The set of weights of a module in the $\mathcal O$ category has a maximal element First, some definitions: 

Given a Kac-Moody algebra $\mathfrak g$, the category $\mathcal O$ of $\mathfrak g$ is the category whose objects are $\mathfrak g$-modules $V$ which are weight modules  ($V = \bigoplus_{\lambda \in \mathfrak h^*} V_\lambda),$ every weight-space $V_\lambda$ is finite dimensional and exists $\{\lambda_1,\cdots,\lambda_s\}\subset \mathfrak h^*$ such that $V_\lambda \neq 0 \implies \lambda \leq \lambda_k$ for some $k \in \{1,\cdots, s\}$.

The question:
As in the first sentence of Proposition 9.3's proof of Kac's book Infinite Dimensional Lie Algebras, I want to justify that the set of weights $P(V) = \{\lambda \in \mathfrak h^*: V_\lambda \neq 0\}$ has a maximal element.
I tried the obvious: use Zorn's Lemma. The set $P(V)$ is by assumption non-void and we may consider $\{\mu_i\}_{i\in I}$ a chain. The first try, of course, is to say that $\sum_{i\in I} \mu_i$ is a bound for this chain, but the problem is: it may be the case that $\sum_{i\in I} \mu_i \not \in P(V)$. The next try then was to work with the set $\{\lambda_1,\cdots, \lambda_s\}$. I tried then to suppose that $\{\lambda_1,\cdots,\lambda_s\}\cap P(V) \neq \emptyset$ and see what happens. Well, the set $\{\lambda_1,\cdots,\lambda_s\}\cap P(V)$ is finite and must have a maximal element $\lambda^*$. But it is not true that any $\mu_i$ in the chain is such that $\mu_i\leq \lambda^* $ and even if $\mu_i \leq \lambda_k$ for some $\lambda_k \in \{\lambda_1,\cdots, \lambda_s\},$ it may not be the case that $\lambda_k\leq \lambda^*$. So i'm stuck here and out of ideas. Any help to solve this problem?
 A: Rather than giving a short answer as to why this is true, I'll give a long answer as to why it should be true intuitively.
Recall that we have the root lattice $Q = \bigoplus_{i \in I} \mathbb{Z} \alpha_i \subseteq \mathfrak{h}^*$, we set $Q^+ = \sum_{i \in I} \mathbb{N} \alpha_i$, and the definition of the partial order is that $\lambda \leq \mu$ if and only if $\mu - \lambda \in Q^+$. This, together with the fact that the $\alpha_i$ are linearly independent in $\mathfrak{h}^*$, means that for any $\lambda$, the set of weights $\mu$ such that $\mu \leq \lambda$ is a discrete set. In particular, we should not expect any kind of limiting behaviour (in the analytic sense). We should also expect there to be a great many $\lambda$ and $\mu$ which are completely unrelated in this partial order.
The point of seeing the discrete nature of that ordering is this: suppose that $\lambda \leq \mu$. Then the set $\{ \nu \in \mathfrak{h}^* \mid \lambda \leq \nu \leq \mu \}$ is finite. This is easy to show: suppose $\mu = \lambda + \sum_{i \in I} n_i \alpha_i$, then the set of possible $\nu$ is just $\lambda + \sum_{i \in I} m_i$ where $0 \leq m_i \leq n_i$. Therefore for any two weights $\lambda, \mu$, the set $\{\nu \mid \lambda \leq \nu \leq \mu\}$ is finite (in many cases, empty).
Now you can use finite induction to prove the claim. Suppose we have $V \in \mathcal{O}$, with weights are bounded from above by some finite collection $\lambda_1, \ldots, \lambda_r$. Choose any weight $\lambda \in P(V)$. If $\lambda$ is not maximal, i.e. there is $\mu \in P(V)$ with $\mu \leq \lambda$, then replace $\lambda$ by $\mu$ and continue. This decreases the size of the set $\{ \mu \in P(V) \mid \mu \geq \lambda\}$, a set which we know to be finite since it is contained inside $\{ \mu \in \mathfrak{h}^* \mid \lambda \leq \mu \leq \lambda_k \text{ for some } k \}$.
