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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 $C(K)$?

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Is this homework? – André Caldas Nov 23 '11 at 20:01
first, do you have an example of compact space whith dense isolated points that has non-isolated points ? – Glougloubarbaki Nov 23 '11 at 20:07
What is the topology of $C(K)$? – user17090 Nov 23 '11 at 20:47
$C(K)$ is regarded as a Banach space with the supremum norm. – BartoszTryk Nov 23 '11 at 20:48
Typically, a ball $\{g\in C(K): \sup_{x\in K}|f(x)-g(x)|<\varepsilon\}$. – BartoszTryk Nov 23 '11 at 21:13
  1. Try a very simple space for an example. Can you find one which has only one non-isolated point?

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

  3. Find a sequence of functions $f_n \in A$ such that $f_n \to f$ uniformly. This will show that $A$ is not closed.

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I cannot give a full solution to your problem, but here is what I think I found out.

  1. 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 $f \in A$.

  2. 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 $K\setminus{D}$.

  3. 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 $g \in \overline{A}$.

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

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