J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 25: Components and Local Connectedness

Definition

Let \(X\) be a topological space. Define an equivalence relation \(\sim\) on \(X\) by: \(x \sim y\) iff there exists a connected subspace \(C\) of \(X\) such that \(x \in C\) and \(y \in C\). The (connected) components of \(X\) are the equivalence classes of \(X\) under this equivalence relation.

Theorem 25.1

The components of \(X\) are the maximal connected subspaces of \(X\).

Definition

Let \(X\) be a topological space. Define an equivalence relation \(\sim\) on \(X\) by: \(x \sim y\) iff there exists a path in \(X\) from \(x\) to \(y\). The path components of \(X\) are the equivalence classes of \(X\) under this equivalence relation.

Theorem 25.2

The path components of \(X\) are the maximal path connected subspaces of \(X\).

Proposition

Every component is closed.

Proof

Since the closure of a connected subspace is connected.

Definition

Let \(X\) be a topological space. For \(x \in X\), \(X\) is locally (path) connected at \(x\) iff, for every neighbourhood \(U\) of \(x\), there exists a (path) connected neighbourhood \(V\) of \(x\) such that \(V \subseteq U\). \(X\) is locally (path) connected iff it is locally (path) connected at every point.

Example 3

  • \(\mathbb{R}\) is connected and locally connected.
  • \(\mathbb{R} - \{0\)\) is disconnected but locally connected.
  • The topologist's sine curve is connected but not locally connected.
  • \(\mathbb{Q}\) is neither connected nor locally connected.

Theorem 25.3

Let \(X\) be a topological space. Then \(X\) is locally connected if and only if, for every open set \(U\) in \(X\), every component of \(U\) is open in \(X\).

Proof

  • If \(X\) is locally connected then, for every open set \(U\) in \(X\), every component of \(U\) is open in \(X\).
    • Assume \(X\) is locally connected.
    • Let \(U\) be an open set in \(X\).
    • Let \(C\) be a component of \(U\). Prove: \(C\) is open in \(X\).
    • Let \(x \in C\). Prove: there exists a neighbourhood \(V\) of \(x\) such that \(V \subseteq C\).
    • Pick a connected neighbourhood \(V\) of \(x\) such that \(V \subseteq U\).
    • \(V \subseteq C\)
  • If, for every open set \(U\) in \(X\), every component of \(U\) is open in \(X\), then \(X\) is locally connected.
    • Assume that, for every open set \(U\) in \(X\), every component of \(U\) is open in \(X\).
    • Let \(x \in X\). Let \(U\) be a neighbourhood of \(x\).
    • Let \(V\) be the component of \(U\) that contains \(x\).
    • \(V\) is open in \(X\).
    • \(V\) is a connected neighbourhood of \(x\) such that \(V \subseteq U\).
  • \(\Box\)

Theorem 25.4

Let \(X\) be a topological space. Then \(X\) is locally path connected if and only if, for every open set \(U\) in \(X\), every path component of \(U\) is open in \(X\).

Proof

Similar.

Theorem 25.5

In a locally path connected space, the components and the path components are the same.

Proof

  • Let \(X\) be a locally path connected space.
  • Let \(P\) be a path component of \(X\).
  • Let \(C\) be the component of \(X\) that includes \(P\). Prove: \(P = C\).
  • Let \(Q\) be the union of all the other path components included in \(C\).
  • \(P\) and \(Q\) are open.
    • Theorem 25.4
  • \(Q = \emptyset\)
    • Otherwise \(P\) and \(Q\) would form a separation of \(C\).
  • \(\Box\)

Exercises

Exercise 1

What are the components and path components of \(\mathbb{R}_l\)? What are the continuous maps \(\mathbb{R} \rightarrow \mathbb{R}_l\)?

Solution

The components of \(\mathbb{R}_l\) are the one-point sets (i.e. \(\mathbb{R}_l\) is totally disconnected). For let \(A \subseteq \mathbb{R}_l\) have more than one point. Pick \(a,b \in A\) with \(a < b\). Then \(A \cap (-\infty, b)\) and \(A \cap [b, + \infty)\) form a separation of \(A\).

Therefore, the path components of \(\mathbb{R}_l\) are the one-point sets.

Therefore, the only continuous maps \(\mathbb{R} \rightarrow \mathbb{R}_l\) are the constant maps. For, if \(f : \mathbb{R} \rightarrow \mathbb{R}_l\) is continuous and we have \(a,b \in \mathbb{R}\) with \(a < b\) and \(f(a) \neq f(b)\), then \(f \restriction [a,b]\) would be a path from \(a\) to \(b).

Exercise 2

(a)

What are the components and path components of \(\mathbb{R}^\omega\)?

Solution

Since \(\mathbb{R}^\omega\) is connected (Example 23.7), it has only one component \(\mathbb{R}^\omega\).

We also have \(\mathbb{R}^\omega\) is path connected. Let \(a,b \in \mathbb{R}^\omega\). Define \(f : [0,1] \rightarrow \mathbb{R}^\omega\) by \(f(t)_n = (1-t)a_n + tb_n\). Then \(f\) is continuous (Theorem 19.6) and hence is a path from \(a\) to \(b\). Thus, \(\mathbb{R}^\omega\) has only one path component.

(b)

Consider \(\mathbb{R}^\omega\) in the uniform topology. Show that \(\vec{x}\) and \(\vec{y}\) lie in the same component of \(\mathbb{R}^\omega\) if and only if the sequence \(\vec{x} - \vec{y}\) is bounded.

Solution

  • The component containing \(\vec{0}\) is the set of bounded sequences.
    • The set of bounded sequences is connected.
      • It is path connected, with the straight line path joining any two points.
    • It is a maximal connected subspace.
      • If \(A\) is any set containing both bounded and unbounded sequences, then \(\{\vec{x} \in A \mid \vec{x} \text{ is bounded} \}\) and \(\{\vec{x} \in A \mid \vec{x} \text{ is unbounded} \}\) form a separation of \(A\).
  • For any \(\vec{x} \in \mathbb{R}^\omega\), the component containing \(\vec{x}\) is the set of all sequences \(\vec{y}\) such that \(\vec{y} - \vec{x}\) is bounded.
    • Since the function mapping any sequence \(\vec{y}\) to \(\vec{y} + \vec{x}\) is a homeomorphism of \(\mathbb{R}^\omega\) uniform with itself.
  • \(\Box\)

(c)

Give \(\mathbb{R}^\omega\) the box topology. Show that \(\vec{x}\) and \(\vec{y}\) lie in the same component of \(\mathbb{R}^\omega\) if and only if the sequence \(\vec{x} - \vec{y}\) is eventually zero.

Solution

  • The component containing \(\vec{0}\) is the set of all sequences that are eventually zero.
    • The set of all sequences that are eventually zero is connected.
      • It is path connected, with the straight line path joining any two points.
    • If a set contains both \(\vec{0}\) and a sequence that is not eventually zero, then it is disconnected.
      • Let \(A\) be a set with \(\vec{0} \in A\) and \(\vec{x} \in A\) where \(\vec{x}\) is not eventually zero.
      • Define \(h : \mathbb{R}^\omega \rightarrow \mathbb{R}^\omega\) by \(h(\vec{y})_n = y_n\) if \(x_n = 0\), and \(h(\vec{y})_n = n y_n / x_n\) if \(x_n \neq 0\).
      • \(h\) is a homeomorphism of \(\mathbb{R}^\omega\) with itself.
      • \(h(\vec{x})\) is unbounded.
      • \(h(A)\) is disconnected.
        • The set of bounded sequences in \(h(A)\) and the set of unbounded sequences in \(h(A)\) form a separation.
      • \(A\) is disconnected.
  • For any \(\vec{x} \in \mathbb{R}^\omega\), the component containing \(\vec{x}\) is the set of all sequences \(\vec{y}\) such that \(\vec{y} - \vec{x}\) is eventually zero.
    • Since the function mapping any vector \(\vec{y}\) to \(\vec{y} + \vec{x}\) is a homeomorphism of \(\mathbb{R}^\omega\) with itself.
  • \(\Box\)

Exercise 3

Show that the ordered square is locally connected but not locally path connected. What are the path components of this space?

Solution

For any point \(x\) and any neighbourhood \(U\) of \(x\), pick an interval \(I\) such that \(x \in I \subseteq U\). Then \(I\) is a linear continuum, hence connected. (This in fact shows that every linear continuum in the order topology is locally connected.)

If the ordered square is locally path connected then it would have only one path component (Theorem 25.5) and so be path connected, which it is not.

The path components are the subsets of the form \(\{a\} \times [0,1]\) for \(a \in [0,1]\). These subsets are clearly path connected. There is no path from \((a,s)\) to \((b,t)\) if \(a \neq b\) by the same argument as Example 24.7.

Exercise 4

Let \(X\) be locally path connected. Show that every connected open subspace of \(X\) is path connected.

Solution

  • Let \(U\) be a connected open subspace of \(X\).
  • The path components of \(U\) are open in \(X\).
    • Theorem 25.4
  • The path components of \(U\) are open in \(U\).
  • \(U\) has only one path component.
    • Otherwise, for any path component \(P\), we would have \(P\) and the union of the other path components form a separation of \(U\).
  • \(\Box\)

Exercise 8

A quotient of a locally connected space is locally connected.

Solution

  • Let \(p : X \rightarrow Y\) be a quotient map.
  • Assume \(X\) is locally connected.
  • Let \(U\) be open in \(Y\) and \(C\) a component of \(U\). Prove: \(C\) is open in \(Y\).
  • \(p^{-1}(U)\) is open in \(X\).
  • \(p^{-1}(C)\) is a union of components of \(p^{-1}(U)\).
    • If \(D\) is a component of \(p^{-1}(U)\) that intersects \(p^{-1}(C)\), then \(p(D)\) is connected and intersects \(C\), hence \(p(D) \subseteq C\) and so \(D \subseteq p^{-1}(C)\).
  • \(p^{-1}(C)\) is open in \(X\).
    • Since every component of \(p^{-1}(U)\) is open by Theorem 25.3.
  • \(C\) is open in \(Y\).
  • \(\Box\)
    • By Theorem 25.3

Exercise 9

Let \(G\) be a topological group. Then the component that contains \(e\) is a normal subgroup of \(G\).

Solution

  • Let \(C\) be the component that contains \(e\).
  • For all \(x \in G\), we have \(xC\) is the component that contains \(x\).
    • \(xC\) is connected.
      • Since the function that maps \(y\) to \(xy\) is a homeomorphism of \(G\) with itself.
    • Let \(D\) be the component such that \(xC \subseteq D\).
    • \(C \subseteq x^{-1}D\)
    • \(C = x^{-1}D\)
      • Since \(x^{-1}D\) is connected because the function that maps \(y\) to \(xy\) is a homeomorphism of \(G\) with itself.
    • \(D = xC\)
  • For all \(x \in G\), we have \(Cx\) is the component that contains \(x\).
    • Similar.
  • \(\forall x \in G. xC = Cx\)
  • \(\Box\)

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