Proof. Let \(\zeta \) be an open covering of \(A\). The limit \(x\) belongs to some set \(G_0\) in \(\zeta \). Since \(G_0\) is open, there is an \(\varepsilon > 0\) with \((x- \varepsilon , x + \varepsilon ) \subseteq G_0\), and since \(x_n\longrightarrow x\) there exists \(m\in \mathbb {N}\) such that \(x_n\in G_0\) for every \(n > m\). Each of the remaining terms \(x_1,\dots ,x_m\) belongs to some set \(G_i \in \zeta \). Hence \(A\) is covered by the finitely many sets \(G_0, G_1, \dots , G_m\). □

Proof. Let \(S\) be the set of all \(x\in [a,b]\) such that the interval \([a,x]\) is covered by finitely many sets of \(\zeta \). Then \(a\in S\), because \([a,a] = \{a\}\) and \(a\) belongs to some set in \(\zeta \). So \(S\) is non-empty and bounded above by \(b\); let \(M= \sup S\). Then \(a\leq M \leq b\). We show that \(M\in S\) and \(M = b\).

Since \(M\in [a,b]\), there is a set \(G^*\) in \(\zeta \) with \(M \in G^*\), and since \(G^*\) is open there is an \(\varepsilon > 0\) such that \((M - \varepsilon , M + \varepsilon ) \subseteq G^*\). As \(M\) is the supremum of \(S\), there exists \(x \in S\) with \(M - \varepsilon < x \leq M\). Since \(x\in S\), there are finitely many sets \(G_1, G_2, \dots , G_r\) in \(\zeta \) with \([a,x] \subseteq G_1 \cup G_2 \cup \dots \cup G_r\). Moreover \([x,M+\frac {\varepsilon }{2}] \subseteq (M - \varepsilon , M + \varepsilon ) \subseteq G^*\). Hence \[\left [a, M+\tfrac {\varepsilon }{2}\right ] \subseteq G^*\cup G_1 \cup G_2 \cup \dots \cup G_r .\] In particular \([a,M]\) is covered by finitely many sets of \(\zeta \), so \(M\in S\). If \(M<b\), the same display shows that every point of \(\bigl [M, \min \{M+\frac {\varepsilon }{2}, b\}\bigr ]\) lies in \(S\), contradicting \(M=\sup S\). Hence \(M = b\in S\), and \([a,b]\) has a finite subcovering. □

Proof. Suppose \(A\) is compact. The open intervals \((-n,n)\), \(n \in \mathbb {N}\), have union \(\mathbb {R}\), so they cover \(A\). By hypothesis finitely many of them cover \(A\), and since they increase with \(n\), \(A\subseteq (-m, m)\) for some \(m\in \mathbb {N}\). Consequently \(A\) is bounded.

To show that \(A\) is closed we show that no point outside \(A\) belongs to \(\overline {A}\). Let \(x\notin A\). For each \(a\in A\) put \(r_a = \frac {1}{2}|a - x| > 0\), and let \[U_a = (a-r_a, a+r_a),\qquad V_a = (x-r_a, x+r_a).\] Then \(U_a\) and \(V_a\) are disjoint open intervals with \(a\in U_a\) and \(x\in V_a\).

As \(a\) varies over \(A\), the sets \(U_a\) form an open covering of \(A\), so by compactness \(A\subseteq U_{a_1} \cup U_{a_2} \cup \dots \cup U_{a_r}\) for some \(a_1,\dots ,a_r\in A\). Let \[V = V_{a_1} \cap V_{a_2} \cap \dots \cap V_{a_r} = (x-r, x+r),\qquad r = \min \{r_{a_1},\dots ,r_{a_r}\} > 0 .\] Then \(V\) is disjoint from each \(U_{a_j}\), hence from their union, and consequently \(V\cap A = \emptyset \). Thus \(x\notin \overline {A}\). Hence \(\overline {A}\subseteq A\), and \(A\) is closed.

Conversely, suppose \(A\) is bounded and closed, and let \(\zeta \) be an open covering of \(A\). Since \(A\) is bounded, \(A\subseteq [a,b]\) for some closed interval \([a,b]\), and since \(A\) is closed, the set \(V = \mathbb {R} \setminus A\) is open by Theorem 1.3.6. The sets of \(\zeta \) cover the points of \([a,b]\) that lie in \(A\), and \(V\) covers the rest. So \(\zeta \) together with \(V\) is an open covering of \([a,b]\), and by the previous theorem finitely many of these sets, say \(V\) and \(U_1, U_2, \dots , U_r\) from \(\zeta \), cover \([a,b]\). Since \(A\) is disjoint from \(V\), it follows that \(A \subseteq U_1 \cup U_2\cup \dots \cup U_r\). Hence \(A\) is compact. □

Proof. By the Heine–Borel Theorem, \(A\) is bounded and closed. Let \(M = \sup A\). For each \(n\), \(M-\frac {1}{n}\) is not an upper bound of \(A\), so there is \(x_n\in A\) with \(M-\frac {1}{n}<x_n\leq M\). Then \(x_n\longrightarrow M\), and since \(A\) is closed, \(M \in A\). So \(M\) is the largest element of \(A\). Similarly \(\inf A\) belongs to \(A\) and is its smallest element. □

Proof. For \(n\in \mathbb {N}\) let \(G_n = \left (\frac {1}{n + 1}, 1\right )\), and let \(\zeta = \{G_n : n\in \mathbb {N}\}\). If \(x\in (0,1)\), then by the Archimedean property there exists \(n \in \mathbb {N}\) with \(n > \frac {1}{x}\), so that \(\frac {1}{n+1} < \frac {1}{n} < x\) and \(x \in G_n\). Hence \(\zeta \) is an open covering of \((0,1)\).

Now take any finitely many members \(G_{n_1},\dots ,G_{n_k}\) of \(\zeta \), and let \(N = \max \{n_1,\dots ,n_k\}\). Since the sets \(G_n\) increase with \(n\), their union is \(G_N = \left (\frac {1}{N+1}, 1\right )\), which does not contain the point \(\frac {1}{N + 2}\) of \((0,1)\). So no finite subfamily of \(\zeta \) covers \((0,1)\), and \((0,1)\) is not compact. (This agrees with the Heine–Borel Theorem, since \((0,1)\) is not closed.) □

Solution. We use the Heine–Borel Theorem 1.4.6.

(a) Closed (a finite union of closed sets) and bounded, so compact.

(b) Bounded, and closed because it equals \(\overline {A}\) for \(A = \left \{\frac {1}{n}\right \}\) (see the example in the section on closed sets). So it is compact.

(c) Not bounded, so not compact. Directly: the open intervals \((-1,n)\), \(n\in \mathbb {N}\), cover \([0,\infty )\), but any finitely many of them lie in \((-1,N)\) for the largest \(N\) used, which misses the point \(N\).

(d) Bounded but not closed (its closure is \([0,1]\)), so not compact. A concrete open cover without finite subcover: since \(\frac {1}{\sqrt {2}}\) is irrational, the sets \(\left (-1,\frac {1}{\sqrt {2}}-\frac {1}{n}\right )\) and \(\left (\frac {1}{\sqrt {2}}+\frac {1}{n},2\right )\), \(n\in \mathbb {N}\), cover \(\mathbb {Q}\cap [0,1]\), but finitely many of them miss the rationals close enough to \(\frac {1}{\sqrt {2}}\).

Solution. Let \(K_1\) and \(K_2\) be compact, and let \(\zeta \) be an open covering of \(K_1\cup K_2\). Then \(\zeta \) covers \(K_1\), so finitely many members of \(\zeta \) cover \(K_1\), and finitely many cover \(K_2\). Together these finitely many sets cover \(K_1\cup K_2\).

If \(\{K_i\}\) is any family of compact sets, each \(K_i\) is closed and bounded, so \(\bigcap _i K_i\) is closed (Theorem 1.2.4) and bounded (it lies in any one \(K_i\)). By the Heine–Borel Theorem it is compact.

Solution. \(K\cap F\) is closed, as an intersection of two closed sets, and bounded, as a subset of the bounded set \(K\). By the Heine–Borel Theorem it is compact. The closed set \(\mathbb {Z}\) is not compact, since it is unbounded; the open intervals \(\left (n-\frac {1}{2},n+\frac {1}{2}\right )\), \(n\in \mathbb {Z}\), cover it, and each contains exactly one integer, so no finite subfamily covers \(\mathbb {Z}\).