J. Munkres. Topology (2013) Chapter 3: Connectedness and Compactness. 28: Limit Point Compactness

Definition

A topological space is limit point compact iff every infinite subset has a limit point.

Theorem 28.1

Every compact space is limit point compact.

Proof

  • Let \(X\) be a compact space.
  • Let \(A \subseteq X\) have no limit points. Prove: \(A\) is finite.
  • \(A\) is closed.
  • \(A\) is compact.
  • For all \(a \in A\), there exists an open set \(U\) such that \(U \cap A = \{a\}\).
  • \(\{U \text{ open in } X \mid U \cap A \text{ is a singleton} \}\) is a covering of \(A\) by sets open in \(X\).
  • Pick a finite subcovering \(\{U_1, \ldots, U_n\}\).
  • \(A = (U_1 \cap A) \cup \cdots \cup (U_n \cap A)\) has cardinality \(\leq n\).
  • \(\Box\)

Example 1

Let \(Y\) be a two-point set with the indiscrete topology. Then \(\mathbb{Z}_+ \times Y\) is limit point compact but not compact.

Example 2

Let \(\Omega\) be the first uncountable ordinal in the order topology. Then \(\Omega\) is limit point compact but not compact.

Definition

A topological space is sequentially compact iff every sequence has a convergent subsequence.

Theorem 28.2

Let \(X\) be a metrizable space. Then the following are equivalent.

  1. \(X\) is compact.
  2. \(X\) is limit point compact.
  3. \(X\) is sequentially compact.

Proof

  • \(1 \Rightarrow 2\)
    • Theorem 28.1.
  • \(2 \Rightarrow 3\)
    • Assume \(X\) is limit point compact.
    • Let \((a_n)\) be a sequence in \(X\). Prove: \((a_n)\) has a convergent subsequence.
    • Case \(\{a_n \mid n \in \mathbb{Z}_+\}\) is finite.
      • Some element occurs infinitely often. The subsequence of all its occurrences converges.
    • Case \(\{a_n \mid n \in \mathbb{Z}_+\}\) is infinite.
      • Pick a limit point \(l\).
      • For \(n \in \mathbb{Z}_+\), pick \(r_n > r_1, \ldots, r_{n-1}\) such that \(d(a_{r_n},l) < 1/n\).
      • \(a_{r_n} \rightarrow l\) as \(n \rightarrow \infty\).
  • \(3 \Rightarrow 1\)
    • \( \langle 2 \rangle 1 \) Assume \(X\) is sequentially compact.
    • Every open covering of \(X\) has a Lebesgue number.
      • Let \(\mathcal{A}\) be an open covering of \(X\).
      • Assume for a contradiction \(\mathcal{A}\) has no Lebesgue number.
      • \( \langle 3 \rangle 1 \) For \(n \in \mathbb{Z}_+\), pick a set \(C_n\) of diameter \(< 1/n\) that is not included in any member of \(\mathcal{A}\).
      • For \(n \in \mathbb{Z}_+\), pick a point \(x_n \in C_n\).
      • Pick a convergent subsequence \((x_{n_r})\) of \((x_n)\).
        • From \(\langle 2 \rangle 1\).
      • Let \(a = \lim_{r \rightarrow \infty} x_{n_r}\).
      • Pick \(A \in \mathcal{A}\) such that \(a \in A\).
      • Pick \(\epsilon > 0\) such that \(B(a, \epsilon) \subseteq A\).
      • Pick \(r\) such that \(1/n_r < \epsilon / 2\) and \(d(x_{n_r},a) < \epsilon / 2\).
      • \(C_{n_r} \subseteq B(x_{n_r}, \epsilon/2)\)
      • \(C_{n_r} \subseteq B(a, \epsilon)\)
      • \(C_{n_r} \subseteq A\)
      • This contradicts \(\langle 3 \rangle 1 \).
    • For all \(\epsilon > 0\), there is a finite covering of \(X\) by \(\epsilon\)-balls.
    • Let \(\mathcal{A}\) be an open covering of \(X\).
    • Pick a Lebesgue number \(\delta\) for \(\mathcal{A}\).
    • Let \(\epsilon = \delta / 3\).
    • Pick a finite covering \(\mathcal{C}\) of \(X\) by \(\epsilon\)-balls.
    • For \(C \in \mathcal{C}\), pick \(U_C \in \mathcal{A}\) such that \(C \subseteq U_C\).
    • \(\{U_C \mid C \in \mathcal{C}\}\) is a finite subcovering of \(\mathcal{A}\).
  • \(\Box\)

Example 3

Let \(\Omega\) be the least uncountable ordinal. \(\Omega + 1\) is not metrizable in the order topology, because \(\Omega\) is a limit point of \(\{\alpha \mid \alpha < \Omega\}\) but there is no sequence of points \(< \Omega\) that converges to \(\Omega\).

\(\Omega\) is not metrizable because it is limit point compact but not compact.

Exercises

Exercise 1

Give \([0,1]^\omega\) the uniform topology. Find an infinite subset of this space that has no limit point.

Solution

The set of all sequences that are all zero except for one 1 has no limit point.

Exercise 2

Show that \([0,1]\) is not limit point compact as a subspace of \(\mathbb{R}_l\).

Solution

The set \(\{1 - 1/n \mid n \in \mathbb{Z}_+\}\) has no limit point.

Exercise 3

Let \(X\) be limit point compact.

(a)

If \(f : X \rightarrow Y\) is continuous, does it follow that \(f(X)\) is limit point compact?

Solution

No. Take \(X\) to be the space from Example 1. Define \(f : X \rightarrow \mathbb{Z}_+\) by \(f((x,a)) = x\). Then \(f\) is continuous but \(f(X) = \mathbb{Z}_+\) is not limit point compact.

(b)

If \(A\) is a closed subset of \(X\), does it follow that \(A\) is limit point compact?

Solution

  • Let \(B\) be an infinite subset of \(A\).
  • Pick a limit point \(l\) of \(B\) in \(X\).
  • \(l \in A\)
  • \(l\) is a limit point of \(B\) in \(A\).
  • \(\Box\)

(c)

If \(X\) is a subspace of the Hausdorff space \(Z\), does it follow that \(X\) is closed in \(Z\)?

Solution

No. \(\Omega\) is limit point compact but not closed in \(\Omega + 1\).

The product of two limit point compact Hausdorff spaces is not necessarily limit point compact, but the counterexample requires Stone-Cech compactification.

Exercise 4

A space \(X\) is said to be countably compact if every countable open covering of \(X\) includes a finite subset that covers \(X\). Show that for a \(T_1\) space \(X\), countable compactness is equivalent to limit point compactness.

Solution

  • Let \(X\) be a \(T_1\)-space.
  • If \(X\) is countably compact then \(X\) is limit point compact.
    • Assume \(X\) is countably compact.
    • Let \(A\) be an infinite subset of \(X\).
    • Pick a countable subset \(B\) of \(A\).
    • Assume w.l.o.g. no point in \(B\) is a limit point of \(A\).
    • For every \(b \in B\), choose a neighbourhood \(U_b\) of \(b\) that does not intersect \(A\) in any point except \(b\).
    • \(\{U_b \cap B \mid b \in B\}\) is a countable open covering of \(B\) that has no finite subcovering.
    • \(B\) is not countably compact.
    • Any closed subspace of a countably compact space is countably compact.
      • Similar to Theorem 26.2.
    • \(B\) is not closed.
    • \(B\) has a limit point.
      • Corollary 17.7
  • If \(X\) is limit point compact then \(X\) is countably compact.
    • Assume \(X\) is limit point compact.
    • Let \(\mathcal{U} = \{U_1, U_2, \ldots \}\) be a countable open covering of \(X\).
    • Assume for a contradiction no finite subset of \(\mathcal{U}\) covers \(X\).
    • Pick \(a_n \in X - (U_1 \cup \cdots \cup U_n)\) for all \(n \in \mathbb{Z}_+\).
    • \(\{a_n \mid n \in \mathbb{Z}_+\}\) is infinite.
    • Pick a limit point \(l\).
    • Pick \(n\) such that \(l \in U_n\).
    • \(U_n - \{a_i \mid i < n, a_i \neq l \}\) is a neighbourhood of \(l\).
    • There exists \(m\) such that \(a_m \in U_n - \{a_i \mid i
    • This is a contradiction.
  • \(\Box\)

Exercise 5

Show that \(X\) is countably compact if and only if every nested sequence \(C_1 \supseteq C_2 \supseteq \cdots\) of closed nonempty subsets of \(X\) has a nonempty intersection.

Solution

  • If \(X\) is countably compact then every nested sequence of closed nonempty subsets of \(X\) has a nonempty intersection.
    • Assume \(X\) is countably compact.
    • Let \(C_1 \supseteq C_2 \supseteq \cdots\) be a nested sequence of closed nonempty subsets of \(X\).
    • Assume for a contradiction \(\bigcap_n C_n = \emptyset\).
    • \(\{X - C_n \mid n \in \mathbb{Z}_+\}\) is a countable open covering of \(X\).
    • Pick a finite subcovering \(\{X - C_{n_1}, \ldots, X - C_{n_k}\}\).
    • Let \(N = \min(n_1, \ldots, n_k)\).
    • \(X - C_N = X\)
    • \(C_N = \emptyset\)
    • This is a contradiction.
  • If every nested sequence of closed nonempty subsets of \(X\) has nonempty intersection then \(X\) is countably compact.
    • Assume every nested sequence of closed nonempty subsets of \(X\) has nonempty intersection.
    • Let \(\{U_n \mid n \in \mathbb{Z}_+\}\) be a countable open covering of \(X\).
    • For \(n \in \mathbb{Z}_+\), let \(C_n = X - (U_1 \cup \cdots \cup U_n)\).
    • \(C_1 \supseteq C_2 \supseteq \cdots\)
    • \(\bigcap_n C_n = \emptyset\)
    • Pick \(n\) such that \(C_n = \emptyset\).
    • \(\{U_1, \ldots, U_n\}\) is a finite subcovering of \(X\).
  • \(\Box\)

Exercise 6

Let \((X,d)\) be a metric space. If \(f : X \rightarrow X\) satisfies the condition

\[ d(f(x),f(y)) = d(x,y) \]

for all \(x,y \in X\), then \(f\) is called an isometry of \(X\). Show that if \(f\) is an isometry and \(X\) is compact, then \(f\) is bijective and hence a homeomorphism.

Solution

  • \(f\) is injective.
    • If \(f(x) = f(y)\) then \(d(f(x),f(y)) = 0\) so \(d(x,y) = 0\) and \(x = y\).
  • \(f\) is surjective.
    • Let \(a \in X\).
    • Assume for a contradiction \(a \notin f(X)\).
    • \(f(X)\) is closed.
      • It is compact and \(X\) is Hausdorff.
    • Pick \(\epsilon > 0\) such that \(f(X) \cap B(a,\epsilon) = \emptyset\).
    • Define the sequence \((x_n)\) by \(x_1 = a\) and \(x_{n+1} = f(x_n)\) for all \(n\).
    • For all \(m,n \in \mathbb{Z}_+\) with \(m \neq n\) we have \(d(x_m,x_n) \geq \epsilon\).
      • For all \(n > 1\) we have \(d(x_1,x_n) \geq \epsilon\).
        • Since \(x_n \in f(X)\).
      • For all \(m > n\) we have \(d(x_m,x_n) \geq \epsilon\).
        • \(d(x_m,x_n) = d(x_{m-1},x_{n-1}) = \cdots = d(x_1,x_{n-m+1}) \geq \epsilon\)
    • \((x_n)\) has no convergent subsequence.
    • This is a contradiction.
  • \(f\) is continuous.
  • \(f\) is a homeomorphism.
    • Theorem 26.6
  • \(\Box\)

Exercise 7

Let \((X,d)\) be a metric space. If \(f\) satisfies the condition

\[ d(f(x),f(y)) < d(x,y) \]

for all \(x,y \in X\) with \(x \neq y\), then \(f\) is called a shrinking map. If there is a number \( \alpha < 1 \) such that

\[ d(f(x), f(y)) \leq \alpha d(x,y) \]

for all \(x,y \in X\), then \(f\) is called a contraction. A fixed point of \(f\) is a point \(x\) such that \(f(x) = x\).

(a)

If \(f\) is a contraction and \(X\) is compact, show \(f\) has a unique fixed point.

Solution

  • \(X\) is bounded.
    • Exercise 26.4
  • Pick \(\alpha < 1\) such that \(\forall x,y \in X. d(f(x),f(y)) \leq \alpha d(x,y)\).
  • For \(n \in \mathbb{Z}_+\), let \(A_n = f^n(X)\).
  • For all \(n \in \mathbb{Z}\), we have \(\operatorname{diam} A_n \leq \alpha^n \operatorname{diam} X\).
  • Let \(A = \bigcap_n A_n\)
  • \(\operatorname{diam} A = 0\)
  • \(A\) is nonempty.
    • Theorem 26.9
  • Let \(A = \{a\}\).
  • \(f(a) = a\)
    • Since \(f(a) \in A\)
  • If \(f(b) = b\) then \(b = a\).
    • If \(f(b) = b\) then \(b = f^n(b) \in A_n\) for all \(n\) and so \(b \in A\).
  • \(\Box\)

(b)

Show more generally that if \(f\) is a shrinking map and \(X\) is compact, then \(f\) has a unique fixed point.

Solution

  • \(f\) is continuous
    • For all \(x,y \in X\) and \(\epsilon > 0\), if \(d(x,y) < \epsilon\) then \(d(f(x),f(y)) < \epsilon\).
  • For \(n \in \mathbb{Z}_+\), let \(A_n = f^n(X)\).
  • Let \(A = \bigcap_n A_n\).
  • \(A = f(A)\)
    • \(f(A) \subseteq A\)
      • By definition of \(A\).
    • \(A \subseteq f(A)\)
      • Let \(x \in A\)
      • For \(n \in \mathbb{Z}_+\), pick \(x_n\) such that \(x = f^{n+1}(x_n)\).
      • Pick \(a\) that is the limit of a convergent subsequence \((f^{n_r}(x_{n_r})\) of \((f^n(x_n))\).
      • \(a \in A\)
        • Assume for a contradiction \(n \in \mathbb{Z}_+\) and \(a \notin A_n\).
        • Pick \(\epsilon > 0\) such that \(B(a,\epsilon) \cap A_n = \emptyset\)
        • Pick \(n' > n\) such that \(d(f^{n'}(x_{n'}),a) < \epsilon\).
        • \(f^{n'}(x_{n'}) \notin A_n\)
        • This is a contradiction.
      • \(f(a) = x\)
        • \(f(f^{n_r}(x_{n_r})) = f^{n_r+1}(x_{n_r}) = x\) and also \(f(f^{n_r}(x_{n_r})) \rightarrow f(a)\) as \(r \rightarrow \infty\).
  • \(A\) has exactly one element \(a\), say.
    • Pick \(x,y \in A\) such that \(d(x,y)\) is maximum.
      • Such a pair exists by the Extreme Value Theorembecause \(d : A^2 \rightarrow \mathbb{R}\) is continuous and \(A^2\) is compact.
    • Pick \(x',y' \in A\) such that \(x = f(x')\) and \(y = f(y')\).
    • If \(x \neq y\) then \(d(f(x'),f(y')) < d(x,y)\) which is impossible.
    • \(x = y\)
    • \(d(x,y) = 0\)
    • For any \(a,b \in A\) we have \(d(a,b) = 0\) and so \(a = b\).
  • \(f(a) = a\)
    • Since \(f(a) \in A\)
  • If \(f(b) = b\) then \(b = a\).
    • If \(f(b) = b\) then \(b = f^n(b) \in A_n\) for all \(n\) and so \(b \in A\).
  • \(\Box\)

(c)

Let \(X = [0,1]\). Show that \(f(x) = x - x^2/2\) maps \(X\) into \(X\) and is a shrinking map that is not a contraction.

Solution

  • For all \(x \in X\) we have \(f(x) \in X\).
    • Let \(x \in X\)
    • \(0 \leq x \leq 1\)
    • \(1 \geq x / 2\)
    • \(x \geq x^2 / 2\)
    • \(f(x) = x - x^2/2 \geq 0\)
    • \(2 - x \leq 2\)
    • \(x (2-x) \leq 2\)
      • Since \(x \leq 1\)
    • \(f(x) = x - x^2/2 \leq 1\)
  • \(f\) is a shrinking map.
    • Let \(x,y \in X\) with \(x \neq y\)
      • \[ \begin{align} |f(x) - f(y)| & = |x - x^2/2 - y + y^2/2| \\ & = | (x-y)(1-(x+y)/2)| & < |x-y| \end{align} \]
  • \(f\) is not a contraction.
    • Let \(\alpha < 1\)
    • Pick \(x \in [0,1)\) such that \((1-x)/2 > \alpha\)
    • \(|f(1)-f(x)| = (1-x)^2/2 > \alpha (1-x)\)
  • \(\Box\)

(d)

The result in (a) holds if \(X\) is a complete metric space, such as \(\mathbb{R}\); see the exercises of section 43. The result in (b) does not: Show that the map \(f : \mathbb{R} \rightarrow \mathbb{R}\) given by \(f(x) = [x+(x^2 + 1)^{1/2}]/2\) is a shrinking map that is not a contraction and has no fixed point.

Solution

  • \(f\) is strictly increasing.
  • \(f\) is a shrinking map.
    • Let \(x,y \in \mathbb{R}\) with \(x \neq y\)
      • \[ \begin{align} \frac{f(y)-f(x)}{y-x) & = \frac{x + y + 1}{2(\sqrt{x^2 + 1} + \sqrt{y^2 + 1})} \\ & < 1 \end{align} \]
  • \(f\) is not a contraction.
    • \(\frac{f(y)-f(x)}{y-x}\) can be made arbitrarily close to 1.
  • \(f\) has no fixed point.
    • Since \(f(x) > (x + |x|)/2 \geq x\) for all \(x\).
  • \(\Box\)

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