MHB How do we see that these are mappings from the definition?

[Guest2]

Definition: If $S$ and $T$ are nonempty sets then a mapping from $S$ to $T$ is a subset, $M$, of $S \times T$ such that for every $s \in S$ there's a unique $t \in T$ such that the ordered pair $(s, t) \in M.$
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Could someone please explain how these are mappings. The notation of the definition and that of the examples is different. How do we see that these are mappings from the definition?

 

[Guest2]

For what it's worth, here's what I understand so far:

Example #1: If $S$ is a non-empty set then a mapping from $S$ to $S$ is a subset, $M$, of $S \times S$ such that the for every $s \in S$ there's unique $s \in S$ such that the ordered pair $(s, s)$ belongs to $M$. More explicitly, $M = \left\{(s, s): s \in S\right\}.$

Example #2: If $S$ and $T$ are non-empty sets then a mapping from $S$ to $T$ is a subset, $M$, of $S \times T$ such that the for every $s \in S$ there's unique $t_0 \in T$ such that the ordered pair $(s, t_0)$ belongs to $M$. More explicitly, $M = \left\{(s, t_0): s \in S, t_0 \in T\right\}.$

Example #3: I can't even begin to comprehend this. (Shake) What's $M$ in this case?

 

[Evgeny.Makarov]

Example #1: $M = \left\{(s, s): s \in S\right\}.$

Example #2: $M = \left\{(s, t_0): s \in S, t_0 \in T\right\}.$

I would write the second formula as $M = \left\{(s, t_0): s \in S\right\}$ because $t_0$ is fixed and does not range over $T$. I have some small problems with your text. For example, when you say "a mapping from $S$ to $T$ is a subset, $M$, of $S \times T$", it sounds like you are giving a definition of what any map from $S$ to $T$ is, but then you describe a specific map. Also, one does not usually say "there exists a unique $t_0$" where $t_0$ is an object introduced earlier. The phrase "there exists" must be followed by a variable, preferably fresh (i.e., not previously used). It's OK to say "there exists a $z$, namely, $z=t_0$".

Example #3: I can't even begin to comprehend this.

$M\subset(J\times J)\times\Bbb Q$ where $\Bbb Q$ is the set of rational numbers. $M=\{((m,n),m/n)\mid m,n\in J,n\ne0\}$.

The book you are reading uses a pretty nonstandard notation by writing function to the right of its argument and by denoting the set of integers as $J$.

 

[Guest2]

Thank you very much! Could you help me with the last example as well, please?

 

[Deveno]

A word about mappings in general:

An element $(a,b)$ of a mapping $f:A \to B$ is a pair. We say: $f$ maps $a$ to $b$. the set $A$ is called the DOMAIN (or souce set) of $f$, and the set $B$ is called the CO-DOMAIN (or target set).

You can think of $f$ as something that "grabs" elements of $A$, and "throws" them into $B$. Two elements of $A$ may hit the same target, but each domain element can only be thrown "once".

The "$b$" of a pair $(a,b) \in f$, is called the IMAGE of $a$ under $f$, and we often write $b = f(a)$ (Apparently your text writes $b = (a)f$. This is non-standard, but some authors do it, so that composition of functions occurs in the same order we read: left-to-right). It's important to realize that just listing the RELATIONSHIP of $b$ to $a$ is "not enough", for example it is bad to write:

"the function $x^2$"

and better to say, the function $f:A \to B$ such that $f(a) = a^2$ (or in your "style," $(a)f = a^2$) for every $a \in A$ (of course, this pre-supposes that $a^2$ "makes sense" for $B$, that is, we can multiply elements of $A$ together, and such multiples are, in fact, in $B$).

Another way mappings are often indicated is like so:

$f: a \mapsto f(a)$ (or, again, $a \mapsto (a)f$), but this again, is "not enough", we have to say WHAT SETS (what kinds of "things") $a$ and $f(a)$ (or...$(a)f$...) live in.

With your fourth example, we have the domain: $J \times (J-\{0\})$, whose elements are pairs of integers, with the second element of the pair non-zero. For example, one such element is $(-4,3)$.

Under the mapping $\tau: J \times (J - \{0\}) \to \Bbb Q$ we have:

$(k,m)\tau = \dfrac{k}{m}$, for example, with the pair above, we obtain $(-4,3)\tau = -\dfrac{4}{3}$.

Note that this mapping takes the pairs $(2,2)$ and $(3,3)$ (which are clearly "different" pairs) to the same rational number, $1$.

 

[Guest2]

Many thanks for the explanation.

(Apparently your text writes $b = (a)f$. This is non-standard, but some authors do it, so that composition of functions occurs in the same order we read: left-to-right).

I think this is largely where my confusion comes from. I'm very much used to $b = f(a)$, not $b = (a)f$, which to me is very strange.

 

[Evgeny.Makarov]

Thank you very much! Could you help me with the last example as well, please?

In fact, I looked at the wrong place and my last example was example #4 from the picture. And yes, the domain is $J\times(J-\{0\})$, so the function is a subset of $J\times(J-\{0\})\times\Bbb Q$.

In example #3 one can argue that the function is not well defined because, for example, $\dfrac{2}{3}=\dfrac{-2}{-3}$, so a single positive rational number can be mapped to two pairs of integers: $(2,3)$ and $(-2,-3)$. If we stipulate that both elements of the resulting pair are positive integers, then we have a function. Its domain is $\Bbb Q^+=\{q\in\Bbb Q\mid q>0\}$ and the codomain is $\Bbb Z^+\times\Bbb Z^+$ where $\Bbb Z^+=\{z\in\Bbb Z\mid z>0\}$ and $\Bbb Z=J$ is the set of integers. The function is
\[
\{(q,m,n)\mid m/n=q\text{ and the greatest common divisor of }m,n\text{ is 1}\}.
\]

Note that a single rule for mapping inputs into outputs and a single domain corresponds to many functions, which have different codomains. All images of the domain elements must be present in the codomain, but the codomain may also contain other elements. So the function from example 3 (with the stipulation that $m,n>0$ may be viewed as a function from $\Bbb Q^+$ to $\Bbb Z^+\times\Bbb Z^+$ or from $\Bbb Q^+$ to $\Bbb Z\times\Bbb Z$ or in infinitely many other ways.

 

[Deveno]

Would this happen to be from Herstein, by any chance?

 

[Guest2]

Would this happen to be from Herstein, by any chance?

It's indeed from Herstein! :D