What are the connected open subsets in Smirnov's deleted sequence topology? Let $X$ be set of real numbers with Smirnov's deleted sequence topology (sometimes also called the $K$-topology).
Can anybody help me find precisely what are the connected open subsets of $X$?

Description of the topology on  $X$
Let $A = \{ \frac 1n : n \in \mathbb N \}$. Then the open subsets of $X$ are exactly those of the form $U \setminus B$ where $U$ is an open subset of $\mathbb R$ with the usual (metric) topology, and $B \subseteq A$.
 A: We will first show the following.
Fact. $A \subseteq \mathbb{R}$ is connected in $\mathbb R$ (with the usual topology) iff it is connected in $X$.
Proof. 
(⇒) Suppose that $C \subseteq \mathbb R$ is connected (with respect to the usual topology). If $C = \{ x \}$ then $C$ is clearly connected in $X$, so assume that $C$ contains at least two points. Then $C$ is a line, line segment, or ray (i.e., is of one of the forms $(a,b)$, $(a,b]$, $[a,b)$, $[a,b]$, $(-\infty,b)$, $(-\infty,b]$, $(a,+\infty)$, $[a,+\infty)$, $(-\infty,+\infty)$). Note that for each $x \in C$ there is a $\varepsilon > 0$ such that either $[x,x+\varepsilon) \subseteq C$ or $(x-\varepsilon , x] \subseteq C$, or both.
Suppose that $U,V$ are open subsets of $X$ such that


*

*$U \cup V \supseteq C$, and

*$U \cap C \neq \emptyset \neq V \cap C$.


We will show that $U \cap V \cap C \neq \emptyset$.
Write $U = U_0 \setminus B_U$ and $V = V_0 \setminus B_V$ where $U_0 , V_0$ are open in $\mathbb R$, and $B_U, B_V \subseteq A$. Note that


*

*$U_0 \cup V_0 \supseteq U \cup V \supseteq C$, and

*$U_0 \cap C \supseteq U \cap C \neq \emptyset$ and $V_0 \cap C \supseteq V \cap V \neq \emptyset$.


Since $C$ is connected in $\mathbb R$, it follows that $U_0 \cap V_0 \cap C \neq \emptyset$. Picking $x \in U_0 \cap V_0 \cap C$ (which must belong to $A$), by the fact that $U_0, V_0$ are open in $\mathbb R$, and the property of $C$ desribed above, there is an $\varepsilon > 0$ such that $( x - \varepsilon , x + \varepsilon ) \subseteq U_0 \cap V_0$ and (without loss of generality) $[ x , x + \varepsilon ) \subseteq C$. Since $[ x , x + \varepsilon ) \setminus A \neq \emptyset$, it follows that $U \cap V \cap C \supseteq ( U_0 \cap V_0 \cap C ) \setminus A \supseteq [ x , x+\varepsilon) \neq \emptyset$.
Therefore $C$ is connected in $X$.
(⇐) If $C \subseteq \mathbb R$ is not connected (in $\mathbb R$), then clearly $C$ cannot be connected in $X$. (Just use the same witnesses.)

Therefore the open connected subsets of $X$ are precisely the connected subsets of $\mathbb R$ which are open in $X$. Since every open set of $X$ which is not open in $\mathbb R$ is not connected in $\mathbb R$ (since the are "missing" at least one of the points from $A$), it follows that the open connected subsets of $X$ are the open connected subsets of $\mathbb R$. That is, they are of the form $(a,b)$, $(-\infty , b )$, $(a,+\infty)$ or $(-\infty , + \infty )$ for some $a < b \in \mathbb R$.
