Discrete dense subset Let $K$ be a compact Hausdorff space such that the set $D$ of isolated points in $K$ is countable and dense in $K$. Consider the linear subspace $A$ of $C(K)$ consisting of those functions $f\in C(K)$ such that $f$ is constant on $K\setminus D$ and the set $\{x\in D\colon f(x)= 0\}$ is finite or its complement in D is finite.
Is $A$ a closed subspace of $C(K)$?
 A: *

*Try a very simple space for an example.  Can you find one which has only one non-isolated point?

*On this space, find a continuous function $f \notin A$.

*Find a sequence of functions $f_n \in A$ such that $f_n \to f$ uniformly.  This will show that $A$ is not closed.
A: I cannot give a full solution to your problem, but
here is what I think I found out.


*

*Let $y_1, \dots, y_n \in \mathbb{R}$ be any real numbers
then for any $x_1,\dots,x_n \in D$ define
$f(x_i) = y_i$ and $f(x) = 0$ everywhere else.
For every closed $A \subset R$ you have $f^{-1}(A)$
closed in $K$ hence $f$ is continuous and so $f \in A$.

*Any $g \in C(K)$ which is not constant on $K\setminus{D}$ 
lies in the interior of $C(K)\setminus{A}$, as all the $h \in C(K)$
with $\|g-h\|< \frac{1}{2} (\max_{x \in K\setminus{D}}{g(x)} - \min_{x \in K\setminus{D}}{g(x)})$
cannot be constant on $K\setminus{D}$. 

*This leaves us considering $g \in C(K)$ with $g(x) = y_0$ 
for all $x \in K\setminus{D}$ and some $y_0$ and $g(x) = 0$ 
for infinitely many $x \in D$ and $g(x) \neq 0$ for infinitely many $x \in D$. 
Note that for any open neighborhood $U$ of $y_0$ we have 
$g^{-1}(U)$ containig all but finitely many elements of $D$. 
Hence we conclude $y_0 = 0$.
Now for every $\varepsilon > 0$ you can choose $f \in A$
with $f(x) = g(x)$ for all $x \in D$ with $|g(x)| \geq \varepsilon$
and $f(x) = 0$ everywhere else.
It follows that $\|f-g\| < \varepsilon$ and as $\varepsilon >0$ was
arbitrary we conclude $g \in \overline{A}$.

*So all in all $A$ is not closed as long as the second condition is
not superfluous.
