Open cover, subcover, compact subset of $\mathbb{R}$ (every open cover has a finite subcover), and sequentially compact subset

Let $K\subseteq \mathbb{R}$, with open sets as in Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

• An open cover of $K$ is a family $\mathcal{U}$ of open subsets of $\mathbb{R}$ with $K\subseteq \bigcup \mathcal{U}$.
• A subcover of $\mathcal{U}$ is a subfamily $\mathcal{V}\subseteq \mathcal{U}$ that is still an open cover of $K$.
• A subfamily $\mathcal{V}\subseteq \mathcal{U}$ is finite when $\mathcal{V}=\mathrm{\varnothing }$ or there are $n\in \mathbb{N}$ and members $U_0,\dots ,U_n$ of $\mathcal{U}$ with $\mathcal{V}=\{U_0,\dots ,U_n\}$; repetitions in the list are allowed and harmless.
• $K$ is compact when every open cover of $K$ has a finite subcover: for every open cover $\mathcal{U}$ of $K$, either $K=\mathrm{\varnothing }$ and the empty subfamily covers it, or there are $n\in \mathbb{N}$ and $U_0,\dots ,U_n\in \mathcal{U}$ with $K\subseteq U_0\cup \cdots \cup U_n\mathrm{.}$
• $K$ is sequentially compact when every sequence $(x_k)$ of reals with $x_k\in K$ for all $k\in \mathbb{N}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences) has a subsequence converging (Limits and Cauchy sequences of reals) to some point of $K$; equivalently, when every such sequence has a subsequential limit (Subsequential limit of a real sequence, and the subsequential limit set) that lies in $K$.

Compactness is a property of $K$ alone. The covering families range over open subsets of $\mathbb{R}$, not over sets open in some other ambient space, so the notion defined here is compactness of $K$ as a subset of $\mathbb{R}$. Nothing below relativises it to a smaller ambient field; where an ordered field other than $\mathbb{R}$ is meant, as in FALSE: in every ordered field a closed bounded set is compact, so Heine-Borel needs no completeness, the whole vocabulary is set up again there for that field.

$\mathrm{\varnothing }$ is compact and sequentially compact. The empty subfamily covers it, and there is no sequence with all terms in $\mathrm{\varnothing }$, so both conditions hold vacuously.

• Why "finite" is spelled out by listing. A finite subfamily is described here as one that can be written $\{U_0,\dots ,U_n\}$ with $n\in \mathbb{N}$, which is exactly the form every proof on this page produces or consumes: the bisection argument of Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact produces a one-member list, and the arguments of A compact subset of $\mathbb{R}$ is closed and bounded consume a list by taking a maximum over it (Every nonempty finite set of reals has a maximum and a minimum). Since $\mathbb{N}$ contains $0$, the shortest nonempty list is $\{U_0\}$.

• The two notions are not defined to be equivalent, and their equivalence is a theorem. For subsets of $\mathbb{R}$ it is A subset of $\mathbb{R}$ is compact iff it is sequentially compact; both of its implications run through the characterisation of compactness by closed and bounded, and its forward implication additionally uses Bolzano-Weierstrass. Neither implication is formal.

• Compactness is not inherited by subsets, but by closed subsets. A closed subset of a compact set is compact, which is immediate from A subset of $\mathbb{R}$ is compact if and only if it is closed and bounded once that is available, whereas $(0,1)\subseteq [0,1]$ shows that an arbitrary subset of a compact set need not be compact.

• The empty cover. If $K\mathrm{\ne }\mathrm{\varnothing }$ then no open cover of $K$ is empty, so the case distinction in the definition of compactness only ever matters for $K=\mathrm{\varnothing }$; it is written out so that the definition does not quietly assume $K$ nonempty.