J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 24: Connected Subspaces of the Real Line

Definition

A linear continuum is a linearly ordered set with more than one element that is dense and complete.

Proposition

A convex subset of a linear continuum that has more than one element is a linear continuum.

Theorem 24.1

A linear continuum under the order topology is connected.

Proof

  • Let \(L\) be a linear continuum.
  • Assume for a contradiction \(U\) and \(V\) form a separation of \(L\).
  • Pick \(a \in U\) and \(b \in V\).
  • Assume w.l.o.g. \(a < b\).
  • Let \(c = \sup (U \cap [a,b])\)
  • Case \(c \in U\)
    • Pick \(e > c\) such that \([c,e) \subseteq U\).
    • \(e \leq b\)
    • Pick \(f\) such that \(c < f < e\).
    • \(f \in U\)
    • This contradicts the fact that \(c\) is an upper bound for \(U \cap [a,b]\).
  • Case \(c \in V\)
    • Pick \(e < c\) such that \((e,c] \subseteq V\).
    • \(e\) is an upper bound for \(U \cap [a,b]\).
      • Let \(x \in U \cap [a,b]\)
      • \(x \leq c\)
      • \(x \notin (e,c]\)
      • \(x \leq e\)
    • This contradicts the leastness of \(c\).
  • \(\Box\)

Corollary 24.2

The real line \(\mathbb{R}\) is connected, and so are intervals and rays in \(\mathbb{R}\).

Theorem 24.3 (Intermediate Value Theorem)

Let \(X\) be a connected space. Let \(Y\) be a linearly ordered set under the order topology. Let \(f : X \rightarrow Y\) be continuous. Let \(a,b \in X\) and \(r \in Y\). If \(f(a) < r < f(b)\), then there exists \(c \in X\) such that \(f(c) = r\).

Proof

Otherwise \(\{y \in f(X) \mid y < r\}\) and \(\{y \in f(X) \mid y > r\}\) would form a separation of \(f(X)\).

Example 1

The ordered square is a linear continuum.

Definition

Let \(X\) be a topological space and \(a,b \in X\). A path from \(a\) to \(b\) is a continuous function \(p : [0,1] \rightarrow X\) such that \(p(0) = a\) and \(p(1) = b\). The space \(X\) is path connected iff, for any two points \(a,b \in X\), there exists a path from \(a\) to \(b\).

Proposition

Every path connected space is connected.

Proposition

The continuous image of a path connected space is path connected.

Example 6

The ordered square is connected but not path connected. It is connected by Theorem 24.1, but there is no path from \((0,0)\) to \((1,1)\).

Definition

The unit ball \(B^n\) in \(\mathbb{R}^n\) is \(\{\vec{x} \in \mathbb{R}^n \mid \| x \| \leq 1 \}\).

Example 3

Every open ball and every closed ball in \(\mathbb{R}^n\) is path connected. Any two points are connected by the straight line path.

Definition

Punctured Euclidean space is \(\mathbb{R}^n - \{\vec{0}\}\).

Example 4

If \(n > 1\) then \(\mathbb{R}^n - \{\vec{0}\}\) is path connected.

Definition

The unit sphere \(S^{n-1} \subseteq \mathbb{R}^n\) is \(\{ \vec{x} \in \mathbb{R}^n \mid \| x \| = 1 \}\).

Example 5

If \(n > 1\) then \(S^{n-1}\) is path connected. It is the continuous image of \(\mathbb{R}^n - \{\vec{0}\}\) under the function that maps \(\vec{x}\) to \(\vec{x} / \| \vec{x} \|\).

Definition

The topologist's sine curve is \( \{(x, \sin (1/x)) \mid 0 < x \leq 1\} \cup (\{0\} \times [-1,1])\).

Example 6

The topologist's sine curve is connected but not path connected.

It is connected because it is the closure of \(\{(x, \sin (1/x)) \mid 0 < x \leq 1\}\), which is the continuous image of \((0,1]\). But there is no path from \((0,0)\) to \((1, \sin 1)\).

Definition

The long line is \(\Omega \times [0,1)\) under the dictionary order, where \(\Omega\) is the first uncountable ordinal.

Proposition

The long line is path connected and locally homeomorphic to \(\mathbb{R}\) (i.e. every point has a neighbourhood homeomorphic with an open subset of \(\mathbb{R}\)), but is not embeddable in \(\mathbb{R}\).

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