Poles of a sum of functions The other question is here.

Let $F$ be a function field in one variable over a field $k$. Let $S$ a nonempty finite subset of the set of all places of $F$. Prove that if $P \in S$, there is an element $f$ of $F$ such that $P$ is the only pole of $f$.

For $P \in S$, let $f_P$ be an element of $F$ such that $P$ is the only pole of $f_P$. Let $$f:= \sum_{P \in S} f_P.$$How do I prove that $S$ coincides with the set of all poles of $f$?
 A: For every place $Q$, the function $\text{ord}_Q$ is a discrete valuation of $F$, and satisfies the strong triangle inequality
$$\text{ord}_Q(h+g) \geq \min \{\text{ord}_Q(h), \text{ord}_Q(g)\}.$$
The key point is that :

When $\text{ord}_Q(h) \neq \text{ord}_Q(g)$, the triangle inequality
  is an equality:
$$\text{ord}_Q(h+g) = \min \{\text{ord}_Q(h), \text{ord}_Q(g)\}.$$

It is easy to see from this, by induction, that if $g_1, \dots, g_n \in F$, and $\text{ord}_Q(g_1) < \text{ord}_Q(g_i)$ for $i>1$, then
$$\text{ord}_Q(\Sigma h_i) = \text{ord}_Q(h_1).$$
Now let us look at your $f = \sum_{P \in S} f_P$. Choose a place $Q$ of $F$. If $Q \notin S$, then $\text{ord}_Q(f_P) \geq 0$ for all $P \in S$ because $P$ is the only pole of $f_P$. In that case, 
$$\text{ord}_Q(f) \geq \min \{\text{ord}_Q(f_P) : P \in S\} \geq 0$$
so $Q$ is not a pole of $f$. If $Q=P\in S$, then $\text{ord}_P(f_P) < 0$ and $\text{ord}_P(f_{P'})  = 0$ for all other $P'\in S$, by definition of the $f_P$'s. Therefore, the discussion above shows that
$$\text{ord}_P(f) = \text{ord}_P(f_P) < 0,$$
so that $P$ is really a pole of $f$, with the same order as the pole of $f_P$. 
