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Problem Sheet 7

Question 1

Which of the following sets \(Y\) are compact in the given space \((X, \tau)\)?

  1. \(Y = \mathbb{Z}\) in \((\mathbb{Q}, |\cdot|)\).
  2. \(Y = \mathbb{Z}\) in \(\mathbb{R}\) with the co-countable topology.
  3. \(Y = \{(\sin 3t, \cos 2t, \sin 5t) \in \mathbb{R} : t \in [0, 2\pi]\}\) in \((\mathbb{R}, |\cdot|)\).
  4. \(Y = \{e^{ik} : k \in \mathbb{Z}\}\) in \((\mathbb{C}, |\cdot|)\).
  5. \(Y = \{A \in M_{44}(\mathbb{R}) : det(A) = 1\}\) in \((M_{44}(\mathbb{R}), \lVert \cdot \rVert_{op})\).
  6. \(Y = \{f \in C^1[0, 1] : f(0) = 0, \lVert f' \rVert_\infty \leq 3\}\) in \((C[0, 1], \lVert \cdot \rVert_\infty)\).

Solution

  1. No, since it is not bounded.
  2. No, take \(\{ \mathbb{R} \setminus \mathbb{Z} \cup \{n\}\}_{n \in \mathbb{Z}}\).
  3. Yes.
  4. No, as it is not closed.
  5. No, as it is unbounded. Consider upper-triangular matrices with diagonal entries all equal to 1.
  6. No, since it is not closed.

Question 2

Give an example of a noncompact topological space in which a subset is compact if and only if it is finite.

Solution

Let \(X\) be an infinite set. An example of a noncompact space in which the compact sets are exactly the finite ones would be any infinite space with the discrete topology on \(X\). Also, we could cocountable topology on \(X\). Suppose that \(Y\) is an infinite subset of \(X\) with countably infinite subset \(C = \{c_n\}_{n=1}^\infty\). Then \(\{ (X \setminus C) \cup \{c_n\}\}_{n = 1}^\infty\) is an open cover of \(Y\) without a finite subcover.

Question 3

Let \(X = \mathbb{R}\) with the left-ray topology

\[ \tau = \{\emptyset, \mathbb{R}\} \cup \{(-\infty, a) : a \in \mathbb{R}\}. \]

What are the compact subsets of \(X\)?

Solution

A set is compact in \(\tau\) precisely when it is bounded from above and contains its supremum.

If a set is not bounded from above, then \(\{(-\infty, n)\}_{n \in \mathbb{N}}\) is an open cover without a finite subcover. If it is bounded from above but does not contains its supremum, then letting \(s\) denote the supremum of the set, we have that \(\{(-\infty, s - \frac{1}{n})\}_{n \in \mathbb{Z}^+}\) is an open cover with a no finite subcover.

Conversely, suppose that a set is bounded from above and contains its supremum. Let \(\{(-\infty, a_i)\}_{i \in I}\) be an open cover of such a set with supremum \(M\). Then \(I' = \{i \in I: a_i > s\}\) is non-empty, and so we can take \(\{(-\infty, a_i)\}\) as a finite subcover for some \(i \in I'\).

Question 4

Let \(X = \mathbb{Q} \cap [0, 1]\) with the usual metric topology. Prove directly from the definition that \(X\) is not compact.

Solution

Let \(\{ q_n \}_{n = 1}^\infty\) be an enumeration of the rationals, and \(I_n\) an interval containing \(q_n\) of length at most \(2^{-n}\). Then \(\{I_n\}_{n = 1}^\infty\) is an open cover of \(\mathbb{Q} \cap [0, 1]\) but any finite subcover \(\{I_{n_k}\}_{k = 1}^m\) satisfies

\[\mathrm{Length}\left(\bigcup_{k = 1}^m I_{n_k}\right) \leq \sum_{k = 1}^m \mathrm{Length}(I_{n_k}) \leq \sum_{k = 1}^m 2^{-n_k} < \sum_{n = 1}^\infty 2^{-n} = 1, \]

and thus cannot cover all of \(\mathbb{Q} \cap [0, 1]\).

Question 6

Let \(X\) be a set and let \(\tau_1\) and \(\tau_2\) be two topologies on \(X\) such that \(\tau_1 \subseteq \tau_2\). Does compactness of \(X\) in either \(\tau_1\) or \(\tau_2\) imply compactness in the other?

Solution

\(\tau_1 \subseteq \tau_2\). If \(X\) is compact in \(\tau_2\), then it is also compact in \(\tau_1\). Indeed, any open cover in \(\tau_1\) is also an open cover in \(\tau_2\), upon which a finite subcover can be extracted.

Question 7

Give an example of a family \(\{F_\alpha\}_{\alpha \in A}\) of subsets of \(\mathbb{R}\) which has empty intersection, but for which every finite subfamily has non-empty intersection. Do this in the open unit disk in the plane.

Solution

\(\{(-\frac{1}{n}, \frac{1}{n}) \setminus \{0\}\}_{n = 1}^\infty\) has empty intersection, but any finite subcollection has non-empty intersection.

Question 9

Let \((X, d)\) be metric space, and let \(Y \subseteq X\) be a subset. Show that the following are equivalent.

  1. For any \(\epsilon > 0, Y\) can be covered by finitely many \(\epsilon\)-balls with centers in \(Y\).
  2. For any \(\epsilon > 0, Y\) can be covered by finitely many \(\epsilon\)-balls with centers in \(X\).

Solution

Clearly i) implies ii), so given \(\epsilon > 0\), let \(n \in \mathbb{Z}^+\) and \(\{x_i\}_{i = 1}^n\) be such that \(Y \subseteq \bigcup_{i = 1}^n B(x_i, \frac{\epsilon}{2})\). Without loss, assume that \(Y \cap B(x_i, \frac{\epsilon}{2}) \neq \emptyset\) for all \(i\) and let \(z_i \in Y \cap B(x_i, \frac{\epsilon}{2})\). Let \(f: Y \to \{1, 2, \dots, n\}\) be a function such that \(y \in B(x_{f(y)}, \frac{\epsilon}{2})\). Then

\[ d(y, z_{f(y)}) \leq d(y, x_{f(y)}) + d(x_{f(y)}, z_{f(y)}) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \]

So \(Y \subseteq \bigcup_{i = 1}^n B(z_i, \epsilon)\).

Question 12

Give an example of a closed and bounded subset of \(C[0, 1]\) which is not compact. Prove your answer!

Solution

The closed unit ball of \(C[0, 1]\) is not compact. Considering spike functions with mutually disjoint supports will produce a bounded sequence with no convergent subsequence since \(\lVert f_{n_k} - f_{n_\ell} \rVert_\infty = 1\).

The closed unit ball of a normed space if compact if and only if it the space is finite dimensional.