V.7 · Compactness & Connectedness — Problem 7 #

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Problem. (a) Show that every finite subset of a metric space is compact. (b) In the discrete metric, show that a set is compact iff it is finite. (c) In \((\mathbb{R}, |\cdot|)\), let \(T = \{\tfrac{1}{n} \mid n \in \mathbb{N}\}\). Decide whether \(T\) and \(T \cup \{0\}\) are compact, directly from the definition.

Hint #

For (a), each point of the set lies in some member of the cover; how many members do you need? For (b), in the discrete metric every singleton \(\{p\} = B_{1/2}(p)\) is open, so the singletons form an open cover. For (c), to show \(T\) is not compact, build a cover in which each open set captures exactly one point of \(T\). To show \(T \cup \{0\}\) is compact, note that whichever open set covers \(0\) must also swallow all but finitely many of the points \(\tfrac{1}{n}\).

Solution #

(a) Let \(T = \{p_1, \ldots, p_n\}\) and let \(\{A_i\}_{i \in I}\) be an open cover of \(T\). For each \(j\) choose an index \(i_j\) with \(p_j \in A_{i_j}\). Then \(A_{i_1}, \ldots, A_{i_n}\) is a finite subcover. (The empty set is compact for the same reason, with no sets needed.)

(b) Finite sets are compact by (a). Conversely, suppose \(T\) is compact in the discrete metric. Each singleton \(\{t\} = B_{1/2}(t)\) is open, so \(\{\{t\} \mid t \in T\}\) is an open cover of \(T\). A finite subcover \(\{t_1\}, \ldots, \{t_k\}\) covers \(T\) only if \(T \subseteq \{t_1, \ldots, t_k\}\), so \(T\) is finite.

(c) \(T\) is not compact. For \(n \geq 2\) let \(A_n = \big(\tfrac{1}{n+1}, \tfrac{1}{n-1}\big)\) and let \(A_1 = \big(\tfrac{1}{2}, 2\big)\). Each \(A_n\) is open, contains \(\tfrac{1}{n}\), and contains no other point of \(T\), since \(\tfrac{1}{m} \leq \tfrac{1}{n+1}\) for \(m > n\) and \(\tfrac{1}{m} \geq \tfrac{1}{n-1}\) for \(m < n\). So \(\{A_n\}\) is an open cover of \(T\), but any finite subcollection \(A_{n_1}, \ldots, A_{n_k}\) covers only the \(k\) points \(\tfrac{1}{n_1}, \ldots, \tfrac{1}{n_k}\), not all of the infinite set \(T\).

\(T \cup \{0\}\) is compact. Let \(\{A_i\}_{i \in I}\) be an open cover of \(T \cup \{0\}\). Some \(A_{i_0}\) contains \(0\), and since it is open there is an \(\epsilon > 0\) with \((-\epsilon, \epsilon) \subseteq A_{i_0}\). Choose \(N\) with \(\tfrac{1}{N} < \epsilon\); then \(\tfrac{1}{n} \in A_{i_0}\) for all \(n \geq N\). For each of the finitely many remaining points \(1, \tfrac{1}{2}, \ldots, \tfrac{1}{N-1}\) choose a set \(A_{i_n}\) containing it. Then \(A_{i_0}, A_{i_1}, \ldots, A_{i_{N-1}}\) is a finite subcover.

Long solution #

Strategy. Compactness is a statement about every open cover, so to prove a set is compact we take an arbitrary cover and manufacture a finite subcover, while to prove a set is not compact we exhibit one cover with no finite subcover. Part (a) is the simplest possible instance of the first kind of argument. Part (b) shows the discrete metric is the extreme case: the only compact sets are the ones that are compact for trivial reasons. Part (c) contrasts a set that is “almost” compact with the set obtained by adding its single missing limit point, and the proof shows exactly how that one point repairs the failure.

Definition. \(T \subseteq S\) is compact if every open cover \(\{A_i\}_{i \in I}\) of \(T\) (a collection of open sets whose union contains \(T\)) has a finite subcover: finitely many \(A_{i_1}, \ldots, A_{i_k}\) whose union still contains \(T\).

Part (a). Let \(T = \{p_1, \ldots, p_n\}\) and let \(\{A_i\}_{i \in I}\) be any open cover of \(T\). Because the cover’s union contains \(T\), each \(p_j\) belongs to at least one \(A_i\); pick one and call its index \(i_j\). Then

a subcover with at most \(n\) members. Since the cover was arbitrary, \(T\) is compact. \(\blacksquare\)

Note that the argument does not use anything about the metric: finite sets are compact in every metric space.

Part (b). Recall that in the discrete metric \(\rho(p,q) = 1\) when \(p \neq q\), so \(B_{1/2}(p) = \{p\}\): every singleton is an open ball, hence open, and therefore every subset is open (as a union of singletons).

(\(\Leftarrow\)) If \(T\) is finite it is compact by part (a).

(\(\Rightarrow\)) Suppose \(T\) is compact. The family \(\{\{t\} \mid t \in T\}\) is a collection of open sets whose union is \(T\), so it is an open cover of \(T\). By compactness some finite subfamily \(\{t_1\}, \ldots, \{t_k\}\) covers \(T\), i.e. \(T \subseteq \{t_1, \ldots, t_k\}\). Hence \(T\) is finite. \(\blacksquare\)

Consequence. In \((\mathbb{R}, \text{discrete})\) the whole space is closed (every set is) and bounded (all distances are at most \(1\)), yet not compact. So “closed and bounded” does not imply compact in general: the Heine–Borel theorem is a special feature of \(\mathbb{R}\) with its usual metric, and of its proof via the least upper bound property.

Part (c): \(T\) is not compact. We need one bad cover. The idea is to surround each point \(\tfrac{1}{n}\) by an open interval so small that it contains no other point of \(T\). The neighbours of \(\tfrac{1}{n}\) in \(T\) are \(\tfrac{1}{n+1}\) below and \(\tfrac{1}{n-1}\) above, so set

Each \(A_n\) is an open interval containing \(\tfrac{1}{n}\), so \(\{A_n\}_{n \in \mathbb{N}}\) is an open cover of \(T\). We claim \(A_n \cap T = \{\tfrac{1}{n}\}\). Indeed, if \(m > n\) then \(\tfrac{1}{m} \leq \tfrac{1}{n+1}\), which is the left endpoint of \(A_n\) and is excluded; if \(m < n\) then \(\tfrac{1}{m} \geq \tfrac{1}{n-1}\), the right endpoint, also excluded (for \(n = 1\) there is no such \(m\)). Now any finite subcollection \(A_{n_1}, \ldots, A_{n_k}\) covers exactly the \(k\) points \(\tfrac{1}{n_1}, \ldots, \tfrac{1}{n_k}\) of \(T\), and \(T\) is infinite, so no finite subcover exists. \(\blacksquare\)

Part (c): \(T \cup \{0\}\) is compact. Let \(\{A_i\}_{i \in I}\) be an arbitrary open cover of \(T \cup \{0\}\). The point \(0\) lies in some \(A_{i_0}\). Because \(A_{i_0}\) is open, it contains a ball about \(0\): there is \(\epsilon > 0\) with \((-\epsilon, \epsilon) \subseteq A_{i_0}\). By the Archimedean property choose \(N \in \mathbb{N}\) with \(\tfrac{1}{N} < \epsilon\). Then for every \(n \geq N\),

Thus the single set \(A_{i_0}\) covers \(0\) together with all but finitely many points of \(T\). The leftover points \(1, \tfrac{1}{2}, \ldots, \tfrac{1}{N-1}\) are finitely many; for each choose \(A_{i_n} \ni \tfrac{1}{n}\). Then

a finite subcover. As the cover was arbitrary, \(T \cup \{0\}\) is compact. \(\blacksquare\)

Why the one point matters. In the bad cover for \(T\), the intervals \(A_n\) shrink toward \(0\), and \(0\) itself is never covered because it is not in \(T\). Once \(0\) is added to the set, any cover must include an open set containing \(0\), and openness forces that set to contain a whole interval about \(0\), which swallows the tail of the sequence. This is exactly the mechanism behind the Heine–Borel theorem: compare the two sets with Heine–Borel in mind. \(T\) is bounded but not closed (its limit point \(0\) is missing, see Problem IV.6), while \(T \cup \{0\}\) is closed and bounded. The direct proofs above agree with the theorem, as they must.