vector spaces - interpretation of a part of a text There is a part of a text that I don't understand - it is not homework, so I don't have to give any answer, it is just for understanding what I am reading.
The text, after introducing the definition of number field and its properties and the definition of a vector space and its properties, says: "Let M be any set, and let F(M,R) the set of all real-valued functions on M. (Think of an interval I ⊂ R for M.) Deﬁne addition and multiplication with scalars by (f + g)(m) = f(m) + g(m), (af)(m) = af(m). Then F(M,R) is a real vector space.
Check this! What is M in that case?".
I find what is written here a bit abstract and I don't get it...
 A: Since the set of the real functions on $M$ is $F(M,\mathbb{R}):=\{f:M\rightarrow\mathbb{R}\mid f \text{ a function}\}$, $M$ can be any set over which the functions are defined. This doesn't give problems to the definition of a vector space, since sum and product by scalars work anyway: you can choose the neutral element as $f_0$ defined by $f_0(m)=0$, for every $m\in M$, and the negative of $f_1$ as $-f_1$ (defined, as always, by $-f_1(m)=(-1)f_1(m)$, for every $m\in M$; it can be done because $(\mathbb{R},+)$ is a group). 
A: In order to understand what's going on, the first thing you need to understand is what a function is.

What a function is
Put simply, to have a function, you need two sets of things.  Call them $A$ and $B$.  Choose one of the sets as the set of inputs.  Let's choose $A$ as the set of our inputs.  Then a function is just something that "sends" each $a$ in $A$ to some $b$ in $B$.  If the function is named $f$ and the input $a$ is sent to $b$, then we write $f(a) = b$ as notation for that.  The important property of a function is that no input is sent to more than one output.  So we can't have one $a$ in $A$ being sent to more than one $b$ in $B$.
$A$, the set of inputs, is called the domain of the function.  The subset of $B$ that has things from $A$ being mapped to them is called the range of the function.
You can always come up with random/arbitrary functions given two sets.  Let $A = \{1,2,3\}$ and $B = \{4,5,6,7\}$.  You can write a random function by taking each element of $A$ and "sending" it to something in $B$.  I will send $1$ from $A$ to $6$ from $B$.  I will send $2$ from $A$ to $4$ from $B$.  I will send $3$ from $A$ to $6$ in $B$.  I can call this function $f$, so now I have $f(1) = 6$, $f(2) = 4$, and $f(3) = 6$.  Notice that every input has no more than one output, which is a requirement of functions.  But the output $6$ has two inputs ($1$ and $3$) being sent to it -- which is completely OK.  The domain of $f$ is $A$, and the range is $\{4,6\}$, since this is the subset of $B$ with things from $A$ being sent to them.

The next thing you need to understand is what a vector space is.
A vector space consists not of one set, but of two sets.  You specify one set which has elements that you call "vectors".  We can name this set $V$.  The other set has elements which you call "scalars".  We can name this set $S$.
So, to have a vector space, you need two sets.  Now, $S$ has to be a special type of set.  It has to be what's called a field, so it has to satisfy some properties listed in your textbook.  The main one it has to satisfy is that under the operation of multiplication (*), every non-zero element has a multiplicative inverse.  What this means is every non-zero element $a$ in $S$ has an element $b$ in $S$ with $ab = 1$.  We can write $\frac{1}{a}$ instead of $b$.  One obvious example of a field is $\Bbb R$, the set of real numbers.  $\pi$, for example, has $\frac{1}{\pi}$ as its multiplicative inverse, since $\pi * \frac{1}{\pi} = 1$.  Also, the number $1$ must be in the field $S$ (it's in every field).
Once you have the two sets $V$ and $S$, we think of the vector space as the set of elements of $V$ -- but you can multiply these elements of $V$ by elements of $S$.  For example, if $V$ is $\Bbb R^{2}$, and $S$ is $\Bbb R$, then of course we can multiply $\begin{bmatrix} a \\ b \end{bmatrix}$ in $\Bbb R^{2}$ by, for example, $2$ in $\Bbb R$.  The result is $2\begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 2a \\ 2b \end{bmatrix}$.
Anyway, to be a vector space, this new set of stuff from $V$ multiplied with $S$ has to satisfy the vector space axioms written in your textbook.  If the new set does satisfy all of those, it is called a vector space.

Now that you hopefully understand all of the above, we can discuss what the book is saying.
The book says "Let $M$ be any set."  What we mean by that is exactly what it says.  Let $M$ be any set of things.  It can be a set of numbers, or even the set $\{\text{book}, \text{shoe}, \text{car}\}$.  Given any two sets, we can always define a function between them as we did above, so it doesn't matter what set $M$ is.  It is just a random set.  No need to specify which one for the theorem that they will state to hold.
Now, if $M$ is any set, we can think about all of the functions between the set $M$ and $\Bbb R$.  Think about all of the possible functions you can define between these two sets.  The set of all of these possible functions is denoted $F(M, \Bbb R)$, and it is a set of functions.
Now, for a vector space, something I did not mention above is that you need two operations on it.  You need to be able to add any two elements in it, and as we said above you need to be able to multiply things in $V$ with things in $S$ (called scalar multiplication).  In the definition you are asking about, they are specifying how the addition and scalar multiplication works.
If you have two functions in $F(M, \Bbb R)$ (which is our $V$ here), you can define the new function $f + g$ as the function that, for each input $m$ from $M$, $f + g$ sends $m$ to $f(m) + g(m)$ (the sum of where $m$ is sent to under $f$ and $g$).  It looks like the scalar set $S$ here is $\Bbb R$.  Since $V$ is $F(M, \Bbb R)$, we should be able to talk about multiplication between $V$ and $S$ here.  That's what they define next: if $a$ is a scalar (i.e., an element of $\Bbb R$) and $f$ is a vector (i.e., an element of $F(M, \Bbb R)$), then define $af$, the product, to be the function that sends each input $m$ to $a*f(m)$, the product of $a$ and the place where $f$ sends $m$ to.  $a$ and $f(m)$ are both just real numbers, so multiplication between them makes sense.
Now, armed with these definitions, the book claims that the sets $V = F(M, \Bbb R)$ and $S = \Bbb R$ form a vector space, which is something you should check.  It's actually really easy, even though it looks hard or intimidating at first.  Just check that each of the vector space axioms from the book holds in this case.
