$[0,1]$ is compact

We prove that \([0, 1]\) is compact directly using the least-upper-bound property of \(\RR\).

Let \(\mathcal{U} = \{U_\alpha\}_{\alpha \in A}\) be an open cover of \([0, 1]\) by open subsets of \(\RR\). Define the set: \[ S = \{x \in [0, 1] \mid [0, x] \text{ can be covered by finitely many sets from } \mathcal{U}\}. \]

1. \(S\) is non-empty and bounded:

   • Since \(0 \in [0, 1]\), there is some \(U_0 \in \mathcal{U}\) with \(0 \in U_0\). Thus \([0, 0] = \{0\} \subseteq U_0\), so \(0 \in S\), and \(S \neq \emptyset\).

   • By definition, \(S \subseteq [0, 1]\), so \(S\) is bounded above by \(1\).

2. Supremum of \(S\): By the completeness (least upper bound property) of \(\RR\), \(c = \sup S\) exists, and \(0 \leq c \leq 1\).

3. Show \(c \in S\): Since \(c \in [0, 1]\), there exists some \(U_c \in \mathcal{U}\) containing \(c\). Since \(U_c\) is open in \(\RR\), there exists \(\varepsilon > 0\) such that \((c - \varepsilon, c + \varepsilon) \cap [0, 1] \subseteq U_c\). By definition of the supremum, there exists \(x_0 \in S\) with \(c - \varepsilon < x_0 \leq c\). Since \(x_0 \in S\), \([0, x_0]\) is covered by a finite subcollection \(\{U_{\alpha_1}, \ldots, U_{\alpha_k}\} \subseteq \mathcal{U}\). Then: \[ [0, c] \subseteq [0, x_0] \cup [x_0, c] \subseteq \left(\bigcup_{i=1}^k U_{\alpha_i}\right) \cup U_c. \] This is a finite subcover of \([0, c]\), so \(c \in S\).

4. Show \(c = 1\): Suppose towards a contradiction that \(c < 1\). Since \(c \in U_c\) and \(U_c\) is open, we can choose \(x_1 \in (c, c + \varepsilon) \cap [0, 1]\) (since \(c < 1\)). Then \([0, x_1] = [0, c] \cup [c, x_1] \subseteq \left(\bigcup_{i=1}^k U_{\alpha_i}\right) \cup U_c\). Thus \([0, x_1]\) is also covered by finitely many sets from \(\mathcal{U}\), meaning \(x_1 \in S\). Since \(x_1 > c\), this contradicts that \(c = \sup S\).

Therefore, \(c = 1\), and since \(c \in S\), we have \(1 \in S\). Hence \([0, 1]\) is covered by a finite subcollection of \(\mathcal{U}\), proving that \([0, 1]\) is compact.