V. Compactness & Connectedness #
Problems on compact sets and their stability properties, the failure of the Heine–Borel theorem outside \(\mathbb{R}\), and connectedness.
Problem 1 #
Show that the union of two compact sets is compact.
For this problem, you may consider being inspired by the theorem and its proof detailed in this short lecture.
Solutions: Hint · Solution · Long solution
Problem 2 #
Suppose that \(T\) is closed and \(U\) is compact. Show that \(T \cap U\) is compact.
For this problem, you may consider being inspired by the theorem and its proof detailed in this short lecture.
Solutions: Hint · Solution · Long solution
Problem 3 #
Show that if \(T_i\) is compact for every \(i \in I\), then so is the intersection \(\bigcap_{i \in I} T_i\).
For this problem, you may consider being inspired by the theorem and its proof detailed in this short lecture.
Solutions: Hint · Solution · Long solution
Problem 4 #
The Heine–Borel Theorem shows that all closed and bounded subsets of the metric space \(\mathbb{R}\) are compact. Its proof uses the least upper bound property of \(\mathbb{R}\) in an essential way. Hence it is a natural question to examine whether such a theorem is still true in a metric space which does not satisfy the least upper bound property.
Question: Consider the metric space \(\mathbb{Q}\) with the usual metric. Let \(T = (a,b) \cap \mathbb{Q}\) where both \(a\) and \(b\) are irrational numbers. Show that \(T\) is both closed and bounded, but that it is not compact.
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Problem 5 #
Let \((S,\rho)\) be a metric space endowed with the discrete metric. Describe all of its connected subsets.
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Problem 6 #
Let \(T\) be an open subset of \(\mathbb{R}\). An open interval \(I\) is called a component interval if \(I \subset T\) and there is no open interval \(J \neq I\) such that \(I \subset J \subset T\). The following statement can be verified by using a proof by contradiction:
Lemma. Every point of a nonempty open set \(T \subset \mathbb{R}\) belongs to one and only one component interval of \(T\).
Using this lemma and the fact that a collection of disjoint open subsets of \(\mathbb{R}\) must be countable, gleaned from the last exam, it is possible to prove the following characterization of all open subsets of \(\mathbb{R}\):
Theorem. Every non-empty open set \(T\) in \(\mathbb{R}\) is the union of a countable collection of disjoint component intervals of \(T\).
Problem: Using the facts discussed above freely, prove that every open interval in \(\mathbb{R}\) is connected.
Solutions: Hint · Solution · Long solution
Problem 7 #
Let \((S,\rho)\) be a metric space.
(a) Show that every finite subset of \(S\) is compact.
(b) Now suppose that \(\rho\) is the discrete metric. Show that a subset of \(S\) is compact if and only if it is finite.
(c) Let \((S,\rho) = (\mathbb{R}, |\cdot|)\) and consider the set
$$T = \{\, \tfrac{1}{n} \mid n \in \mathbb{N} \,\}.$$Is \(T\) compact? Is \(T \cup \{0\}\) compact? Prove your answers directly from the definition of compactness, without appealing to the Heine–Borel theorem.
Solutions: Hint · Solution · Long solution
Problem 8 #
Suppose that \(K_1 \supseteq K_2 \supseteq K_3 \supseteq \ldots\) is a nested sequence of non-empty compact subsets of a metric space \((S,\rho)\).
(a) Show that \(\displaystyle\bigcap_{n=1}^\infty K_n\) is non-empty.
(b) Show by example that the conclusion of part (a) can fail if the sets \(K_n\) are only assumed to be closed.
(c) Show by example that it can also fail if the sets \(K_n\) are only assumed to be bounded.
Hint for (a): Argue by contradiction and suppose the intersection is empty. Let \(U_n = S \setminus K_n\). Why is each \(U_n\) open? Explain why the collection \(\{U_n\}\) is an open cover of \(K_1\), and notice that the \(U_n\) are nested in the opposite direction: \(U_1 \subseteq U_2 \subseteq U_3 \subseteq \ldots\). What does a finite subcover of a nested collection look like?
Hint for (b) and (c): Work in \((\mathbb{R}, |\cdot|)\) and look for sequences of intervals. You do not need the same example for both parts. Make sure to say in each case which hypothesis of part (a) has been dropped and why your sets are nested.
Solutions: Hint · Solution · Long solution