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. This is exactly what (3) forbids. In such a partition, each piece is also closed, being the complement of the other open piece. Hence (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).

Now let \(U\) be any proper non-empty subset of \(X\) and consider its 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. It follows that 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 preimages \(f^{-1}(\{0\})\) and \(f^{-1}(\{1\})\), which partition \(X\) and are themselves clopen because continuity is equivalent to openness of preimages. 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 criteria for connectedness). We may assume \(X \neq \emptyset\), since otherwise \(f(X) = \emptyset\) is connected. Let \(g: f(X) \to \{0,1\}\) be any continuous function into the discrete two-point space. Viewed as a map onto the metric subspace \(f(X)\), the function \(f\) is still continuous, so \(g \circ f : X \to \{0,1\}\) 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, we conclude that \(f(X)\) is connected.

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 \(a \lt b\). 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 \(\delta \gt 0\) with \((x - \delta, x + \delta) \cap S \subseteq W\).

Case \(c \in U\).
Openness of \(U\) in \(S\) gives \(\delta \gt 0\) 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 \(c \lt b\), 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 \(\delta \gt 0\) 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 \(a \lt c\). 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 \(a \lt c \lt b\). 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 \(S\).

Proof:

By the theorem that continuity preserves connectedness, \(f(X) \subseteq \mathbb{R}\) is connected. The hypothesis on \(\alpha\) forces \(\inf f(X) \lt \sup f(X)\), so \(f(X)\) is non-empty, and the theorem classifying connected subsets of \(\mathbb{R}\) as intervals shows that \(f(X)\) 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 \(a \lt \alpha \lt b\). The interval \(f(X)\) then contains \(\alpha\), so there exists \(z \in X\) with \(f(z) = \alpha\).

Proof:

Suppose for contradiction that the union admits a separation \(\bigcup_i C_i = U \sqcup V\) into disjoint non-empty sets open in \(\bigcup_i C_i\). 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 \((C_{i_0} \cap U,\, C_{i_0} \cap V)\) therefore consists of disjoint non-empty sets open in \(C_{i_0}\) whose union is \(C_{i_0}\), so it is a separation of \(C_{i_0}\), contradicting the connectedness of \(C_{i_0}\).

Proof:

Suppose for contradiction that \(D = U \sqcup V\) is a separation into non-empty disjoint sets open in \(D\). Then \(C \cap U\) and \(C \cap V\) are disjoint, their union is \(C\), and they are open in \(C\), being intersections of \(C\) with sets open in \(D\). Both are non-empty. Given \(q \in U\), openness of \(U\) in \(D\) yields \(\varepsilon \gt 0\) with \(B(q, \varepsilon) \cap D \subseteq U\), a ball of \(D\) being the trace on \(D\) of the ball of \(X\). Since \(q \in D \subseteq \overline{C} = C \cup \partial C\), either \(q \in C\) or \(q\) is a boundary point of \(C\), and in both cases \(\operatorname{dist}(q, C) = 0\), so the ball \(B(q, \varepsilon)\) contains a point of \(C\). That point lies in \(D\), hence in \(U\), hence in \(C \cap U\). Symmetrically \(C \cap V\) is non-empty. This gives a separation of \(C\), contradicting its connectedness. Taking \(D = \overline{C}\) gives the final statement.

Proof:

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 lemma on unions with a common point 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 \(X = \bigcup_{x \in X} C(x)\). Conversely, every component arises this way. When \(X \neq \emptyset\) a component \(D\) cannot be empty, since any singleton would be a strictly larger connected superset, and choosing \(z \in D\) gives \(D \subseteq C(z)\), so maximality of \(D\) forces \(D = C(z)\).

Disjointness.
Suppose two components \(C(x)\) and \(C(y)\) intersect at a point \(p\). Both contain \(p\), so the same 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 \(C(x) \cup C(y) = C(x)\), that is, \(C(y) \subseteq C(x)\). The symmetric argument gives \(C(x) \subseteq C(y)\), hence \(C(x) = C(y)\).

Closedness.
Let \(C\) be a component. By the lemma on sets between a connected set and its closure, its closure \(\overline{C}\) is connected, and it contains \(C\), so maximality forces \(\overline{C} = C\), making \(C\) closed.

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.

Proof:

Let \(U \subseteq \mathbb{R}^n\) be open and connected. There is nothing to prove if \(U = \emptyset\), so fix \(p \in U\) and let \(A\) be the set of points of \(U\) joined to \(p\) by a path in \(U\). It contains \(p\), by the constant path. Two paths can be run one after the other: if \(f\) is a path from \(a\) to \(b\) and \(g\) a path from \(b\) to \(c\), then the map equal to \(f(2t)\) on \([0, \tfrac12]\) and to \(g(2t - 1)\) on \([\tfrac12, 1]\) is a path from \(a\) to \(c\). It is continuous: away from \(t = \tfrac12\) each half is continuous, and at \(t = \tfrac12\), where the halves agree, each half supplies a \(\delta\) for a given \(\varepsilon\) and the smaller of the two works. Reversing the parameter, \(t \mapsto f(1 - t)\) is a path from \(b\) to \(a\).

Now let \(x \in U\). Since \(U\) is open there is \(r \gt 0\) with \(B(x, r) \subseteq U\). Every \(y \in B(x, r)\) is joined to \(x\) by the segment \(t \mapsto (1 - t)x + ty\), which is continuous and stays in the ball because \(|(1 - t)x + ty - x| = t\,|y - x| \lt r\), and back to \(x\) by its reversal. If \(x \in A\), running a path from \(p\) to \(x\) and then the segment to \(y\) shows \(y \in A\), so \(B(x, r) \subseteq A\). If some \(y \in B(x, r)\) lay in \(A\), running a path from \(p\) to \(y\) and then the segment back to \(x\) would put \(x\) in \(A\), so \(B(x, r)\) misses \(A\) when \(x \notin A\). Thus \(A\) and \(U \setminus A\) are both open, and they are disjoint with union \(U\). Since \(U\) is connected and \(A\) is non-empty, the Open Union Criterion forces \(U \setminus A = \emptyset\), that is, \(A = U\). Finally, any two points \(a, b \in U\) are joined by running the reversal of a path from \(p\) to \(a\) and then a path from \(p\) to \(b\).