Is $\mathbb Q \subset \mathbb R$ the subspace topology, discrete?, and questions about showing discreteness I'm looking for subspace topologies of $\mathbb R$ that are discrete. 
In order to show that a subspace $U$ of $\mathbb R$ is discrete, I'm trying to show that the singleton set in $U$ is open. Since arbitrary unions of open sets are open, this way I could generate all of the subsets of $U$, hence discrete.
What do you think about this method? Does it sound right?
It can be shown easily that each singleton set in $\mathbb Z$ is open, so $\mathbb Z$ is discrete. I have no problem with that, however I'm looking for a subspace of $\mathbb R$ other than $\mathbb Z$ that is discrete. 
$\mathbb Q$ came into my mind, but couldn't show that it is discrete preciously. Every rationals are surrounded by irrational numbers, can I find an open interval $I \subset \mathbb R$ such that $I \cap \mathbb Q$ is equal to a singleton set in $\mathbb Q$?
What do you think about my method of showing discreteness? Can an $I$ be found? What else could be a discrete subspace of $\mathbb R$? 
Are my questions. 
Thanks. 
 A: $\mathbb{Q}$ is definitely not discrete in $\mathbb{R}$. Suppose that we have some open set $O$ of the reals such that $O \cap \mathbb{Q} = \{q\}$ for some $q \in \mathbb{Q}$ (which should exist if $\{q\}$ were open). Then as $q \in O$ we have some open interval $(q-t, q+t) \subseteq O$. Between any two distinct reals lies a rational number (and an irrational number too), so we have some $q' \in \mathbb{Q}$ with $q < q' < q+t$. But then this $q' \in O \cap \mathbb{Q}$ too, so this set cannot be equal to just $\{q\}$, contradiction. No singleton is thus open in the subspace topology on $\mathbb{Q}$, every point of $\mathbb{Q}$ is a limit point of $\mathbb{Q}$. The same holds for the irrationals $\mathbb{P}$ in the subspace topology.
Besides the classical $\mathbb{N}$ and $\mathbb{Z}$ which are discrete we have $A = \{\frac{1}{n}: n=1,2,\ldots \}$ as a discrete subspace (although $0$ is a limit point of $A$ so $A$ is not closed, no point of $A$ itself is a limit point of $A$). In any metric space all finite subspaces are discrete.
A: You can show a (metric) space is not discrete by producing a sequence that converges to a point that doesn't appear in the sequence.
This relies on the fact that a sequence in a discrete space is convergent if and only if it is eventually constant.
The example for $\mathbb{Q}$ is the sequence $1/n$.
A: It is not difficult to show that any non-empty interval in the real numbers contains a rational, and therefore infinitely many (you can divide an interval into any number of sub-intervals, and each contains at least one rational number). So the inherited topology on $\Bbb Q$ is not discrete. That being said, any interval of rational numbers that do not have (rational) end points are both open and closed ("clopen"), so it is a very disconnected topology.
