Show that \([0, 1]\) is connected.
Solution
1. Suppose for contradiction that \(I = [0,1]\) is disconnected, so \(I = A \union B\) with \(A, B\) nonempty, disjoint, and separated: \(\cl_I(A) \intersect B = A \intersect \cl_I(B) = \emptyset\).
definition of a disconnection.
2. Relabel so that \(0 \in A\), and set \(s \da \sup A\).
\(0\) lies in exactly one of \(A, B\); swap the labels if needed. The supremum exists because \(A \subseteq [0,1]\) is bounded above and \(\RR\) has the least-upper-bound property.
3. \(s \in A\).
3.1. Suppose instead \(s \in B\).
\(s \in [0,1] = A \union B\), so if \(s \notin A\) then \(s \in B\).
3.2. Since \(\cl_I(A) \intersect B = \emptyset\), the point \(s \in B\) is not in \(\cl_I(A)\), so there is a neighborhood \(U\) of \(s\) with \(U \intersect A = \emptyset\).
a point outside the closure of \(A\) has a neighborhood disjoint from \(A\).
3.3. But \(s = \sup A\) means every neighborhood of \(s\) meets \(A\): for any \(\eps > 0\) there is \(a \in A\) with \(s - \eps < a \le s\).
if some neighborhood \((s-\eps, s+\eps)\) missed \(A\), then \(s - \eps\) would be an upper bound of \(A\) smaller than \(s\), contradicting that \(s\) is the least upper bound.
3.4. 2 and 3 contradict each other, so \(s \notin B\), hence \(s \in A\).
1.
4. \(s = 1\).
4.1. Since \(A \intersect \cl_I(B) = \emptyset\), the point \(s \in A\) is not in \(\cl_I(B)\), so there is \(\eps > 0\) with \((s - \eps, s + \eps) \intersect I \subseteq A\).
a point outside the closure of \(B\) has a neighborhood disjoint from \(B\); within \(I\) that neighborhood lies entirely in \(A\).
4.2. If \(s < 1\), then \((s, s + \eps) \intersect I\) contains points of \(A\) strictly larger than \(s\).
take any \(t\) with \(s < t < \min(s + \eps, 1)\); then \(t \in (s-\eps, s+\eps) \intersect I \subseteq A\).
4.3. This contradicts \(s = \sup A\), so \(s = 1\).
2 produces an element of \(A\) exceeding the supremum \(s\).
5. Contradiction: \(1 = s \in A\), but \(B\) is nonempty.
5.1. Since \(B \neq \emptyset\), pick \(b \in B \subseteq [0,1]\).
\(B\) is nonempty by hypothesis.
5.2. \(b \le 1 = s = \sup A\), so for every \(\eps > 0\) there is \(a \in A\) with \(b - \eps < a \le b\).
\(b\) is below the supremum of \(A\), so \(A\) meets every neighborhood of \(b\).
5.3. Hence \(b \in \cl_I(A)\), contradicting \(\cl_I(A) \intersect B = \emptyset\).
2 shows every neighborhood of \(b\) meets \(A\), which is the definition of \(b \in \cl_I(A)\).
6. Therefore no disconnection of \([0,1]\) exists, so \([0,1]\) is connected.
1 through 5.