# Show that every point of $\mathbb{Z}$ is isolated, no point of $\mathbb{Q}$ is isolated

somebody to help me, consider $\mathbb{Z}$ and $\mathbb{Q}$ as subspaces of $\mathbb{R}$ with the usual topology. The question is to show that every point of $\mathbb{Z}$ is isolated,and to show that no point of $\mathbb{Q}$ is isolated.

The definition for isolated point ; a point $x \in X$ is said to be isolated point of if the one -point set $\{x\}$ is open in $X$

THANKS!

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Well, what makes a set open in the subspace topology? Can you show that a single point in the integers is open? What about the rationals? What have you tried so far, and where are you stuck? – MartianInvader May 31 '11 at 21:05

Given $n \in \mathbb{Z}$, we have: $n = n \cap (n-1 , n+1)$ wich is open in the subspace topology by definition.Given $x \in \mathbb{Q}$ ,no interval on the line can contain only one rational x, because $\mathbb{Q}$ is dense in $\mathbb{R}$
For $\mathbb{Z}$ you can just use the definition of the subspace topology by picking a small enough open set of $\mathbb{R}$ and intersecting it with $\mathbb{Z}$ to get the singleton set $\{ n \}$ for an integer $n$ to see that it is open.
For example for $1$, the interval $(0.5, 1.5)$ is open in $\mathbb{R}$ and $(0.5, 1.5) \cap \mathbb{Z} = \{ 1 \}$ so $\{ 1 \}$ is open in $\mathbb{Z}$.
For the rationals you would have to use the fact that they are dense in the reals, and this would cause the trouble when intersecting open sets in $\mathbb{R}$ with $\mathbb{Q}$.