Connectedness

Proof Sketch:

All five criteria say the same thing in different dialects. A separation of \(X\) is a partition \(X = U \cup V\) into two disjoint non-empty open sets — exactly what (3) forbids. In such a partition, each piece is also closed, being the complement of the other open piece; so (3) and (4) are the same statement with the roles of open and closed interchanged. A subset that is both open and closed (clopen), proper and non-empty, is precisely one half of a separation, giving (2). And for the boundary \(\partial U\): the conditions "\(U\) is open" and "\(U\) is closed" together force \(\partial U = \emptyset\), and conversely \(\partial U = \emptyset\) makes \(U\) clopen; so having a proper non-empty subset with empty boundary is the same as having a clopen proper non-empty subset — giving (1).

The continuity criterion (5) is the same idea through a different lens. A continuous function \(f: X \to \{0, 1\}\) into the two-point discrete space (where every subset is clopen, since the metric \(d(0,1) = 1\) makes each singleton open) is determined by the clopen preimages \(f^{-1}(\{0\})\) and \(f^{-1}(\{1\})\), which partition \(X\). Constant functions correspond to the trivial case where one preimage is empty; non-constant continuous functions correspond to genuine separations. So the absence of non-constant continuous \(\{0,1\}\)-valued functions is the absence of clopen partitions, which is (2).

Proof:

We use the Continuity Criterion (item 5 of the previous theorem). Let \(g: f(X) \to \{0,1\}\) be any continuous function into the discrete two-point space. Then the composition \(g \circ f : X \to \{0,1\}\) is continuous as a composition of continuous maps. Since \(X\) is connected, the Continuity Criterion forces \(g \circ f\) to be constant on \(X\), say \((g \circ f)(x) = c\) for all \(x \in X\). Every \(y \in f(X)\) has the form \(y = f(x)\) for some \(x \in X\), so \(g(y) = g(f(x)) = c\); thus \(g\) is constant on \(f(X)\). Applying the Continuity Criterion in the reverse direction, \(f(X)\) is connected. \(\square\)

Proof:

(\(\Leftarrow\)) Suppose \(S\) is an interval, and for contradiction suppose \(S\) admits a separation \(S = U \cup V\) with \(U, V\) disjoint non-empty open subsets of \(S\). Pick \(a \in U\) and \(b \in V\); by swapping \(U\) and \(V\) if necessary, assume \(a < b\). Because \(S\) is an interval containing both \(a\) and \(b\), the entire closed interval \([a, b]\) lies in \(S\).

The idea is to walk from \(a\) toward \(b\) staying inside \(U\) and ask how far we can go. Formally, set \[ c \;=\; \sup\{\, t \in [a,b] : [a,t] \subseteq U \,\}. \] The set on the right is non-empty (it contains \(t = a\), since \([a,a] = \{a\} \subseteq U\)) and bounded above by \(b\), so the supremum exists and \(a \le c \le b\). Since \([a,b] \subseteq S\), we have \(c \in S\), and since \(S = U \cup V\) is a disjoint union, either \(c \in U\) or \(c \in V\). Each case leads to a contradiction.

We will use the fact that a subset \(W \subseteq S\) is open in \(S\) precisely when \(W = S \cap \tilde{W}\) for some open \(\tilde{W} \subseteq \mathbb{R}\), so that for any \(x \in W\) there exists \(\delta > 0\) with \((x - \delta, x + \delta) \cap S \subseteq W\).

Case \(c \in U\). Openness of \(U\) in \(S\) gives \(\delta > 0\) with \((c - \delta, c + \delta) \cap S \subseteq U\). By the definition of \(c\) as a supremum, there exists \(t_0 \in (c - \delta, c]\) with \([a, t_0] \subseteq U\). Combining this with \([t_0, c] \subseteq (c - \delta, c + \delta) \cap S \subseteq U\) gives \[ [a, c] \;=\; [a, t_0] \cup [t_0, c] \;\subseteq\; U. \] If \(c = b\), then \(b \in U\); but \(b \in V\) by choice, contradicting \(U \cap V = \emptyset\). If \(c < b\), shrink \(\delta\) if needed so that \(c + \delta \le b\); then \([c, c + \delta/2] \subseteq (c - \delta, c + \delta) \cap S \subseteq U\), and together with \([a, c] \subseteq U\) we obtain \([a, c + \delta/2] \subseteq U\). This makes \(c + \delta/2\) an element of \(\{t \in [a,b] : [a,t] \subseteq U\}\) strictly larger than \(c\), contradicting the fact that \(c\) is the supremum.

Case \(c \in V\). Openness of \(V\) in \(S\) gives \(\delta > 0\) with \((c - \delta, c + \delta) \cap S \subseteq V\). Note \(a \ne c\) (since \(a \in U\), \(c \in V\), and \(U \cap V = \emptyset\)), so \(a < c\). By the definition of \(c\) as a supremum, there exists \(t_0 \in (c - \delta, c]\) with \([a, t_0] \subseteq U\); in particular \(t_0 \in U \subseteq S\). Combining \(t_0 \in (c - \delta, c]\) with \(t_0 \in S\) gives \(t_0 \in (c - \delta, c + \delta) \cap S \subseteq V\), so \(t_0 \in U \cap V = \emptyset\), a contradiction.

Both cases are impossible, so no such separation exists and \(S\) is connected.

(\(\Rightarrow\)) Conversely, suppose \(S\) is connected, and for contradiction suppose \(S\) is not an interval. Then there exist \(a, b \in S\) and \(c \in \mathbb{R} \setminus S\) with \(a < c < b\). The sets \(U = S \cap (-\infty, c)\) and \(V = S \cap (c, \infty)\) are open in \(S\) (intersections of \(S\) with open subsets of \(\mathbb{R}\)), disjoint, and non-empty (\(a \in U\), \(b \in V\)). Their union is \(S \setminus \{c\} = S\) since \(c \notin S\). This is a separation, contradicting connectedness of \(S\). \(\square\)

Proof:

By the theorem that continuity preserves connectedness, \(f(X) \subseteq \mathbb{R}\) is connected. By the corollary classifying connected subsets of \(\mathbb{R}\) as intervals, \(f(X)\) is an interval. Since \(\alpha \in (\inf f(X), \sup f(X))\), the defining properties of infimum and supremum give points \(a, b \in f(X)\) with \(a < \alpha < b\); the interval \(f(X)\) then contains \(\alpha\), so there exists \(z \in X\) with \(f(z) = \alpha\). \(\square\)

Proof Sketch:

The argument rests on one key fact, which we record for reuse:

Union Lemma. If \(\{C_i\}_{i \in I}\) is a family of connected subsets of \(X\) sharing a common point \(p\), then \(\bigcup_{i} C_i\) is connected.

Why it holds: suppose for contradiction that the union admits a separation \(\bigcup_i C_i = U \sqcup V\) into disjoint non-empty open sets. The common point \(p\) lies in exactly one piece, say \(p \in U\). Since \(V\) is non-empty, pick \(q \in V\); then \(q \in C_{i_0}\) for some index \(i_0\), so \(C_{i_0} \cap V\) is non-empty, and \(C_{i_0} \cap U\) contains \(p\) and is also non-empty. The pair \((C_{i_0} \cap U,\, C_{i_0} \cap V)\) is therefore a separation of \(C_{i_0}\), contradicting the connectedness of \(C_{i_0}\).

Cover. For each \(x \in X\), let \(C(x)\) denote the union of all connected subsets of \(X\) containing \(x\). This family is non-empty since \(\{x\}\) is trivially connected, and all its members share the common point \(x\), so the Union Lemma gives \(C(x)\) connected. By construction, every connected set containing \(x\) is a subset of \(C(x)\); hence \(C(x)\) admits no strictly larger connected superset, making it a connected component. Every point lies in its own component, so \(X = \bigcup_{x \in X} C(x)\).

Disjointness. Suppose two components \(C(x)\) and \(C(y)\) intersect at a point \(p\). Both contain \(p\), so the Union Lemma applies to \(\{C(x), C(y)\}\): the union \(C(x) \cup C(y)\) is connected and contains \(C(x)\). Maximality of \(C(x)\) as a component forbids a strictly larger connected superset, so \(C(x) \cup C(y) = C(x)\), i.e. \(C(y) \subseteq C(x)\). The symmetric argument gives \(C(x) \subseteq C(y)\), hence \(C(x) = C(y)\).

Closedness. We claim the closure of any connected set is connected. Suppose for contradiction that \(\overline{C} = U \sqcup V\) is a separation into non-empty disjoint sets open in \(\overline{C}\). Then \(C \cap U\) and \(C \cap V\) are disjoint and open in \(C\). Moreover, both are non-empty: given \(q \in U\), openness of \(U\) in \(\overline{C}\) yields \(\varepsilon > 0\) with \(B(q, \varepsilon) \cap \overline{C} \subseteq U\); since \(q \in \overline{C}\) means \(\text{dist}(q, C) = 0\), the ball \(B(q, \varepsilon)\) contains a point of \(C\), which then lies in \(C \cap U\). Symmetrically for \(V\). Their union is \(C\), giving a separation of \(C\) and contradicting connectedness. Applying this to a component \(C\): \(\overline{C}\) is connected and contains \(C\), so maximality forces \(\overline{C} = C\), making \(C\) closed. \(\square\)

Proof:

Suppose \(X\) is a pathwise connected metric space. We assume for the sake of contradiction that \(X\) is disconnected.

According to the Open Union Criterion, there exists a separation \(X = U \cup V\), where \(U\) and \(V\) are disjoint, non-empty open sets. Since \(U\) and \(V\) are non-empty, we can pick arbitrary points \(a \in U\) and \(b \in V\). Because \(X\) is pathwise connected, there exists a continuous path \(f: [0, 1] \to X\) such that \(f(0) = a\) and \(f(1) = b\).

The closed interval \([0,1]\) is connected (connected subsets of \(\mathbb{R}\) are intervals), and \(f\) is continuous, so by the theorem that continuity preserves connectedness, the image \(K = f([0,1])\) is connected. Therefore \(K\) admits no separation. In particular, it cannot be written as \[ K = (K \cap U) \cup (K \cap V) \] with \(K \cap U\) and \(K \cap V\) disjoint non-empty open subsets of \(K\).

However, since \(a \in K \cap U\) and \(b \in K \cap V\), both intersections are non-empty. Moreover \(K \cap U\) and \(K \cap V\) are open in \(K\) (intersections of \(K\) with open subsets of \(X\)) and disjoint. Their union is \(K\) since \(K \subseteq X = U \cup V\). This gives a separation of \(K\), contradicting the connectedness of \(K\) established above. Hence no separation of \(X\) can exist, and \(X\) must be connected. \(\square\)