Is singleton set open in $\Bbb Q$ with subspace topology and what are the connected sets of this topology? I think {$x$} is not open in $\Bbb Q$ as if I take any open set $G$ in $\Bbb R$, then $G$ intersection $\Bbb Q$ is never the singleton set {$x$} as {$x$} is not open in $\Bbb R$ and any open set in $\Bbb R$ has infinitely many rationals in it.
Am I correct?
One more question what are the connected sets in this subspace topology on $\Bbb Q$ ?
 A: Your reasoning is correct.
$\{x\}$ is not open in $\Bbb{Q}$ .
Now I claim only connected subsets of $\Bbb{Q}$ are singleton subsets.
Suppose, $A\subset \Bbb{Q}$ is connected.
Let, $x\neq  y \in \Bbb{Q} $
Then $\exists r\in \Bbb{R} \setminus \Bbb{ Q} $ such that $x<r<y$
And $A = A\cap (-\infty , r) \cup A\cap (r, \infty) $
Implies $A$ is not connected.
Hence, all connected components are one point sets.
A: To see that the singletons are not open, note that for each $x\in\mathbb{Q}$, there is a sequence $q_n\in\mathbb{Q}\setminus\{x\}$ with $q_n\to x$, so that $x$ is contained in the closure of $\mathbb{Q}\setminus\{x\}$. Since $\mathbb{Q}\setminus\{x\}$ is not equal to its closure, it is not closed and therefore its complement $\{x\}$ is not open.
$\mathbb{Q}$ is totally disconnected, which means that its only connected subsets are the singletons. To see this, let $A\subseteq\mathbb{Q}$ be connected and suppose that $A$ is not a singleton. Then there exists $a,b\in A$ with $a\neq b$. Without loss of generality, assume that $a<b$.
Since $A$ is connected and contained in $\mathbb{R}$, it is path-connected, so there exists continuous $p:[0,1]\to A$ with $p(0)=a$ and $p(1)=b$. Choose $v\in(a,b)\setminus\mathbb{Q}$. It then follows from the Intermediate Value Theorem that there exists $c\in[0,1]$ with $p(c)=v\notin\mathbb{Q}$, a contradiction. So the connected components of $\mathbb{Q}$ are the singletons.
