Is the rational sequence topology extremely disconnected? Is it true that the rational sequence topology is extremely disconnected?
A space $X$ is said to be extremely disconnected if it is $T_2$ and the closure of any open set is open.
 A: It's not extremely disconnected. Take $x \in \mathbb{R} \setminus \mathbb{Q}$ and let $(x_n)_{n=1}^\infty$ be rational sequence associated with $x$, for clarity of proof we may assume that $(x_n)$ does not contain repetitions then $X_0 = \{x_i : 2 | i \}$, $X_1 = \{x_i : 2 \not | i \}$
(on other hand we may do something a little messy like:
$X = \{x_i : i \in \mathbb{N}\}$. Let $X_{0,1} = \{x_i : x_i \in X, d_e(x_i,x) \text{ is maximal}\}$ (supremum is achieved in relation to convergance of sequence), $X_{1,1} = \{x_i : x_i \in X \setminus X_{0,1}, d_e(x_i,x) \text{ is maximal}\}$, $X_{0,2} = \{x_i : x_i \in X \setminus \left(X_{0,1} \cup X_{1,1}\right), d_e(x_i,x) \text{ is maximal}\}$ and so on... then $X_0 = \bigcup_{i=1}^{\infty}X_{0,i}$, $X_1 = \bigcup_{i=1}^{\infty}X_{1,i}$).
So $X_0, X_1$ are pairwise disjoint open sets such that $x \in \mathrm{cl}X_0, \mathrm{cl}X_1$, contradicition.
A: Seems to me that it is. Proof that "any closer of set A is open" goes that way:
Take any $x \in cl(A)\setminus A$, we aim to proof that there exists open set $B$ such that $x \in B \subset A$:


*

*If $x$ is rational then put $B=\{x\}$

*If $x$ is irrational then we take a look on the definition of closer: it mean that any base set have common point with A. From that it is easy to construct sequence $x_{k}$ of points of $A$ that converge to $x$. Put $B=\Cup_{k\in N}\{x_{k}\} \cup\{x\}   $

