# Showing a set $V$ is a vector space over the real numbers

Let $V=\{f : S \to \mathbb{R} \}$ be the set of all the functions $f : S \to \mathbb{R}$. Show that $V$ is a vector space over $\mathbb{R}$ with addition and scalar multiplication defined by: $(f_1 + f_2)(x) = f_1(x) + f_2(x)$ and $(cf)(x) = cf(x)$ for all $f_1,f_2 \in S$ and $c \in \mathbb{R}$. I am not to sure what he wants me to show exactly, could it be to show all the other axioms (commutative, distributive, etc.) follow true if addition and scalar multiplication are defined as above?

• LaTeX formatting would make your question much more readable... Jan 19, 2017 at 1:59
• you have a set of axioms that define a vector space. i.e. closure under addition, zero vector, additive inverse, distributive property of scalar multiplication over vector addition. Show that all hold. Jan 19, 2017 at 2:00

Here's an example of something you should write:

We need to show that there's a "zero vector", i.e., a vector $z$ with the property that for all $v \in V$, we have $z + v = v + z = v$.

In our case, such a vector must be a function from $S$ to $\Bbb R$. We define $$z(s) = 0$$ for every $s \in S$.

Now suppose that $f \in V$. Then for any $s \in S$, we have $$(z+f)(s) = z(s) + f(s) = 0 + f(s) = f(s)$$ and similarly, $$(f+z)(s) = f(s)$$ Since functions are equal when their values are equal for every element of the domain, we see that $f+z$ and $z + f$ and $f$ are all equal, as required.

Thus the purported vector space $V$ has an identity for addition, namely, $z$.

Now: Why don't you edit your original question to include a proof of one of the other axioms, like $1\cdot f = f$ for any $f \in V$?

• I ended up showing all the axioms, like additive identity, additive inverse and other axioms like f+(-1)f = 0, etc. Thanks! Jan 19, 2017 at 23:19