Show that \(\beta(\mathbb{Z}_+)\) has cardinality at least as great as \(I^I\), where \(I = [0, 1]\). [Hint: The space \(I^I\) has a countable dense subset.]
Solution
1. Properties of the product space \(I^I\):Proof:
1.1. By Tychonoff’s Theorem, \(I^I = [0, 1]^{[0, 1]}\) is a compact Hausdorff space.
1.2. The cardinality of the index set and factor space is \(|I| = \mathfrak{c} = 2^{\aleph_0}\), so:\[|I^I| = \mathfrak{c}^\mathfrak{c} = (2^{\aleph_0})^\mathfrak{c} = 2^{\aleph_0 \cdot \mathfrak{c}} = 2^\mathfrak{c}.\]
1.3. By the Hewitt-Marczewski-Pondiczery Theorem on product separability, a product of at most \(2^{\aleph_0}\) separable spaces is separable.
1.4. Since \(I\) is separable, \(I^I\) is separable, meaning it possesses a countable dense subset \(D \subseteq I^I\).
2. Construction of continuous surjection \(\beta(\mathbb{Z}_+) \twoheadrightarrow I^I\):Proof:
2.1. Enumerate the countable dense subset as \(D = \{y_n \mid n \in \mathbb{Z}_+\}\).
2.2. Define the map \(f: \mathbb{Z}_+ \to I^I\) by \(f(n) = y_n\).
2.3. Because \(\mathbb{Z}_+\) has the discrete topology, \(f\) is continuous.
2.4. By the universal extension property of the Stone-Čech compactification \(\beta(\mathbb{Z}_+)\), since \(I^I\) is compact Hausdorff, there exists a unique continuous extension:\[\beta f: \beta(\mathbb{Z}_+) \to I^I\]such that \(\beta f|_{\mathbb{Z}_+} = f\).
2.5. The image \(\beta f(\beta(\mathbb{Z}_+))\) is a compact subset of \(I^I\) (hence closed, as \(I^I\) is Hausdorff) containing the dense set \(f(\mathbb{Z}_+) = D\).
2.6. Therefore:\[\beta f(\beta(\mathbb{Z}_+)) \supseteq \overline{D} = I^I.\]
2.7. Thus \(\beta f: \beta(\mathbb{Z}_+) \to I^I\) is surjective.
3. Cardinality comparison:Proof:
3.1. Since \(\beta f: \beta(\mathbb{Z}_+) \to I^I\) is surjective by step 2,\[|\beta(\mathbb{Z}_+)| \ge |I^I| = 2^\mathfrak{c}.\]
3.2. Furthermore, since \(\beta(\mathbb{Z}_+)\) embeds into \([0, 1]^{C(\mathbb{Z}_+, [0, 1])}\) and \(|C(\mathbb{Z}_+, [0, 1])| = \mathfrak{c}^{\aleph_0} = \mathfrak{c}\), we have \(|\beta(\mathbb{Z}_+)| \le \mathfrak{c}^\mathfrak{c} = 2^\mathfrak{c}\), so \(|\beta(\mathbb{Z}_+)| = |I^I| = 2^\mathfrak{c}\).
4. Q.E.D.Step 3 gives \(|\beta(\mathbb{Z}_+)| \ge |I^I|\).