# Thread: any open set of Q?

1. ## any open set of Q?

Is there any one point set of Q which is open?

2. Originally Posted by szpengchao
Is there any one point set of Q which is open?
A set $S$ is open in $\mathbb{Q}$ iff for any $a\in S$ there is $\epsilon > 0$ such that $B(a,\epsilon) \subseteq S$.

Where $B(a,\epsilon) = \{ x\in \mathbb{Q} : |x - a| < \epsilon \}$.

If $S = \{ a\}$ is a single point set then for this set to be open we require that there is $\epsilon > 0$ so that $B(a,\epsilon) \subseteq S$. But this is impossible since for any $\epsilon > 0$ we can find other rational points in $|x-a|<\epsilon$.

3. Originally Posted by szpengchao
Is there any one point set of Q which is open?
You have asked several questions on this or simpler concepts.
But there is really no way to answer unless you specify the topology.
Even if you say “with the usual topology”, it is not clear if you mean the set is to be viewed as a subset of some other space or as a space in and of itself.

Suppose we specify the superset is the set of real numbers with the usual topology.
Now, the subset of rational numbers, $\mathbb{Q}$, we can ask if $\mathbb{Q}$ is an open subset of $\mathbb{R}$.
Here is a hint. Every point of an open set is an interior point of the set. But every point of $\mathbb{Q}$ is a boundary point.

4. Originally Posted by szpengchao
Is there any one point set of Q which is open?
Under the discrete topology on $\mathbb{Q}$ every subset of $\mathbb{Q}$ is open, so in particular $\forall x \in \mathbb{Q};\ \{x\}$ is open under the discrete topology.

RonL