Let $\mathbb{R}_K$ be the topological space $(\mathbb{R}, \tau)$ where $\tau$ is the topology generated by the family of open sets $$\{(a,b), \, (a,b) \setminus K \mid a \leq b \in \mathbb{R} \}$$ where $$K = \{\frac{1}{n} \mid n \in \mathbb{N} \}$$
Prove that the usual topology of $\mathbb{R}$ is strictly contained in $\tau$.
Prove that $\mathbb{R}_K$ is connected but not arc connected.
$\textbf{My attempt}$
Clearly the euclidean topology is contained strictly in $\tau$ because one can consider the open set $(-1,1) \setminus K$ in the topology $\tau$ as a neighborhood of $0$, and there is no usual open set $A$ such that $0 \in A$ and $A \subset (-1,1) \setminus K$
It's not arc connected , in fact consider any continuous map $\gamma : [0,1] = I \to R_K$ joining $0$ and $1$. $A := \gamma^{-1}((-1,1) \setminus K)$ must be open in $I$. So there must be an interval like $[0,\varepsilon) \subset A$ with $\gamma(\varepsilon) >0$. But this can't be true, since if it was, we would have that $\gamma([0,\varepsilon)) \subset (-1,1) \setminus K $. Now since $\gamma$ is also continuous in the usual topology, intervals are mapped into intervals, so all the points $\frac{1}{n}$ for $n$ big should be in $\gamma([0,\epsilon))$ and this is not possible.
Now I can't prove that it has to be connected.
Thanks in advance for anyone who wants to give me some help!
Best Answer
$A = \{x \in \Bbb R: x>0\}$ is connected as it has the same topology as a subspace of $\Bbb R$ and of $\Bbb R_K$, and the same holds for $B=\{x \in \Bbb R: x < 0\}$.
Now, taking the closures in $\Bbb R_K$, it's easy to see that $\overline{A}=[0,+\infty)$ and $\overline{B}=(-\infty,0]$ and as closures of connected sets, they are connected too in $\Bbb R_K$. They intersect in $0$ so their union, i.e. $\Bbb R_K$ is connected as well.