Can you define a metric on $(-\infty,0)\cup\mathbb Z^+$?

Question

Can you define a metric on $X=(-\infty,0)\cup\mathbb Z^+$ (positive integers)? Would every element of $\mathbb Z^+$ be an isolated point of $X$?

Answer 1

Whether $X$ admits a metric and whether $x\in X$ is isolated depends solely on the topology of $X$:

• $X$ with the discrete topology admits a metric and every $x\in X$ is isolated.
• $X$ with the standard relative topology admits a metric and every $x\in\mathbb Z^+$ is isolated.
• $X$ with the trivial topology does not admit a metric.

Answer 2

I'll interpret $\mathbb{Z}^+$ as the set $\{1,2,3,\dots\}$. Now, as $X$ is a subset of $\mathbb{R}$, it of course inherits a metric from $\mathbb{R}$. That is, $X$ is a metric space when given the metric $$ d(x,y) := |x-y|, $$ and I leave it to you to check that $d$ indeed defines a metric on $X$. With respect to this metric, every $x \in \mathbb{Z}^+$ would be an isolated point of $X$, and every $x \in (-\infty, 0)$ would not be isolated:

• If $x \in (-\infty,0)$, then any open ball $B(x,\epsilon)$ will contain other points of $(-\infty,0) \subseteq X$ (convince yourself of this)
• If $x \in \mathbb{Z}^+$, then the ball $B(x,1/2) = \{x\}$ cannot contain any other points in $X$.

However, whether or not points in $X$ are isolated depends heavily on the metric we are giving it. The metric described above is the "natural one" inherited from the parent space $\mathbb{R} \supset X$. Were we to instead give $X$ the discrete metric $$ d(x,y) := \begin{cases} 1 & \text{if } x \neq y,\\ 0 & \text{if } x =y, \end{cases} $$ then every point in $X$ would be isolated as $B(x,1)$ would simply be equal to $\{x\}$ for every $x \in X$.

Answer 3

There is a bijection $f$ from your space to $\Bbb R$. Define a metric $d$ on $X$ by $$ d(x,y)=|f(x)-f(y)| $$ and $f$ becomes an isometry. This makes your space essentially equal to $\Bbb R$ with the standard metric, only with different names for all the points. In particular, it will have no isolated points.