Compact-open limit of continuous functions is continuous? Let $X$ be a topological space and $Y$ a metric space. 
A classical result states that compact-open topology on the space $C(X,Y)$ of continuous functions is the same as the topology of uniform convergence on compact sub-sets.
In general one may define compact-open topology on the whole space $Y^X$ of all functions from $X$ to $Y$. 
Is it true that if $f_n\to f$ with respect to the compact-open topology, and all $f_n$ are continuous, then the limit $f$ is continuous? 
I never seen any theorem stated that way, and I suspect that it may be false.
If so, is there some condition to put on $X$ so to make the above statement true?
 A: All you can say about the limit function is that its restrictions to compacta are continuous. In general, this does not imply global continuity.
In fact spaces which DO have the required property have been much studied. They are called $k$– spaces (sometimes Kelley spaces).  The defining property is that a subset is open if its intersection with each compact subset is open in the corresponding induced topology.  You can find a great deal of information on such spaces in standard textbooks on general topology (I would recommend Engelking‘s monograph with precisely this title).  Examples are metric or locally compact spaces.
A: It seems to me that Ycor example is correct, but I propose another counterexample:
$X=[0,1]$, $Y=[-1,1]$, $f=\sin(1/x)$ for $x\neq 0$ and $f(0)=0$.
$f_n=\sin(1/x)$ for $x>1/2\pi n$ and $f_n(x)=0$ for $x\in[0,1/2\pi n]$
The functions $f_n$ are continuous, $f$ is not, but $f_n\to f$ with respect to the compact open topology.
Proof.
Given a compact $K$ and an open $U$ let $V(K,U)=\{g:g(K)\subseteq U\}$. We prove that for any $V(K,U)$ if it containis $f$, then it contains also $f_n$ for $n$ big enough. 
There are two cases: 
1) $0\notin K$. In this case $f_n$ and $f$ coincide on $K$ for $n$ big enough.
2) $0\in K$. In this case, if $f\in V(K,U)$ then $0\in U$. Since $f_n(x)=0$ for $x\in[0,1/2\pi n]$ and $f_n=f$ elsewhere, if follows that $f_n\in V(K,U)$ (for any $n$). 
So, for any open set $A$ in the compact-open topology containing $f$, then $A$ must eventually contain $f_n$. I.e. $f_n\to f$.
