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I am working on the following exercises from the appendix of Lee's Topological Manifolds.

Is $S$ open, closed, both or neither in $M$?

  1. $M=\mathbb{Z}$ with the metric topology and $S=\mathbb{N}$

I think that $\mathbb{N}$ is both open and closed in $\mathbb{Z}$. It is closed since it contains all of its limit points: every convergent sequence of natural numbers must converge to a natural number (such a sequence could not converge to an element of $\mathbb{Z}\setminus\mathbb{N}$. It is open since for any natural number $n$, there is an open ball around $n$ of arbitrarily small radius $\epsilon>0$ in $\mathbb{N}$. This open ball is just $\{n\}$.

  1. $M=\mathbb{R}^3$ with the usual topology and $S=\{(x,y,z)\in\mathbb{R}^3 : z=0 \text{ and } x^2+y^2<1\}$.

$S$ is neither. $S$ cannot be closed since it does not contain its boundary $\{(x,y,0)\in\mathbb{R}^3 : x^2+y^2=1\}$. $S$ cannot be open since for each point of $S$, open balls around each point are not contained in $S$. This is because open balls in $\mathbb{R}^3$ cannot be contained in a plane, and $S$ is a subset of the plane $z=0$.

Are these correct?

A. Goodier
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1 Answers1

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These are correct. Regarding the first one, note that while limit points are the same thing as limits of sequences in metric spaces, that isn't true in a general topological space. Another way to see that $\mathbb{N}$ is closed in $\mathbb{Z}$ is by observing that its complement is open, for exactly the same reasons that $\mathbb{N}$ is open (in fact, every subset of $\mathbb{Z}$ is open, since, as you have observed, singletons $\{n\}$ are open balls).

Justin Barhite
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