I am reading Principles of Mathematical Analysis by Walter Rudin.
In chapter two, it has the following definitions:
A neighborhood of $p$ is a set $N_r(p)$ consisting of all $q$ such that $d(p,q) \lt r$, for some $r \gt 0$. The number $r$ is called the radius of $N_r(p)$.
A point p is a limit point of the set $E$ if every neighborhood of $p$ contains a point $q \neq p$ such that $q \in E$.
$E$ is closed is every limit point of $E$ is a point of $E$.
By an open cover of a set $E$ in a metric space $X$ we mean a collection ${G_\alpha}$ of open subsets of $X$ such that $E \subset \cup_\alpha G_\alpha$.
A subset $K$ of a metric space $X$ is said to be compact if every open cover of $K$ contains a finite subcover. More explicitly, the requirement is that if ${G_\alpha}$ is an open cover of K, then there are finitely many indices $\alpha_1, \ldots, \alpha_n$ such that $$K \subset G_{\alpha_1} \cup \cdots \cup G_{\alpha_1}$$
I was pretty sure I understood these until I found this theorem:
Theorem $\;$ Compact subsets of metric spaces are closed.
I know that the proof involves showing that the complement is open, and I don't have any problems with that, but I found a set that is compact but not closed, at least according to what I understood the definitions to be.
My logic is as follows:
The set $\{x \in \mathbb R^2 \;|\; |x| \lt 1\}$ is not closed (any point on the circle surrounding it is a limit point but is not a member), but it is compact (it is a subset of the open set $\{x \in \mathbb R^2 \;|\; |x| \lt 2\}$), which contradicts the theorem.
I assume the flaw is in my understanding of the definitions.
Best Answer
That set is not compact. Consider the open sets$$\left\{(x,y)\in\mathbb{R}^2\,\middle|\,\bigl\lVert(x,y)-(1,0)\bigr\rVert>\frac1n\right\},$$with $n\in\mathbb N$. These sets form an open cover of your set without a finite subcover.