If \(X\) is normal and \(y\) is a point of \(\beta(X) - X\), show that \(y\) is not the limit of a sequence of points of \(X\).
Show that if \(X\) is completely regular and noncompact, then \(\beta(X)\) is not metrizable.
Solution
1. Part (a): Points of \(\beta(X) \setminus X\) are not sequential limits of points in \(X\):
1.1. Suppose for contradiction that there exists a sequence \((x_n)_{n=1}^\infty \subseteq X\) such that \(x_n \to y \in \beta(X) \setminus X\).Each point \(x \in X\) occurs only finitely often in the sequence: otherwise a constant subsequence converges to \(x \neq y\), which contradicts uniqueness of limits in the Hausdorff space \(\beta(X)\). Passing to a subsequence, assume the \(x_n\) are pairwise distinct.Then the set \(S = \{x_n \mid n \in \mathbb{N}\}\) has no accumulation point in \(X\): an accumulation point \(x \in X\) of \(S\) is a cluster point of the sequence, so \(x = y\) by the Hausdorff property, which contradicts \(y \notin X\). Hence every subset of \(S\) is closed in \(X\).
1.2. Partition \(S\) into two disjoint closed subsets of \(X\):\[ A = \{x_{2k} \mid k \in \mathbb{N}\}, \qquad B = \{x_{2k-1} \mid k \in \mathbb{N}\}. \]Because \(X\) is normal, by Urysohn’s Lemma there exists a continuous function \(f: X \to [0, 1]\) such that:\[ f(a) = 0 \quad \forall a \in A, \qquad f(b) = 1 \quad \forall b \in B. \]
1.3. By the universal property of the Stone–Čech compactification, \(f\) extends to a continuous function \(\beta f: \beta(X) \to [0, 1]\).
1.4. Since \(x_n \to y\) in \(\beta(X)\), continuity of \(\beta f\) implies:\[ \lim_{n \to \infty} f(x_n) = \beta f(y). \]However, along the even subsequence, \(\lim_{k\to\infty} f(x_{2k}) = \lim_{k\to\infty} 0 = 0\), while along the odd subsequence, \(\lim_{k\to\infty} f(x_{2k-1}) = \lim_{k\to\infty} 1 = 1\).Thus \(0 = \beta f(y) = 1\), a contradiction.
1.5. Therefore no point \(y \in \beta(X) \setminus X\) can be the limit of a sequence in \(X\).
2. Part (b): Nonmetrizability of \(\beta(X)\) for noncompact completely regular \(X\):
2.1. Since \(X\) is noncompact, \(X \subsetneq \beta(X)\), so there exists a point \(y \in \beta(X) \setminus X\).
2.2. The subspace \(X\) is dense in \(\beta(X)\), so \(y \in \overline{X}\).
2.3. Suppose for contradiction that \(\beta(X)\) is metrizable.In any metric space, the topological closure coincides with the sequential closure: every point in the closure of a subset is the limit of a sequence of points in that subset.Thus there must exist a sequence \((x_n)_{n=1}^\infty \subseteq X\) such that \(x_n \to y\) in \(\beta(X)\).
2.4. The subspace \(X\) of the metrizable space \(\beta(X)\) is metrizable, hence normal.By step 1 applied to the normal space \(X\), the point \(y \in \beta(X) \setminus X\) is not the limit of a sequence of points of \(X\).This contradicts step
2.3.
2.5. Therefore \(\beta(X)\) is not metrizable.
3. Q.E.D.Step 1 proves part (a) and step 2 proves part (b).