if the original number was negative, we never do).

I hope this doesn't overwhelm you, I get carried away, at times. Also, everything I like Serena wrote is true.  Reply With Quote

5. Domain of $f$ is sometimes denoted by $\operatorname{dom}(f)$. If you are writing a report and you have doubts if the reader knows this notation, you may define it in the beginning.  Reply With Quote

Thanks everyone! That certainly helped me connect some dots. Originally Posted by I like Serena Indeed. This is unconventional.
First of all, allow me to explain the following notation.

$f(x)=2x|x \in \Bbb{R}$

This is something I made up. It is actually a modification of the input syntax of my TI calculator! This is how that syntax works.

$f(x)=2x|x=3$
$f(3)=2\cdot 3$
$f(3)=6$

The $|$ symbol is officially called the "with operator" in the documentation. I believe it works like pipes in computer command line interpreters. It pipes the value of x into the expression, before evaluating the expression. The calculator I use is very programmer friendly, and programming is closely related to math and the concept of functions. That's actually useful for better understanding math.

$f(x)=2x|x \in \Bbb{R}$

While this notation may hold some validity, I can see how it may not be well received by mathematicians. It is unnecessary and unconventional. Also, from a programmer's viewpoint, we can't pipe an infinite set. But in lack of a proper way of describing this I sometimes used this notation. Quote:
Properly we can say that we have the function $f: \mathbb R \to \mathbb R$ given by $f(x)=2x$.
Or alternatively the function $f: \mathbb R \to \mathbb R$ given by $x \mapsto 2x$.
In this notation the first $\mathbb R$ is the domain, and $f$ is required to be defined for every element of the domain.
The first $\Bbb{R}$ is the domain and the second $\Bbb{R}$ is the codomain? We know that because there is an arrow, and it points from the domain to the codomon.? The domain is "imaged" in the codomain?

I just realized that the colon represents the same thing as the vertical bar $|$ that I used earlier. It is read as "such that". So this is read as follows. "The function $f$, such that, domain $\Bbb{R}$ is imaged in codomain $\Bbb{R}$." Correct?

I believe the word "imaged" and "maped" can be exchanged, because they mean almost the same thing. If I am not mistaken, mapping refers to discrete $x$ and $f(x)$ values, while image or imaging refers to the relation between set $X$ to the image subset $\left\{f(x)\right\}$ in set $Y$. I may be wrong...

Imaging:
$\mathbb R \to \mathbb R$

Mapping:
$x \mapsto 2x$

I may be wrong here. But if I am, I hope that someone will correct me. Quote:
$\mathbb R$ is already a set, so $\left\{\Bbb{R}\right\}$ is a set that contains a set, which is not what we want here.

The symbol $D$ has no meaning unless we define it, which we'd typically do with words.
So we might say, that if we define $D$ to be the domain of $f$, that we have $D=\mathbb R$.
From what I understand, what you're saying here is that unless I constrain the set $\Bbb{R}$, there is no need to make a new superset $D$ that contains $\Bbb{R}$. In that case I might as well just use $\Bbb{R}$. Right?

$D=\left\{\Bbb{R}\right\} = \Bbb{R}$

The set $\Bbb{R}$ is as big as a set can get, unless we expand into complex numbers. Quote:
This is legit set builder notation.
It's the same as $D=\mathbb R$ though.
Yes, it's the same. But $\Bbb{R}$ alone doesn't connect to anything. Maybe I used this in a wrong way. But what I meant to say is that the x from set $D$ connects to the x I used in the function rule.

$D=\left\{x|x\in\Bbb{R}\right\}$

This was before I understood the meaning of $f:X \to Y$. I think I understand now how to connect the domain to the codomain using this notation. Therefore I no longer need to make up my own sets. Quote:
A function $f$ is required to be defined for all values $x \in X$.
However, it's not required that every value $y$ in $Y$ has a corresponding $x$.
The image of $f$ is the set of all reachable values and is written as $f(X)$.
So we have that $f(X) = \left\{f(x)\right\}$, which is a subset of (or equal to) the codomain $Y$.
I just realized... this must be the idea behind the following statement.

$Y = f(x)$

This is commonly found in text books when studying functions. If $Y$ is the codomain and $f(x)$ is the image then this would be true. Would it not? The $Y$ is a set, and the $f(x)$ is a subset. But in text books, I think it is assumed that the subset $f(x)$ or the "range", or the "image" is as big as the codomain. But this doesn't have to be true. It would depend on the function rule and conditions or properties of the domain. Right?

Assume we have the following.

$f(x)=x$

This kind of function is called an identity function if I recall correctly. It would have the codomain Y equal to the image f(x).

$f(x)=x$
$Y=f(x)$

But the following function would not have the codomain Y equal to the image f(x). It would be greater than Y.

$f(x)=x^2$
$Y\gt f(x)$

Please correct me if I am wrong. This is not related to the notation problem, but to my conceptual understanding of functions. It is of course important to understand the concept, but in my opinion the conceptual understanding and the notation goes hand in hand.

For me personally, seeing the symbolism we discuss here actually helps me with the understanding of the subject. But in our school system, we learn these things backwards. We first learn to pronounce long fancy words like "domain" and "function" without properly objectifying it. The set and logic symbols are introduced too late in my opinion. They are taught in high level math courses in high school, or in worst case at university level. In other words you learn it when you actually need to use it, which is backwards approach in my opinion. Many students don't have to take these high level math courses in high school, and yet they go on to study at some university programme where the math requirements are relatively low. It's better to be very well prepared for the university. Math is such a universal language and tool that it can be applied to almost any other area of study than pure math alone.

Those are my thoughts anyway... I just wanted to give you a bit of background on why I even pursue this. Surely, to many students there is probably nothing more boring than "math notation". They are just happy if they pass the course. I like math for its own sake, and I try to build a deeper understanding of the concepts and make math my own. What better way to start than to learn to speak the language, i.e. learn the notation? Notation should not be underestimated. Quote:
The domain has to exclude $0$.

Notations for this are:
$\mathbb R^* = \mathbb R^\times = \mathbb R_{x\ne 0} = \left\{x\in\Bbb{R} \mid x\neq 0\right\} = \left\{x \mid x\in\Bbb{R}, x\neq 0\right\} = ...$
So in other words the domain definition should be inclusive, as I thought. The $x \neq 0$ alone is not the domain. If there are exclusion points in the domain, the set notation should be used, or one of those modified R variants, in case it's defined for all real numbers except for zero. I think $\mathbb R_{x\ne 0}$ looks good. The benefit of this one is that my teachers would also understand it, and so would the rest of the students. If they had the chance to see it...  Reply With Quote

7. Originally Posted by samir First of all, allow me to explain the following notation.

$f(x)=2x|x \in \Bbb{R}$

This is something I made up. It is actually a modification of the input syntax of my TI calculator! The $|$ symbol is officially called the "with operator" in the documentation. I believe it works like pipes in computer command line interpreters.

I just realized that the colon represents the same thing as the vertical bar $|$ that I used earlier. It is read as "such that". So this is read as follows. "The function $f$, such that, domain $\Bbb{R}$ is imaged in codomain $\Bbb{R}$." Correct?
In math both $:$ and $\mid$ really simply mean "such that" or "with" or "where".
So that notation does make sense ... it's just not commonly seen like that.
But one thing about math - just explain what you mean with your notation out front (in words ) and you're good to go - for any audience. The nice thing about following a conventional approach is that there's no need to explain.

For the record, the pipe in a computer command line does behave somewhat differently. Quote:
I believe the word "imaged" and "maped" can be exchanged, because they mean almost the same thing. If I am not mistaken, mapping refers to discrete $x$ and $f(x)$ values, while image or imaging refers to the relation between set $X$ to the image subset $\left\{f(x)\right\}$ in set $Y$. I may be wrong...

Imaging:
$\mathbb R \to \mathbb R$

Mapping:
$x \mapsto 2x$

I may be wrong here. But if I am, I hope that someone will correct me.
Imaged and mapped do mean the same thing.
However, it both identifies which set gets mapped to which set, and it identifies how that mapping takes place.
There is no distinction there (regardless of the \mapsto latex directive).  Quote:
From what I understand, what you're saying here is that unless I constrain the set $\Bbb{R}$, there is no need to make a new superset $D$ that contains $\Bbb{R}$. In that case I might as well just use $\Bbb{R}$. Right?

$D=\left\{\Bbb{R}\right\} = \Bbb{R}$

The set $\Bbb{R}$ is as big as a set can get, unless we expand into complex numbers.
There seems to be some confusion here. Let me clarify that $\left\{\Bbb{R}\right\} \ne \Bbb{R}$.
The first is a set that contains the set $\mathbb R$. That is, $\mathbb R \in \left\{\Bbb{R}\right\}$.
The second is just the set $\mathbb R$, and $\mathbb R \notin \mathbb R$.
To clarify further, $2 \notin \left\{\Bbb{R}\right\}$, but $2 \in \Bbb{R}$. Quote:
Yes, it's the same. But $\Bbb{R}$ alone doesn't connect to anything. Maybe I used this in a wrong way. But what I meant to say is that the x from set $D$ connects to the x I used in the function rule.

$D=\left\{x|x\in\Bbb{R}\right\}$
Let's get this clear. The $x$ inside the set builder notation has nothing to do whatsoever with the $x$ in the function definition. They are completely independent.
It's usually smart to write something like $D=\left\{y \mid y\in\Bbb{R}\right\}$ to avoid the confusion that they may refer to the same thing - they don't. Quote:
I just realized... this must be the idea behind the following statement.

$Y = f(x)$

Assume we have the following.

$f(x)=x$

This kind of function is called an identity function if I recall correctly. It would have the codomain Y equal to the image f(x).

$f(x)=x$
$Y=f(x)$

But the following function would not have the codomain Y equal to the image f(x). It would be greater than Y.

$f(x)=x^2$
$Y\gt f(x)$
First off, we can't really mix $Y$ and $x$. The first is a set, and the second is a specific element of a set.
What we do have is that $f(X)$, the image of the set $X$, has to be a subset of $Y$.
That is, $f(X) \subseteq Y$, or we might say that for every $x \in X$ we have that $f(x) \in Y$.

The identity function has an extra restriction.
It requires that the domain is contained in the codomain, otherwise it is not properly defined.
Usually we would define the identity function as $id: X \to X$ given by $id(x)=x$.

In the case of $f(x)=x^2$ we have that the domain can be $\mathbb R$ and the codomain can be $\mathbb R$ as well, but the image is then $\mathbb R_{\ge 0}$.
As a consequence, the image is strictly a subset of the codomain.
Symbolically: $f(\mathbb R) \subsetneq \mathbb R$. Quote:
What better way to start than to learn to speak the language, i.e. learn the notation? Notation should not be underestimated.
Don't underestimate the words. There is a danger in "thinking in symbols". It tends to limit the imagination so that you only try to think of things you can properly express in symbolic notation. This is a pitfall.
It's really the other way around. When you have long sentences, that are mostly the same each and every time, it's just more efficient to use a short hand notation.
Of course it does take some getting used to, and it's smart to learn the symbolic notation early rather than late.
And when symbolic notation "doesn't make sense", it usually helps to spell it out in words. For the record, I expect any math teacher to be familiar with all these notations - whether they teach them (depending on the audience) or not.  Reply With Quote Originally Posted by Deveno The "standard" way to indicate the domain is something like:

$f: D \to B$

Sometimes mathematicians get fussy about "how big you make $B$", it should, at the very least, contain the range of $f$.
The terms range and image refer to the same thing. Do they not?

What you are saying is that B needs to be at least as big as the image?

Example using identity function:

$D=\Bbb{R}$
$B=\Bbb{R}$
$f: D \to B$
$f(x)=x$
$B=f(x)$

$X=\Bbb{R}$
$Y=\Bbb{R}$
$g: X \to Y$
$g(x)=x^{2}$
$Y\neq g(x)$
$Y>g(x)$

In case of a quadratic function, the set Y is bigger than the image. Is this what you meant? Quote:
So, for your first example, we write:

$f: \Bbb R \to \Bbb R$ given by $f(x) = 2x$, or:

$f: \Bbb R \to \Bbb R$ given by $f: x \mapsto 2x$.
What's the difference between the two types of arrows? From what I can tell, the $\mapsto$ arrow is mapping discrete values from first set to discrete values in the second set. The other arrow $\to$ relates the first set to the second set. Quote:
If we just want to exclude a single point from our domain set, like $x_0$ ($x_0 = 0$ in your example $\dfrac{1}{x}$), we can write:

$f: D - \{x_0\} \to B$
So whatever is placed within curly braces is excluded from the domain? Is that a subtraction? Subtracting the set containing zero from the first set $D$, before relating it to the second set $B$? Quote:
If the definition of the domain set $D$ is more complicated, we may have to settle for:

$f: D \to B$, where $D = \{x \in A: \text{some properties hold for }x\}$, hopefully the set $A$ that $D$ is a subset of is "well-understood" (it is quite possible for the chain of subsets to be quite long, if the definition of $D$ is really complicated, we might have $D \subseteq A \subseteq X \subseteq S$, etc.).
By well understood, you mean something like $\Bbb R$ right? Some set that is well known and well defined?

Can you write it like this?

$f: A - \{x \in S: \text{some properties hold for }x\} \to B$

Would that be valid? Quote:
A co-domain set (the type of thing we get "when we're done", or the "target set", or the set the values lie IN).
By "the set the values lie in", you mean the function image? The f(x)? Quote:
There are reasons, often, for making this set larger than it needs to be: the range of values may be difficult to describe, explicitly, but it may be easier to just say "whatever they are, they're all real numbers".
Ok, so this would be something like the example I gave above?

Example using identity function:

$D=\Bbb{R}$
$B=\Bbb{R}$
$f: D \to B$
$f(x)=x$
$B=f(x)$

$X=\Bbb{R}$
$Y=\Bbb{R}$
$g: X \to Y$
$g(x)=x^{2}$
$Y\neq g(x)$
$Y>g(x)$

In case of the identity function, the codomain is equal to the function image? But in case of the quadratic function, the codomain is bigger than the function image? Bigger than what it needs to be. Yet, we do not need to explicitly specify or constrain that codomain? This is what you mean? Quote:
An important distinction: while $f$ takes values IN its co-domain, such values need not be the entire co-domain (in fact, there is a special word for functions where the range = the co-domain, they are called ONTO).
Ok, so this relates to the above example. Would the identity function be an example of one of these "onto" functions? Quote:
A "rule" for pairing a value $f(x)$ (in the co-domain) to any given $x$ in the domain. This is often done using the "=" sign, like so:

$f(x) =$ some expression in $x$.
Yeah, I think I understand this part. The rule is some operation that acts on the input value of a function and transforms it. How this transformation is done is given by the function rule.

Excuse me if I am wrong, but I think you have this upside down. I think you meant to say pairing a value x in the domain to a value f(x) in the codomain. My argument is that x is the independent variable and f(x) is the dependent variable. In other words, the f(x) depends on the x, not the other way around. Quote:
For example, the function $f: \Bbb R \to \Bbb R$ given by the rule $f(x) = x^2$ is actually the set of all pairs of real numbers (a subset of $\Bbb R \times \Bbb R$, which we often represent by a PLANE, or in desperate times, a sheet of paper) of the form:

$(x,x^2)$, so it makes perfect sense, for example, to say:

$(-2,4) \in f$.
I'm not sure what you meant by "plane" here? Every sheet of paper we work with is finite by its nature. While the set of real numbers $\Bbb R$ is infinite. You can never write down something infinite on something finite. What kind of paper do you work with? Infinite paper sheet?

A plane is two dimensional. But so is a sheet of paper! It's a type of surface. So why is it a problem (in desperate times) to represent a number plane on a sheet of paper?

If I understand you correctly, you want to represent multiples of all the real numbers $\Bbb R \times \Bbb R$ on a plane? Well, this is the same idea as the Cartesian coordinate plane. You take two number lines and make one of them perpendicular to the other at some point. But I don't know how you're supposed to represent all real numbers on it. You can express the idea with symbols but you can't actually do it and plot out all the numbers. You can only plot and view a small section of the plane. Quote:
There is actually something a bit subtle going on with part 3: a function, by its very nature, can only take a specific $x$ to one and only one place in the co-domain. This doesn't mean that two DIFFERENT $x$'s cannot have the same "target" (for example, -3 and 3 both map to 9 under a squaring function), but one $x$ in the domain cannot wind up in "two different targets".
Good point! I think most students quickly learn that there can be only one f(x) value for every x. But the fact that different x values can produce the same f(x) value is something we often don't give a much thought about. These are first encountered when studying quadratic functions. We are just happy if we find our "zeros" or the "roots" and we move on. But there are often other sets of x values that give us the same f(x) values. It can occurred at f(x)=0 but it can also occur at f(x)=2. Even though there are no "roots" or "zeros", there can still be multiple x values that give us the same f(x) value. Quote:
If some rule allows this, we call it "ill-defined", and conversely, we require a function NOT be "ill-defined", that it must be "well-defined". An example of an "ill-defined" rule is:

$f(x) = \pm x$.

Such an $f$ is not a function, but merely a RELATION.
Would something like this be considered ill defined?

$\displaystyle f(x)=\begin{cases}1, & x=2 \\x^{2}, & \text{all other real} \, x \\ \end{cases}$ Quote:
So when we say:

$f: A \to B$, given by $f: x \to f(x)$, we mean a lot of things all at once:

1. $x \in A$
2. $f(x) \in B$

For EVERY $x \in A$, $f(x)$ is "defined", uniquely. No ambiguity. We often say: $x$ DETERMINES $f(x)$. (Given $x$, and the rule $f$, everyone should always come up with the same element $f(x)$ of $B$, every time).
All this really says is that for every x value there is an f(x) value. It also says that x is independent of f(x), and that f(x) depends on x. Quote:
It is possible to describe some functions in ways that "look completely different". Here is a "toy example".

Let $D = \{1,2,3\}$, and let $B = \{1,2,3,4,5,6,7,8,9\}$.

We can define $f: D \to B$ by:

(1)

$f(x) = x^2$, or by:

(2)

$f:$
$1 \mapsto 1$
$2 \mapsto 4$
$3 \mapsto 9$

or

(3)

$f = \{(1,1),(2,4),(3,9)\} \subset D \times f(D) \subset D \times B$
Nice! So in example 2, the rule of the pattern is left to be discovered by the reader. Quote:
Not "all" expressions result in a function, but most of the ones you encounter will).
Can you post an example of an expressions that doesn't result in a function? Quote:

$f: X \to Y$, the (usual) reading of this is:

"$f$ such that (this is what the colon means) $f$ maps $X$ into $Y$.

Functions and mappings are often used interchangeably (this changes when you get to abstract algebra, but until then I wouldn't worry about it). Just as maps (the real ones, like in an atlas) don't keep "all the information" of the "real world", functions often "condense" the domain into an "image" (this is sort of like what file compression does, it squeezes out "the repetitious bits"), which keeps SOME, but not necessarily ALL of the original information about the domain.
How about "function $f$, such that subset $x$ in superset $X$ is imaged into subset $f(x)$ in superset $Y$"? Or something like that? I mean it's the set of all discrete values in X that are imaged into Y? Is this the same as mapping? Quote:
For example, the squaring function defined from and to the reals loses "sign information" about the domain element. We can take a square root of a square, but we don't always get back where we started from (if the original number was negative, we never do).
In other words it's a one way street. Quote:
I hope this doesn't overwhelm you, I get carried away, at times. Also, everything I like Serena wrote is true.
It's a lot to take in but I don't think it's that difficult that I would give up. When I encounter a very hard problem or don't understand some complicated subject or idea, I try to break it down to its smallest components and then build up an understanding from the basic blocks.  Reply With Quote Originally Posted by Evgeny.Makarov Domain of $f$ is sometimes denoted by $\operatorname{dom}(f)$. If you are writing a report and you have doubts if the reader knows this notation, you may define it in the beginning.
Yes, I have seen this notation too. But this alone does not define the domain of the function $f$. It would need to be not only described as the "domain of f". It would also need to be defined using set builder notation.

Maybe something like this.

$\operatorname{dom}(f)=\left\{x\in \Bbb{R} | x \gt 7\right\}$

Or something less restrictive.  Reply With Quote

10. Originally Posted by samir The terms range and image refer to the same thing. Do they not?

What you are saying is that B needs to be at least as big as the image?
Yes. The image $f(X)$ is required to be a subset of the co-domain $Y$. This becomes even more important when $X,Y$ have "more structure defined on them" (such as the ability to add elements). Quote:
What's the difference between the two types of arrows? From what I can tell, the $\mapsto$ arrow is mapping discrete values from first set to discrete values in the second set. The other arrow $\to$ relates the first set to the second set.
The $\mapsto$ symbol is usually employed to indicate what happens to a "typical element", $x \in X$ (or $x \in D$). The arrow is a little less concrete, merely indicating "source" and "target" sets, typically. But this isn't "set in stone", some people use the arrow for both meanings, since we're used to the linguistic notion $\to$ means "goes to". Quote:
So whatever is placed within curly braces is excluded from the domain? Is that a subtraction? Subtracting the set containing zero from the first set $D$, before relating it to the second set $B$?
Yes, it is set subtraction:

$A - B = \{x \in A: x \not\in B\}$

This is also written $A\setminus B$, which some people prefer (so it is not confused with *algebraic* subtraction). Quote:
By well understood, you mean something like $\Bbb R$ right? Some set that is well known and well defined?
Yes. Often we aren't looking at a gazillion different domains, but focusing on the different kinds of functions DEFINED on a single domain (such as studying "continuous everywhere" functions in calculus). Quote:
Can you write it like this?

$f: A - \{x \in S: \text{some properties hold for }x\} \to B$

Would that be valid?
Yep. For example, you could define a function defined only on the rationals by:

$f: \Bbb R - \{x \in \Bbb R: x\text{ is irrational }\} \to \Bbb R$ Quote:
By "the set the values lie in", you mean the function image? The f(x)?
Yes. There's some "linguistic" confusion that may arise, $f(D)$ is the *image set* of $D$ under $f$, and is often just called the image of $D$, whereas the individual values $f(x)$ are ALSO called images. This is because we actually have two different (but related) functions:

$f:X \to Y$

$\tilde{f}: \mathcal{P}(X) \to \mathcal{P}(Y)$ (here, $\mathcal{P}(X)$ means the *power set* of $X$, that is the collection of all subsets of $X$).

We define $\tilde{f}$ like so:

For $S \subseteq X$, we have:

$\tilde{f}(S) = \{y \in Y: \exists s \in S, \text{ with } f(s) = y\}$.

Because $f$ and $\tilde{f}$ are so intimately related, often they are both just called "$f$". Quote:
Ok, so this would be something like the example I gave above?

Example using identity function:

$D=\Bbb{R}$
$B=\Bbb{R}$
$f: D \to B$
$f(x)=x$
$B=f(x)$
See the confusion that can arise? What you probably mean is $B = \tilde{f}(D)$, because $\tilde{f}$ is "set-valued", and $f$ is "element-valued". In other words, you have a "type-mismatch" when you write:

$B = f(x)$

since the LHS is a SET, and the RHS is just some element $y \in B$. Quote:

$X=\Bbb{R}$
$Y=\Bbb{R}$
$g: X \to Y$
$g(x)=x^{2}$
$Y\neq g(x)$
$Y>g(x)$
Another type mis-match. What you really mean to say is: $\tilde{g}(X) \subsetneq Y$. Quote:
In case of the identity function, the codomain is equal to the function image? But in case of the quadratic function, the codomain is bigger than the function image? Bigger than what it needs to be. Yet, we do not need to explicitly specify or constrain that codomain? This is what you mean?
Sometimes you have to be careful about "naming a function".

To call $f(x) = x$ the "identity function", typically ASSUMES that the domain=the co-domain. But you have THIS function:

$f: \Bbb Q \to \Bbb R$, given by $f(x) = x$.

This is not usually called an "identity function", but rather "an inclusion function". Changing the co-domain can change everything. Quote:
Ok, so this relates to the above example. Would the identity function be an example of one of these "onto" functions?
Only if the co-domain = the domain (see my example above). In somewhat loose terms, the function rule $f(x) = x$ does not map the rationals onto the reals. Quote:
Yeah, I think I understand this part. The rule is some operation that acts on the input value of a function and transforms it. How this transformation is done is given by the function rule.

Excuse me if I am wrong, but I think you have this upside down. I think you meant to say pairing a value x in the domain to a value f(x) in the codomain. My argument is that x is the independent variable and f(x) is the dependent variable. In other words, the f(x) depends on the x, not the other way around.
Yes, the domain houses the "independent" variables, and the co-domain the "dependent" values. We pick the domain element first, though, and THEN assign the "image". For example, suppose we have $f: \Bbb R \to \Bbb R$ where $f(x) = x^2$, and I tell you to "pair something with the image $4$". Am I going to be able to be 100% sure of what you pick? Quote:
I'm not sure what you meant by "plane" here? Every sheet of paper we work with is finite by its nature. While the set of real numbers $\Bbb R$ is infinite. You can never write down something infinite on something finite. What kind of paper do you work with? Infinite paper sheet?
Yes, I have infinite paper, and I tear off finite pieces when I need to draw something. Of course. Quote:
A plane is two dimensional. But so is a sheet of paper! It's a type of surface. So why is it a problem (in desperate times) to represent a number plane on a sheet of paper?
Imagine my tongue poking out the surface of my cheek. Quote:
If I understand you correctly, you want to represent multiples of all the real numbers $\Bbb R \times \Bbb R$ on a plane? Well, this is the same idea as the Cartesian coordinate plane. You take two number lines and make one of them perpendicular to the other at some point. But I don't know how you're supposed to represent all real numbers on it. You can express the idea with symbols but you can't actually do it and plot out all the numbers. You can only plot and view a small section of the plane.
Correct. But what we CAN do, is indicate part of the GRAPH of a function, by drawing "darker" on a blank sheet of paper the points on the graph (a curious procedure called "plotting", possibly because many monarchies have been ended this way). Quote:
Good point! I think most students quickly learn that there can be only one f(x) value for every x. But the fact that different x values can produce the same f(x) value is something we often don't give a much thought about. These are first encountered when studying quadratic functions. We are just happy if we find our "zeros" or the "roots" and we move on. But there are often other sets of x values that give us the same f(x) values. It can occurred at f(x)=0 but it can also occur at f(x)=2. Even though there are no "roots" or "zeros", there can still be multiple x values that give us the same f(x) value.
Yep. Constant functions are *really* bad this way-it's impossible to tell how we got there, since "all roads lead to the same place". Quote:
Would something like this be considered ill defined?

$\displaystyle f(x)=\begin{cases}1, & x=2 \\x^{2}, & \text{all other real} \, x \\ \end{cases}$
No, it's fine...you give me any real $x$, and I can tell you what $f(x)$ is, no problems there. Quote:
All this really says is that for every x value there is an f(x) value. It also says that x is independent of f(x), and that f(x) depends on x.
It says a "little" more. Not much though..it tells us "what kinds of things" $x$ and $f(x)$ can be. Functions are pretty straight-forward, compared to a lot of stranger things, like set-measures, for example. Quote:
Nice! So in example 2, the rule of the pattern is left to be discovered by the reader.
Yep. Part of the fun in mathematics, is occasionally, one discovers "cross-references" to places one has already been. It's a happy thing. Quote:
Can you post an example of an expressions that doesn't result in a function?
I did. Here is another one, the so-called "unit circle function" (which isn't one):

$f(x) = \pm\sqrt{1 - x^2}$ Quote:
How about "function $f$, such that subset $x$ in superset $X$ is imaged into subset $f(x)$ in superset $Y$"? Or something like that? I mean it's the set of all discrete values in X that are imaged into Y? Is this the same as mapping?
Pretty much any rule that returns a SET of images from a single element, rather than just an image element, is an example. Sometimes, we can "get around this" by redefining our domain (see the discussion of $\tilde{f}$ above) and co-domain. Equivalence classes are one way we've discovered as a way of "binding sets up into elements". Quote:
In other words it's a one way street.
Not always. But when it IS a two-way street (such functions are called "bijections" or "one-to-one correspondences") it's a very special event. Quote:
It's a lot to take in but I don't think it's that difficult that I would give up. When I encounter a very hard problem or don't understand some complicated subject or idea, I try to break it down to its smallest components and then build up an understanding from the basic blocks.
Judging by the astute questions you are asking, you're doing great!  Reply With Quote

Thank you guys! I am really impressed by your display of knowledge here! This is very helpful. I will surely revisit this thread often as I explore these topics.

Let me get back to some examples again. I will jump back to the linear function again.

$y=2x$

$y: X\to Y$

$X=\Bbb{R}$

$Y=\Bbb{R}$

First of all, is this a valid notation? Maybe it's just me, but I find this confusing. I see x and y appear all over the place. I know they are not all the same. Maybe I just need to accept this, and make sure that whoever is reading my work understands that $x$ and $X$ are not the same, and that $y$ and $Y$ are not the same.

I used the lower case $y$ instead of $f(x)$ this time. Am I wrong if I say that the $y$ is not a function name? Because my understanding is that $f(x)$ notation was introduced so that we could name our functions.

$y=2x$

I would like an explanation of this notation. What is this notation saying? Is it referring to the domain and codomain? Or the function image? If so, then what is the name of the function? Is it unnamed? It just relates values in set X to values in set Y?

$f(x)=2x$

$f: X\to Y$

$X=\Bbb{R}$

$Y=\Bbb{R}$

I used the $f(x)$ notation here and it makes it more clear to me what is what. But I still have the x appearing in two places. When written by hand, $x$ and $X$ tend to look the same. So one could still be mistaken for the other.

$f(x)=2x$

$f: D\to C$

$D=\Bbb{R}$

$C=\Bbb{R}$

Now I got rid of the repeating x. I use "D" for domain and "C" for codomain. But I see a problem now! How does the x connect to the domain? If I were to graph this, would my X-axis be labeled as D-axis, and my Y-axis as C-axis?

Then there is a slightly more verbose version of this.

$f(x)=2x$

$f: Dom\to Cod$

$Dom(f)=\Bbb{R}$

$Cod(f)=\Bbb{R}$

Is this valid? I am doubtful about the "Dom(f)" and "Cod(f)" notation. Perhaps "Dom" and "Cod" alone, without the "of f" part would be better. The use of "Dom(f)" to say "the domain of f" appears to be a function in its own right, rather than a set name. Such that, input is the name of the function, and output is the domain set of the function.

That may be useful if we have several functions defined, with each having its own domains and codomains, and we want to call for or refer to the domain or codomain of one of them. We can call it with Dom(function name) and Cod(function name) and have the output be a set. Function name is the only required argument (like in programming we have arguments and parameters).

$f(x)=2x$

$g(x)=x^2$

$Dom(f)=\Bbb{R}$

$Cod(f)=\Bbb{R}$

$Dom(g)=\Bbb{R}$

$Cod(g)=\left\{y|y\in \Bbb{R}, y \geq 0\right\}$

But as a set name, we can't use "Dom(f)" or "Cod(f)"? I would argue that if it looks like a function then it probably is a function, and we can't have a function as the name of a set. So I would probably change this to say:

$f(x)=2x$

$f: Dom\to Cod$

$Dom=\Bbb{R}$

$Cod=\Bbb{R}$

But then it's not much different than using just "D" and "C", like in the example above it.

The domain set needs to have a name? If so, I don't see how a function can be used as the name of a set. Maybe it can, I don't know. But it seems odd to me. But using "Dom(f)" and "Cod(f)" as a shorthand notation for referring to the domain and codomain of a defined function could be a viable option.  Reply With Quote

domain, function, learn, real, write #### Posting Permissions 