I am trying to solve the following exercise:
Let $(\mathbb Q, \tau_1)$ be a subtopology of the euclidean topology $\mathbb R$. Is the topological space $(\mathbb Q, \tau_1)$ connected?
I answered affirmative and made a proof for this, however, I searched this online and I read that the topological space $(\mathbb Q, \tau_1)$ is not connected, meaning that my proof is incorrect.
My proof
A topological space $(X,\tau)$ is connected if the only clopened subsets are $X$ and $\emptyset$.
So, let $A \in \tau_1$ such that $A \neq \mathbb Q$ and $A \neq \emptyset$.
Let $\mathcal{B}_1 = \{(a,b) \cap \mathbb Q: (a,b) \subset \mathbb R\}$, with $a < b$, be a basis for $\tau_1$.
Then we have that
$$\begin{align}
&A = \bigcup \limits _i (a_i ,b_i)\cap \mathbb{Q}=
\\
\\
&= \mathbb Q \cap \bigcup_i (a_i,b_i)
\end{align}
$$
Let's now consider the set $\mathbb Q \setminus A$:
$$\begin{align}
&\mathbb Q \setminus A =\mathbb Q \setminus \mathbb Q \cup \mathbb Q \setminus \bigcup_i (a_i,b_i)
\\
\\
&= \bigcap _i \mathbb Q \setminus (a_i,b_i)
\end{align}
$$
We have that $\mathbb Q \setminus (a_i,b_i)= ((-\infty, a_i] \cup [b_i, \infty)) \cap \mathbb Q$. So we conclude that:
$$\mathbb Q \setminus A = \mathbb Q \cap \bigcap _i ((-\infty, a_i] \cup [b_i, \infty))$$.
We have that $\bigcap _i ((-\infty, a_i] \cup [b_i, \infty))$ is never an open set of $\mathbb R$, so we conclude that $\mathbb Q \cap \bigcap _i ((-\infty, a_i] \cup [b_i, \infty)) \notin \tau_1 \to \mathbb Q \setminus A \notin \tau_1$, so $A$ is not clopen, proving that $(\mathbb Q, \tau_1)$ is a connected set.
What did I do wrong in my proof?
Best Answer
You assume that from the fact that $(-\infty, a] \cup [b, \infty)$ is not open in $\Bbb R$, that it follows that its intersection with $\Bbb Q$ cannot be open. Well, if you take $a=\sqrt{2}, b= \sqrt{5}$, say, then in fact
$$\left((-\infty, \sqrt{2}] \cup [\sqrt{5}, \infty)\right) \cap \Bbb Q = \left((-\infty, \sqrt{2}) \cup (\sqrt{5}, \infty)\right) \cap \Bbb Q$$
as the "boundary points" are not rational, and so the set does have an open intersection with $\Bbb Q$ after all.