A compact subset of a metric space is always closed

Question

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.

Answer 1

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.

Answer 2

A space $S$ is $T_2$ (a.k.a. a Hausdorff space) when for any distinct $x,y\in S$ there are disjoint open $U,V \subset S$ with $x\in U$ and $y\in V.$

A metric space $(S,d)$ is Hausdorff. For if $x,y$ are distinct members of $S,$ let $r=d(x,y)/2$ and let $U=N_r(x),V=N_r(y).$ The triangle inequality implies that $U,V$ are disjoint.

Theorem: If $S$ is Hausdorff and $T$ is a compact subset of $S$ then $T$ is closed. Equivalently, if $T$ is not closed then $T$ is not compact.

Proof: Suppose $T$ is not closed. Take $y\in \bar T$ \ $T.$ For each $x\in T$ let $U_x, V_x$ be a disjoint pair of open sets with $x\in U_x$ and $y\in V_x$. Consider the open cover $C=\{U_x:z\in T\}$ of $T.$

Suppose $n\in \Bbb Z^+$ and $D=\{U_{x_1},...,U_{x_n}\}$ is any finite subset of $C.$ Then $N=\cap_{j=1}^nV_{x_j}$ is an open set containing the point $y,$ and $y\in \bar T,$ so there exists $z\in N\cap T.$

But $N$ is disjoint from $\cup_{j=1}^nU_{x_j}=\cup D,$ so $D$ is not a cover of $T.$ ( That is, $z\in T$ \ $\cup D).$ So $C$ is an open cover of $T$ with no finite sub-cover.