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Question about Sets and Functions

LeibnizIsBetter

MHB Donor
Aug 28, 2013
3
I know this is probably the most basic question imaginable so please bear with me. I did google it but I still couldn't figure it out.

Say you have a function \(\displaystyle f(x, y, z)\) and a point \(\displaystyle (x_0, y_0, z_0)\) that satisfies the equation \(\displaystyle f(x, y, z) = 0\).

Does that imply that \(\displaystyle f(x_0, y_0, z_0) \in f(x, y, z)\) ? Where \(\displaystyle \in\) means "is an element of".

Or, what's the relationship between \(\displaystyle (x_0, y_0, z_0)\) and \(\displaystyle f(x, y, z)\) if \(\displaystyle (x_0, y_0, z_0)\) is a point on the surface defined by \(\displaystyle f(x, y, z) = 0\)?

Thanks so much. I'm new to this and didn't drink enough coffee today.
 
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Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Hi LeibnizIsBetter!

I know this is probably the most basic question imaginable so please bear with me. I did google it but I still couldn't figure it out.

Say you have a function \(\displaystyle f(x, y, z) = 0\) and a point \(\displaystyle (x_0, y_0, z_0)\) that satisfies the equation \(\displaystyle f(x, y, z) = 0\).
Formally, f(x,y,z) is not a function. It's an output of a function for some input (x,y,z).
The function itself is simply called "f".

To write it properly, we can say:

Let f be a function $\mathbb R^3 \to \mathbb R$ and let $(x_0, y_0, z_0)$ be a point such that $f(x_0, y_0, z_0) = 0$.


Does that imply that \(\displaystyle f(x_0, y_0, z_0) \in f(x, y, z)\) ? Where \(\displaystyle \in\) means "is an element of".

Or is there a better way of stating what I'm trying to say?

Thanks so much. I'm new to this and didn't drink enough coffee today.
Since $f(x,y,z)$ is not a set but an output value, it does not have elements.

I think you mean to say that $f(x_0, y_0, z_0) \in \text{codomain of }f$.
We might also refer to the "range" or "image" of f instead of the "codomain".

Alternatively, we can also write that $f(x_0, y_0, z_0) \in f(\mathbb R^3)$, where the latter represents the so called "image" of f.
 

LeibnizIsBetter

MHB Donor
Aug 28, 2013
3
Wow! Thank you so much!
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
To underscore what ILikeSerena said, to have a function we need 3 things:

1. A set that the function $f$ acts upon (the "source set" or "set of input values", although the term "domain" is currently the most fashionable).

2. A set that the function $f$ maps to (the "target" set or "set where output values live").
Typically, this is called the "co-domain", and it gets confused with "range".

Let me give an example, to underscore the subtle difference:

A commonly encountered function is the "squaring function" usually written:

$f: \Bbb R \to \Bbb R, f(x) = x^2$.

When we say the co-domain is the real numbers, all we are saying is that for any real number $x$, $f(x) = x^2$ is also a real number. The range is often a PROPER subset of the co-domain (in this case, it is the set of all non-negative real numbers).

Technically, the function:

$f: \Bbb R \to \Bbb R_0^{+}, f(x) = x^2$ is a different function, because the graph of one includes the plane below the x-axis, while in the other there's nothing there. In practice, the difference between:

$f:A \to B$

and

$f: A \to f(A)$

often is irrelevant, but I wish to point out that the SECOND function is ONTO, whereas the first one may not be. Sometimes, this is important (like when you are finding inverse functions).

Finally you need:

3. A "rule of assignment" that tells you WHICH element of $B$ (the co-domain) $f(a)$ is, for each and every element $a \in A$. This assignment may be via a formula, such as:

$f(x) = x^2$

or it may be explicitly defined, such as:

$f:\{1,2,3\} \to \{1,2,3\}$

$f(1) = 1, f(2) = 3, f(3) = 2$

(indeed, functions can be finite).

FORMALLY, the definition of function is this:

$f \subseteq AxB: \forall (a_1,b_1),(a_2,b_2) \in f, a_1 = a_2 \implies b_1 = b_2$

that is, a function can only have one value $f(a)$ for every $a \in A$.

So, for example, "circles in the plane" are not functions, because for every point $x \in (-r,r)$ there are TWO values of $y$ such that:

$x^2 + y^2 = r^2$, namely:

$y = \sqrt{r^2 - x^2}$
$y = -\sqrt{r^2 - x^2}$

whereas the semi-circle obtained by consistently choosing either the positive, or the negative square root *is* a function.

*******

It is often typical in many textbooks to find the author confusing a function with its value.

You may read something like:

"Consider the function "$p(x) = x^2 + ax + b$".

This OUGHT to be:

"Consider the function $p$ defined by $p(x) = x^2 + ax + b$".

However, in speaking of functions, we often have to give them a "name", and while a "technically correct" name for $p$ might be:

"$[\ \ ]^2 + a[\ \ ] + b$"

such an arrangement gets to be cumbersome.
 
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