Proving that the set $\{f \in \mathcal{F}(A,\mathbb{F}): f(a_0) = 0\}$ contains the zero vector. Suppose that $A$ is a nonempty set and $\mathbb{F}$ is a field. I would like to show that for any $a_0 \in A$, the set $\{f \in \mathcal{F}(A,\mathbb{F}): f(a_0) = 0\}$ is a subspace of $\mathcal{F}(A,\mathbb{F})$, where $\mathcal{F}(A,\mathbb{F})$ is defined to be the set of all functions from $A$ to $\mathbb{F}$. Furthermore, two functions $f$ and $g$ in $\mathcal{F}(A,\mathbb{F})$ are equal if $f(a) = g(a)$ for every $a \in A$. 
The closed under addition and multiplication part for showing it is a subspace is trivial for me. However, it is actually the zero vector condition that is confusing me. I understand that since we have that the set contains all functions where $f(a_0) = 0$, then it appears trivial that the condition is satisfied. However, I was thinking what if the set $A$ consisted of just one element and that the field were $\mathbb{R}$. In that case, there would be only one function, and so would it necessarily be the case that there is a function that maps to zero?
In other words, how do I know that $\{f \in \mathcal{F}(A,\mathbb{F}): f(a_0) = 0\}$ is non-empty? Thanks!

The set $\{f \in \mathcal{F}(A,\mathbb{F}): f(a_0) = 0\}$ is non-empty IFF set $A$ is not empty. In this case the function $f\equiv0$ is always exist and is contained in aforementioned subspace. Thus the subset is (mnemonically trivial) nonempty.
Don't worry :)
