I'm working my way through the examples in Rudin on open sets, closed sets and other related material to get use to working with the definitions.
I'm trying to show that $\mathbb{Z} \subseteq \mathbb{R}^{2}$ is closed, but not open and not bounded. My attempts are below.
Closed:
We want to show that $\mathbb{Z} \subseteq \mathbb{R}^{2}$ contains all of its limit points. Suppose that $p =(p_{1}, p_{2})$ is a limit point of $\mathbb{Z}$ and that $\epsilon = \frac{1}{2}$ (say). Then $B_{\epsilon}(p) = \{(a,b): d((a,b), (p_{1},p_{2})) < \epsilon\}.$
For our given choice of $\epsilon, B_{\epsilon}(p) = p.$ Which contradicts the fact that $p$ is a limit point and therefore $\mathbb{Z}$ doesn't have any limit points. Hence it is closed (it contains all of its limit points, because it doesn't have any!).
NOT open:
Consider the point $(0,0)$, and let $\epsilon = \frac{1}{2}.$ Then $B_{\epsilon}((0,0))$ doesn't contain any other points of $\mathbb{Z}$ other than the point $(0,0)$ itself. Hence it is not open.
Question: What should the distance function actually be for $B_{\epsilon}((0,0))$? I am presuming it is the Euclidean metric in $\mathbb{R}^{2}$ i.e. we would be showing $\sqrt{(a-0)^{2} + (b-0)^{2}} < \epsilon$ for some $(a,b) \in \mathbb{Z}.$
NOT bounded:
If we have a metric space $(X,d)$ and $E \subseteq X$, Rudin defines a bounded set as follows
$E$ is bounded if there is a real number $M$ and a point $q \in X$ such that $d(p,q) < M$ for all $p$ in $E$.
Am I correct that NOT bounded is therefore defined as follows:
$E$ is not bounded is for all real numbers $M$ and all points $q \in X$, $d(p,q) \geq M$ for some $p$ in $E$
(I wasn't sure if its only the quantifiers that I swap)
I'm not too sure how to proceed. I was thinking I could maybe use the Archemedian property in someway, but because I'm in a subset of $\mathbb{R}^{2}$ I'm not sure if that approach will work. Similarly I can't use $\sup$ or $\inf$ as I'm in a subset of $\mathbb{R}^{2}$.
Thanks.
Best Answer
I will assume that we are using a sensible embedding of $\mathbb{Z} \subset \mathbb{R}^2$ (say by $n \mapsto (n, 0)$).
To your question about the metric. From context I would presume that the metric is the usual euclidean one (the claim is false if we take an arbitrary metric - consider the discrete metric which makes all subsets open!).
Your proofs of closedness and non-openness seem fine.
You are right to think the set is not bounded. What you want to show is
To do this, suppose that $p = (x, y)$. Then $d(p, (n,0)) \geq |n - x|$, so if we choose some large $n$ so that $|n - x| > M$ we are done.