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    Cardinality of the Stone--Čech compactification of $\ZZ_+$

    E-7NRBE

    Area
    Topology
    Topics
    Compactness
    Status
    draft

    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 \(|\beta(\mathbb{Z}_+)| \ge |I^I|\).