# Proving that $\mathbb{R}_K$ is connected with the topology $\tau$

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!

$$A = \{x \in \Bbb R: x>0\}$$ 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.
Suppose that $$A$$ and $$B$$ partition $$\mathbb{R}$$ and are clopen disjoint. Suppose that there is an integer $$n$$ such that the interval $$(1/(n+1),1/n)$$ contains points from both $$A$$ and $$B$$: this is a contradiction, since the intervals of that form are clearly connected (the induced topology on them is the standard one). Now suppose that there is $$n$$ such that $$(1/(n+2),1/(n+1))\subset A$$ and $$(1/(n+1),1/n)\subset B$$. Again, this is a contradiction, since the point $$1/(n+1)$$ could not belong to either $$A$$ or $$B$$ (everyone of its basic open neighbourhoods intersects both $$A$$ and $$B$$). So all of the intervals $$(1/(n+1),1/n)$$ belong to, say, A, and so all of the points $$1/n$$. But then, $$0$$ belongs to $$A$$ as well, since every open neighborhoods intersects $$A$$.
This means that the whole interval $$[0,1)$$ is in $$A$$, so we can restrict our attention to $$\mathbb{R}\setminus [0,1)$$. But since on this set the topology is the usual one, we have a contradiction.
Hint: $$(0,+\infty)$$ and $$(-\infty,0)$$ are connected in $$\mathbb{R}_K$$. Consider the connected component which contains $$0$$.