I'm going through my assignments for this week, and I have a problem understanding the (notation of?) this exercise:

Let $S$ be a nonempty set and $F$ a field. Prove that for any $s_0 \in S$, $\{f \in F(S,F):f(s_0)=0\}$, is a subspace of $F(S,F)$. ($F(S,F)$ being the set of all functions).

Now what I'm wondering is what exactly is $s_0$, why the $_0$, and how should I understand the $f(s_0)=0$ part of the subspace?

Thanks for any help :)

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it's actually just one question. The second resolved itself during writing :D – foaly Sep 20 '12 at 15:38
Excellent : ) ${}$ – Rudy the Reindeer Sep 20 '12 at 15:39
It is a convention to give a fixed element a subscript, like $s_0$, to emphasize that it is fixed. In this situation, it would also be common to write $s_1$. But these are conventions, like using $x$ to denote a variable and $a$ to denote a constant. – B R Sep 20 '12 at 16:14

I think they chose the index "$0$" because the functions are zero at $s_0$. In the question, $s_0$ is an element you pick (fix), in $S$.
So if $F(S,F)$ is the space of all functions $S \to F$ then the subspace consists of all functions that are zero at $s_0$. Perhaps one could denote the subspace $F_{s_0}$ to make it clear that it depends on $s_0$.
For example, we could look at the space of all functions $\mathbb R \to \mathbb R$ and consider the subspace of all functions that are zero at $3$.
the zero polynomial? ($a_n = a_{n-1} = ... = a_0 = 0$) – foaly Sep 20 '12 at 16:07
where's the difference? x.x or how would you call it? well, scalar multiplication: where a function is zero, it is still gonna be zero when you multiply it by something ($c\in F$). Not sure how to express that adequately? addition: two functions added to each other will also always result in a function that is zero in the same place, which is $s_0$. Now i really don't know how you would prove / express this other than just with common sense.. ? – foaly Sep 20 '12 at 16:23