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I'm reading Serge Lang's (S.L) linear algebra book. In the beginning, at function spaces section there is such a text:

Let $S$ be a set and $K$ a field. By a function of $S$ into $K$ we shall mean an association which to each element of $S$ associates a unique element of $K$. Thus if f is a function of $S$ into $K$, we express this by the symbols

$$f:S \rightarrow K$$

We also say that $f$ is a $K$-valued function. Let $V$ be the set of all functions of $S$ into $K$. If $f$, $g$ are two such functions, then we can form their sum $f$ + $g$. It is the function whose value at an element $x$ of $S$ is:

$$f(x) + g(x)$$

We write

$$(f + g)(x) = f(x) + g(x)$$

If $c \in K$, then we define $cf$ to be the function such that

$$(cf)(x) = cf(x)$$

Thus the value of $cf$ at $x$ is $cf(x)$. It is then a very easy matter to verify that $V$ is a vector space over $K$. We shall leave this to the reader.

From what I know, set $V$ is a vector space over field $K$ iff it has:

  1. Associative property of addition
  2. Additive inverse of every element equal to additive identity ($0$)
  3. Commutative property of addition
  4. Distributive property for arbitrary scalar multiplied by sum of its elements
  5. Distributive property for element multiplied by sum of arbitrary scalars
  6. Multiplicative associative property
  7. Unaffected elements when multiplied by multiplicative identity ($1$)
  8. Every linear combination of elements belonging to the set

From what I can see, N4 is satisfied, but how can I prove that function space satisfies other properties as well? i.e commutative property

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    $\begingroup$ The commutativity property reads $f+g=g+f$, which means $$f(s)+g(s)=g(s)+f(s) \ \ \ (*)$$ for all $s \in S$. But since $f(s)$ and $g(s)$ are elements in the field $K$, (*) follows. $\endgroup$ – Dzoooks Aug 14 '18 at 22:27
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    $\begingroup$ Equality of functions is checked by their values at each point, and the values lie in the field $K$ which satisfies these properties. $\endgroup$ – Bernard Aug 14 '18 at 22:28
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    $\begingroup$ Well, every member of a field except 0, the additive identity, has a multiplicative inverse. I am not sure I would say that a vector space "inherits" any thing from the field. A vector space is sometimes said to be "over" a field- the scalars defining the "scalar multiplication" in a vector space are members of the field but the vectors themselves are not. $\endgroup$ – user247327 Aug 14 '18 at 22:45
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    $\begingroup$ @ShellRox: More exactly it inherits the vector space properties of $K$ as a vector space over itself. It does not inherits the field properties – only the ring properties.(but that's irrelevant here). $\endgroup$ – Bernard Aug 14 '18 at 22:45
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    $\begingroup$ Every function space i,herits. No this isn't an isomorphism since there's (in general) no bijection. $\endgroup$ – Bernard Aug 15 '18 at 0:47
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According to comments below the question, function space is vector space over the field $K$ due to being over the field $K$.

Any arbitrary field $K$ is a vector space over itself. If $V$ is a set of all functions from a set $S$ to a field $K$, and $f$ and $g$ are functions that belong to such function space, then $f(x)$ and $g(x)$ belong to a field $K$. Functions $g(x)$, $f(x)$ and any function in the function space "inherit" the vector space properties of field $K$ as a vector space over itself, i.e commutative property $f(x) + g(x) = g(x) + f(x)$.

For some reason, S.L only mentions it later in the text book, that any arbitrary field $K$ is a vector space over itself. Thus there perhaps may be other ways to prove that function space is a vector space over field.

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