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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?

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    $\begingroup$ If $X$ is a Cantor space (or any locally compact space with a non-isolated point $x_0$ with a basis $(K_i)$ of clopen neighborhoods) and $Y=\{0,1\}$, the sequence (or net) $(1_{K_i})$ of continuous functions tends to the non-continuous function $1_{\{x_0\}}$ in the compact-open topology. $\endgroup$ Commented Jan 7, 2020 at 15:55
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    $\begingroup$ Sequences of characeristic functions seldom converge uniformly $\endgroup$ Commented Jan 8, 2020 at 7:44
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    $\begingroup$ @YCor Your example seems to me correct. Could you please post it as a (more detailed) answer? $\endgroup$ Commented Jan 8, 2020 at 14:37
  • $\begingroup$ @HennoBrandsma yes $X$ is compact, so what? you might have in mind that for locally compact $X,Y$ the compact-open topology is that of uniform convergence on compact subsets, but this is within continuous functions. $\endgroup$ Commented Jan 8, 2020 at 14:39
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    $\begingroup$ @HennoBrandsma I'm afraid I gave a counterexample to this statement. $\endgroup$ Commented Jan 8, 2020 at 16:41

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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.

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    $\begingroup$ The example of Ycor seems to contradict continuity of the limit function on Compacta $\endgroup$ Commented Jan 7, 2020 at 21:00
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    $\begingroup$ It is a very elementary result of topology that a uniform limit of a sequence of continuous functions is continuous (usualy proved in a beginner‘s analysis course for functions on a subset of the reals). $\endgroup$ Commented Jan 7, 2020 at 22:34
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    $\begingroup$ @HennoBrandsma I checked Kelley book. Theorem 11 at page 230 explicity state the result that compact-open coincides with uniform convergence on compacta, only for continuous functions. $\endgroup$ Commented Jan 8, 2020 at 10:32
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    $\begingroup$ @HennoBrandsma I also checked Engelking book, and I did not found any statement that compact-open convergence imly uniform convergence on compacta, without assuming that the limit function is continuous $\endgroup$ Commented Jan 8, 2020 at 13:48
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    $\begingroup$ As already mentioned, it's false: compact-open convergence to a non-continuous function doesn't imply uniform convergence on compact subsets. Always better to have the proof in mind (or reprove it) to figure out that it uses continuity of the limit. $\endgroup$ Commented Jan 8, 2020 at 14:44
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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. $f_n\to f$.

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    $\begingroup$ Indeed for functions $[0,1]\to\mathbf{R}$, it seems that a compact-open limit of continuous functions maps intervals to intervals (hence it can't be a non-continuous piecewise continuous function, for instance). It's nice you found an example in this setting. $\endgroup$ Commented Jan 8, 2020 at 14:45

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