compact open topology definition I am reading Brendon geometry/topology. I am trying to understand the definition of compact open topology.
Suppose X is locally compact Hausdorff and Y is any Hausdorff space.
We denote $Y^X$ as the set of continous maps $X \rightarrow Y$.
Definition:
The compact-open topology on $Y^X$ is the topology generated by the sets
$M(K,U) = \{f \in Y^X : f(K) \subset U\}$, where $K \subset X$ is compact and $U \subset Y$ is open.
My questions:
First of all why this form a subbases ? Why is the union across the collection equal to $Y^X$ ? 
My thinking:
Since f is continous so given any compact set we have $f(K)$ is compact, but does it mean that $f(K)$ must lie in some open space U ?
Second question:
Why do we need the Hausdorff conditions at all ?
 A: A subbase can be any collection of sets, the fact that their union should equal $Y^X$ is a Munkres "fiction" (I consider the empty intersection to be the whole space, so the finite intersections always form a base). But it's clear as for any $x \in X$ : $ M(\{x\}, Y) = Y^X$, as the condition is only that $f(x) \in Y$ which holds for any $f:X \rightarrow Y$, continuous or not.
The Hausdorffness is not "needed" to make it a topology, but to make it a "nice" space. Suppose that $f \neq g \in Y^X$, then $f(p) \neq g(p)$ for some $p \in X$.
If now $U, V $open in $Y$ with $f(p) \in U, g(p) \in V, U \cap V = \emptyset$, by Hausdorffness of $Y$, then $f \in M(\{p\}, U), g \in M(\{p\}, V), M(\{p\}, U) \cap M(\{p\}, V) = \emptyset$, so $Y^X$ is then Hausdorff. On $X$ he demands locally compact Hausdorff, because then he has "lots of" compact sets (every point has a neighbourhood base of them) to make this topology nicer as well. 
A: For the second question, consider $f:X\rightarrow Y$, $x\in X$ and $U$ an open subset which contains $f(x)$, $f\in M(x,U)$.
