If $X$ is a second countable space and $A$ is a closed subset of $X$, then there exists a perfect set $P$ and a countable set $C$ such that $A=C\cup P$.

I defined $P=\{x\in X: V\cap A \text{ uncountable for all neighborhood V of }x\}$. I could show that $P\subset A$ and closed. To prove $P$ has no isolated point, I argued by contradiction.

Suppose $x$ is the isolated point of $P$, thus $U$ is the neighborhood of $x$ such that $U\cap P=\{x\}$. By definition of $P$, $U\cap A$ is an uncountable set, so is $U\cap A-\{x\}$. Every uncountable set in a second countable space has a limit point. Thus $p$ is the limit point of $U\cap A-\{x\}$. Now if I could show that $p\in P$ I will be done, but even though I have tried various things I couldn't show this. Any help is greatly appreciated!


For any topological space $(X, \tau)$, any point in a closed set is either a limit point or an isolated point.

Since we need to just show that there exists some $P$, just define $P$ to be the set of all limit points of $A$.

Take any point not in $P$. Then either it is in $X - A$ which is open, so an open set not intersecting $P$ around it can be drawn, or it is in $C$, which means an open set around it exists not intersecting $P$ by definition of isolated. Thus, $P$ is closed.

Now, we need to show that the set $C$ of all isolated points of $A$ is countable. For each $c \in C$, let $U_c$ be an open set such that $U_c \cap A = \{c\}$. It follows that $U_c \cap C = \{c\}$.

Associate these through the function $f: C \to \tau $, such that $f(c) = U_c$. Note that $f$ is injective, since $f(x) = f(y) \implies U_x = U_y \implies U_x \cap C = U_y \cap C = x = y$.

By second-countability, there is a countable base $\mathcal B$. There exists some $B_c \in \mathcal{B}$ such that $c \in B_c \subseteq U_c$.

Associate these through the function $g : f(C) \to \mathcal{B}$, such that $g(f(c)) = B_c$. First, note that $c \in g(f(c))$. Then $g$ is also injective, since $g(f(x)) = g(f(y)) \implies x, y \in f(x) \cap f(y)$, and $f(x) \cap f(y)$ is either $\emptyset$ or $f(x) = f(y)$. Only the second option is possible.

Therefore, $g \circ f$ is an injective function from $C \to \mathcal{B}$, and $\mathcal{B}$ is countable, so $C$ is countable.

  • $\begingroup$ I do not see why the sett of all limit points of A is perfect. In fact the part I am stuck is that part. $\endgroup$ – user64066 Jun 21 '18 at 23:35
  • $\begingroup$ By definition it is closed and has no isolated points. $\endgroup$ – Jeffery Opoku-Mensah Jun 21 '18 at 23:49
  • $\begingroup$ I defined it to have no isolated points. I elaborated on why its closed $\endgroup$ – Jeffery Opoku-Mensah Jun 22 '18 at 0:10

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