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What is an open set in the metric space $(\mathbb{Z}^n,d)$, where $d$ is the Euclidean distance in $\mathbb{R}$?

As far as I know, in a metric space an open set $O$ is defined as follows: For each point $x\in O$ it exists an $\varepsilon$ such that that $B_\varepsilon(x)\subset O$ is true.

Since in $\mathbb{Z}^n$ I just have some sort of layer I thought of defining the $\varepsilon$ as follows:

$$ \varepsilon=\sqrt{\sum_{i=1}^{d}x_{i}^{2}} $$

This way I can characterize all the $\varepsilon$-Balls $B_\varepsilon(x)$ for each $x\in\mathbb{Z}^n$. So if $d(x,y)<\varepsilon$ for each $y\in B_\varepsilon(x)$ the set is open. And the same for $d(x,y)\le\varepsilon$ implies a closed set. For compactness I could say that, like in $\mathbb{R}^n$, each closed set which is bounded by a constant (using the euclidean metric here) is compact.

What about the single points in $\mathbb{Z}^n$? Are they open sets, too?

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Single points are open. Everything is open, everything is closed. –  André Nicolas Apr 16 '12 at 16:23

1 Answer 1

up vote 4 down vote accepted

With the euclidean distance, $\mathbb{Z}^n$ is a discrete space. Every subset is open. Every subset is closed. Only the finite subsets are compact.

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