V. Compactness & Connectedness #

Problem 1 #

Show that the union of two compact sets is compact.

Solutions: Hint · Solution · Long solution

Problem 2 #

Suppose that \(T\) is closed and \(U\) is compact. Show that \(T \cap U\) is compact.

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Problem 3 #

Show that if \(T_i\) is compact for every \(i \in I\), then so is the intersection \(\bigcap_{i \in I} T_i\).

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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