J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 23: Connected Spaces

Definition

Let \(X\) be a topological space. A separation of \(X\) is a pair of disjoint nonempty open sets \(U\), \(V\) such that \(U \cup V = X\). The space \(X\) is connected iff there does not exist a separation of \(X\).

Lemma 23.1

Let \(X\) be a topological space. Let \(Y\) be a subspace of \(X\). A separation of \(Y\) is a pair of disjoint nonempty sets \(A\) and \(B\) whose union is \(Y\), neither of which contains a limit point (in \(X\)) of the other.

Example 4

The rationals are not connected. \(\{q \in \mathbb{Q} \mid q^2 < 2\}\) and \(\{q \in \mathbb{Q} \mid q^2 > 2\}\) form a separation.

Lemma 23.2

Let \(X\) be a topological space. If \(U\) and \(V\) form a separation of \(X\), and \(Y\) is a connected subspace of \(X\), then \(Y \subseteq U\) or \(Y \subseteq V\).

Theorem 23.3

Let \(X\) be a topological space. The union of a set of connected subspaces of \(X\) that have a point in common is connected.

Theorem 23.4

Let \(X\) be a topological space. Let \(A\) be a connected subspace of \(X\) and \(B\) be a subspace of \(X\). If \(A \subseteq B \subseteq \overline{A}\) then \(B\) is connected.

Theorem 23.5

The continuous image of a connected space is connected.

Theorem 23.6

The product of two connected spaces is connected.

Exercise 10

The product of a family of connected spaces is connected.

Proof

  • Let \(\{X_\alpha\}_{\alpha \in J}\) be a family of connected spaces.
  • Let \(X = \prod_{\alpha \in J} X_\alpha\).
  • Assume w.l.o.g. each \(X_\alpha\) is nonempty.
  • Pick \(a \in X\).
  • For every finite \(K \subseteq J\), let \(X_K = \{ x \in X \mid \forall \alpha \in J - K. x_\alpha = a_\alpha \} \).
  • For every finite \(K \subseteq J\), \(X_K\) is connected.
    • It is homeomorphic to \(\prod_{\alpha \in K} X_\alpha\).
  • Let \(Y = \bigcup_{K \text{ a finite subset of } J} X_K\).
  • \(Y\) is connected.
    • Theorem 23.3 since \(a\) is a common point.
  • \(X = \overline{Y}\)
    • Let \(x \in X\). Let \(U\) be any neighbourhood of \(x\). Prove: \(U\) intersects \(Y\).
    • Pick a family \(\{U_\alpha\}_{\alpha \in J}\) such that \(x \in \prod_\alpha U_\alpha \subseteq U\), \(U_\alpha\) is an open set in \(X_\alpha\) for all \(\alpha\), and \(U_\alpha = X_\alpha\) for all but finitely many \(\alpha\).
    • Let \(K = \{\alpha \in J \mid U_\alpha \neq X_\alpha\}\).
    • Let \(y\) be the point with \(y_\alpha = x_\alpha\) for \(\alpha \in K\) and \(y_\alpha = a_\alpha\) for \(\alpha \notin K\).
    • \(y \in U \cap X_K\)
    • \(y \in U \cap Y\)
  • \(X\) is connected.
    • Theorem 23.4
  • \(\Box\)

Example 6

\(\mathbb{R}^\omega\) in the box topology is not connected. The set of bounded sequences and the set of unbounded sequences form a separation.

The same argument shows that \(\mathbb{R}^\omega\) in the uniform topology is not connected.

Definition

A topological space is totally disconnected iff the only connected subspaces are the one-point sets.

Examples

Any discrete space is totally disconnected. The rationals are totally disconnected.

Exercise 11

Let \(p : X \rightarrow Y\) be a quotient map. If \(Y\) is connected and \(p^{-1}(y)\) is connected for all \(y \in Y\), then \(X\) is connected.

Comments

Post a Comment

Popular posts from this blog

J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 32: Normal Spaces

Theorem 32.1 Every second countable regular space is normal. Proof Let \(X\) be a second countable regular space. Pick a countable basis \(\mathcal{B}\). Let \(A\) and \(B\) be disjoint closed sets in \(X\). Pick a countable covering \(\{U_1, U_2, \ldots\}\) of \(A\) by open sets whose closures are disjoint from \(B\). Let \(\{U_1, U_2, \ldots\} = \{U \in \mathcal{B} \mid \overline{U} \cap B = \emptyset\}\). Prove: this covers \(A\). Let \(x \in A\). Prove: there exists \(U \in \mathcal{B}\) such that \(x \in U\) and \(\overline{U} \cap B = \emptyset\). Pick disjoint open sets \(U\) and \(V\) such that \(x \in U\) and \(B \subseteq V\). By regularity. Pick \(U' \in \mathcal{B}\) such that \(x \in U' \subseteq U\). \(x \in U'\) and \(\overline{U'} \cap B \subseteq \overline{U} \cap B \subseteq (X - V) \cap B = \emptyset\) Pick a countable covering \(\{V_1, V_2, \ldots\}\) of \(B\) by open sets whose ...
Read more

J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 27: Compact Subspaces of the Real Line

Theorem 27.1 Let \(X\) be a linearly ordered set with the least upper bound property in the order topology. Every closed interval in \(X\) is compact. Proof Let \(a,b \in X\) with \(a Let \(\mathcal{A}\) be a covering of \([a,b]\) by sets open in \(X\). \( \langle 1 \rangle 1 \) For any \(x \in [a,b)\), there exists \(y \in (x,b]\) such that \([x,y]\) can be covered by at most two elements of \(\mathcal{A}\). Let \(x \in [a,b)\). Pick \(U \in \mathcal{A}\) such that \(x \in U\). Pick \(y > x\) such that \([x,y) \subseteq U\). Assume w.l.o.g. \(y \leq b\). Pick \(V \in \mathcal{A}\) such that \(y \in V\). \([x,y]\) is covered by \(U\) and \(V\). Let \(C = \{y \in (a,b] \mid [a,y] \text{ can be covered by finitely many elements of } \mathcal{A} \}\). Let \(c = \sup C\). \(C\) is nonempty. Pick \(y \in (a,b]\) such that \([a,y]\) can be covered by at most two elements of \(\mathcal{A}\). \(\...
Read more

J. Munkres. Topology (2013) Chapter 4: Countability and Separation Axioms. 31: The Separation Axioms

Definition A topological space \(X\) is regular iff it is \(T_1\) and, for any point \(x \in X\) and closed set \(B\) such that \(x \notin B\), there exist disjoint open sets \(U\) and \(V\) with \(x \in U\) and \(B \subseteq V\). A topological space \(X\) is normal iff it is \(T_1\) and, for any disjoint closed sets \(A\) and \(B\), there exist disjoint sets \(U\) and \(V\) with \(A \subseteq U\) and \(B \subseteq V\). Lemma 31.1 Let \(X\) be a \(T_1\) space. \(X\) is regular if and only if, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such that \(\overline{V} \subseteq U\). \(X\) is normal if and only if, for every closed set \(A\) and open set \(U\) including \(A\), there exists an open set \(V\) including \(A\) such that \(\overline{V} \subseteq U\). Proof If \(X\) is regular then, for every point \(x \in X\) and neighbourhood \(U\) of \(x\), there exists a neighbourhood \(V\) of \(x\) such th...
Read more