Proof. Let \(x\) belong to one of the sets in \(\zeta \) say \(\, x \in G\, \) in \(\zeta \). Since \(G\) is open, there is an \(\varepsilon > 0\, \exists \,\, \big (x- \varepsilon \, , \, x + \varepsilon \big ) \subset G\). This mean \(\, \exists m\in \mathbb {N} \ni \, \forall n\, > m, \, x_n\in G\). Now, each of the terms \(\, x_i (i= 1,2 ,\cdots \cdots , m)\) belong to \(G_i \in \zeta \), so \(A\) is covered by the sets \(G_1, \, G_2, \, \cdots \cdots \, G_m\). Therefore \(G\) has a finite subcovering of \(A\). Generally, every open covering of \(A\) has a finite subcovering.

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Proof. Let \(S\) be the set of all \(x\in [a,b]\ni \,\) the closed interval \([a,x]\) is covered by finitely many sets \(G_{\alpha }, \, \alpha \in I\) of \(\zeta \). At least \(a\in S\), because \([a,a] = \big \{a\big \}\) and \(a\) belongs to some set \(G_{\alpha }\) in \(G\). We need to show that \(b\in S\). At any rate, \(S\) is non empty and bounded. Let \(M= \sup S\). Since \(S\subset [a, b]\), we have \(\, a\leq M \leq b\). Here, we want to show that \((1) \,\, M\in S\,\) and \(\, (2)\, \, M = b\).

1.

Since \(M\in [a,b]\subset \bigcup _{\alpha } \, G_{\alpha }\), there is a \(G^*_{\alpha } \in \zeta \ni M \in G^*_{\alpha }\). Since \(G^*_{\alpha }\) is open, \(\big (M - \varepsilon \, , \, M + \varepsilon \big ) \subset G^*_{\alpha }\) for some \(\varepsilon > 0\). Since \(M - \varepsilon < M \) and \(M\) is the sup of \(S\), \(\, \exists \, x \in S \ni M - \varepsilon < x \leq M\). Since \(x\in S,\, \exists \, \) finitely many sets \(G_{\alpha _1}\, , \, G_{\alpha _2}\, , \, \cdots \cdots ,\, G_{\alpha _r}\, \) in \(\zeta \, \ni [a,x] \subset G_{\alpha _1} \cup G_{\alpha _2} \cup \cdots \cdots \cup G_{\alpha _r}\, \) ( i.e the interval \([a,x]\) is covered by finitely many sets in \(\zeta \)). On the other hand, \(\big [x,M\big ]\subset \big (M - \varepsilon \, ,\, M + \varepsilon \big ) \subset G^*_{\alpha } \, \implies \, \big [a, M\big ] = \big [a,x\big ]\, \cup \, \big [ x, M\big ]\,\) is covered by the sets \(\, G^*_{\alpha }\, ,\, G_{\alpha _1}\, ,\, G_{\alpha _2}\, , \, \cdots \cdots , \, G_{\alpha _r}\,\) of \(\zeta \). This means that \(M\in S\).

2.

The proceeding argument that \(\, b - M < \varepsilon \,\) and the argument is valid with \(\varepsilon \) replaced by any positive number smaller than \(\varepsilon \). It follows that \(\, b - M \leq 0\). Thus \(b\leq M\). But \(M\leq b\), from above. Thus \(b = M\in S\). But since \([a,M]\) is covered by finitely many sets in \(\zeta \) it follows that \([a,b]\) has a finite subcovering.

Proof. Suppose \(A(\subset \mathbb {R})\) is compact. The open intervals \((-n,n),\, n \in \mathbb {N}\), have a union \(\mathbb {R}\) and thus they cover \(A\). By hypothesis, \(\exists \,\) a finite number of subintervals which cover \(A\). This means that \(\, A\subset (-m, m)\) for some \(m\in \mathbb {N}\).
Consequently, \(A\) is bounded. To show that \(A\) is closed we need to show that \(\overline {A} \subset A\), equivalently, \(\, A^C \subset \big (\overline {A}\big )^C\). Assuming that \(\, x\not \in A\), lets show that \(\, x\not \in \overline {A}\). We seek a nbd \(V\) of \(\, x\ni V_n A = \emptyset \). If \(\, a\in A\,\) then \(\, x\neq a\,\) since \(x\not \in A\). Thus \(\,\exists \,\) open intervals \(U_{\alpha }\, ,\, V_{\alpha } \ni a\in U_{\alpha }\, , \, x\in V_{\alpha }\,\) and \(\, U_{\alpha } \cap V_{\alpha } = \emptyset \).

As \(a\) varies over \(A\), the set \(U_{\alpha }\) form an open covering of \(A\). Suppose \(\, A\subset U_{\alpha _1} \cup U_{\alpha _2} \cup \cdots \cdots \cup U_{\alpha _r}\). Let \begin {align*} U & = U_{\alpha _1} \cup U_{\alpha _2} \cup \cdots \cdots \cup U_{\alpha _r}\\ V & = V_{\alpha _1} \cup V_{\alpha _2} \cup \cdots \cdots \cup V_{\alpha _r} \end {align*}

Then \(\, A\subset U\, \) and \(V\) is a nbd of \(x\). If \(y \in U_{\alpha _j}\) then \(y\not \in V_{\alpha _j}\). Therefore, \(y \not \in V\). Hence \(V\cap U = \emptyset \) (\(V\) misses every term in the formula for \(U\), so misses their union), and consequently \(V\cap A = \emptyset \). Thus, any point outside \(A\) is not a limit point of \(A\). \(\implies A\) contains all its limit points. Hence \(A\) is closed.
Conversely, suppose \(A\) is bounded and closed and \(\zeta \) is an open covering of \(A\). We must show that \(A\) is compact by finding a finite subcovering of \(\zeta \). By hypothesis the set \(V = \mathbb {R} - A\) is open and \(A\) is contained in some closed intervals, say \(A\subset [a,b]\). The points of \([a,b]\) that are in \(A\) are covered by \(\zeta \). What is left is the set \([a,b]-A\) and is contained in \(V\). Thus, we have an open covering of \([a,b]\) i.e the sets in \(\zeta \) together with the set \(V\) cover \([a,b]\). Therefore, \([a,b] \subset U_1 \cup U_2 \cup \cdots \cdots \cup U_r\) for suitable \(\, U_1,\, U_2, \, \cdots \cdots , U_r\) in \(\zeta \). The set \(A\) is contained in \([a,b]\) but is disjoint from \(V\). \(\implies \, A \subset U_1 \cup U_2\cup \cdots \cdots \cup U_r\), therefore \(A\) is covered by a finite subcovering. Hence \(A\) is compact.

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Proof. By Heine-Borel Theorem, \(A\) is bounded and closed. Let \(M = \sup A\) and choose a sequence \(\, \big \{x_n\big \}\,\) in \( A\, \ni x_n\longrightarrow M\). Then \(M \in A\) (since \(A\) is closed) and \(M\) is the largest element of \(A\). Similarly \(\inf A\) belongs to \(A\) and it is the smallest element of \(A\).

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Proof. Let \(\, \zeta = \big \{ \big (G_n = 1/(n + 1)\, , \,1\big ): \, n\in \mathbb {N}\big \}\). We must show that \(\zeta \) is an open covering of (0,1), but has no finite subcovering which covers (0,1). Let \(\, x\in (0,1)\, \implies x \in (\varepsilon , 1)\,\) where \(0< \varepsilon < 1\). But by the Archimedean property \(\, \exists \, n \in \mathbb {N} \ni n > 1/ \varepsilon \, \implies \, 1/n < \varepsilon \, \implies \, 1/(n + 1) < \varepsilon \). This means that \(\, x \in \big (1/(n + 1) \, , \, 1\big ) = G_n\). Therefore \(\, x \in \bigcup ^{\infty }_{n=1}\, G_n\). Hence \(\, (0,1) \subset \bigcup ^{\infty }_{n = 1}\, G_n\, \implies \, \zeta \,\) is an open covering of (0,1).
Now, we assume that \(G\) is a finite subcovering of \(\zeta \). Then \[G = \Big \{ \Big (\dfrac {1}{n + 1}\, , \, 1\Big )\, , \, n = 1, \, 2, .... , k\in \mathbb {N}\Big \}\] We show that there exist one element in (0,1) which is not covered by \(G\). Clearly, \(\dfrac {1}{(k + 1) + 1} \in (0,1),\, \, \forall k \in \mathbb {N}\). Also, \[\dfrac {1}{k + 2} < \dfrac {1}{k + 1} \leq \dfrac {1}{n + 1}\hspace {0.2cm},\hspace {0.2cm} n = 1,\, 2,\, \cdots \cdots ,\, k\] This means that \(\dfrac {1}{k + 2} \not \in \Big (\dfrac {1}{n + 1}\, ,\, 1 \Big ) = G_n\). Hence \(\displaystyle {\dfrac {1}{k + 2} \not \in \bigcup ^{k}_{n = 1}\, G_n \, \implies \, (0,1)}\) is not covered by \(\,G\, \implies \, G\,\) is not a finite subcovering of \(G\). Therefore (0,1) is not compact.


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