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I've been reviewing properties of compactifications, and I came across this question.

If $cX$ is a compact metric space and $X$ is a dense subset of $cX$, where $X \neq cX$, why is $cX$ not the Stone-Čech compactification of $X$?

Martin Sleziak
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josh
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2 Answers2

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No point $y\in\beta X\setminus X$ is a limit of a sequence of points belonging to $X$, see this question: Stone-Čech compactifications and limits of sequences

If $cX$ is metrizable, and $X$ is dense in $cX$, then every point of $cX\setminus X$ is a limit of a sequence of points of $X$.

Community
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Martin Sleziak
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    In fact one can show that every non-empty closed $G_\delta$-set in $\beta X\setminus X$ contains a copy of $\beta\omega\setminus\omega$ and so has cardinality at least $2^\mathfrak{c}$, while every point of $cX\setminus X$ is of course a closed $G_\delta$ of cardinality $1$. – Brian M. Scott Jan 29 '12 at 20:40
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Consider a completely regular and noncompact space $X$. Then its Stone Cech compactification $\beta X$ is not metrizable.

Proof: Assume $\beta X$ is metrizable. If $X$ is not compact then $X$ is not $\beta X$. Therefore, there exists at least one point $p$ in $\beta X \setminus X$. Also, $\beta X$ is metrizable and $X$ is dense in $\beta X$. So, there is a sequence $x_1, x_2, x_3, x_4, ...$, of distinct points in $X$ converging to $p$ (in $\beta X$). $X$ being a subset of $\beta X$ is also metrizable. Hence, $X$ is a normal space. The sets $O = \{x_i : i \ \text{is odd}\ \}$ and $E =\{ x_i:\ i \ \text{is even}\ \}$ are disjoint closed subsets of the normal space $X$. Therefore, there is a bounded continuous function from $X$ to $\mathbb R$, such that $f[O] = \{0\}$ and $f[E] = \{1\}$. Such a function must extend to a function $g$ from $\beta X$ to $\mathbb R$ such that $g[cl(O)] = \{0\}$ and $g[cl(E)] = \{1\}$ (the closures being taken in $\beta X$). This is a contradiction since $p$ belongs to both closures and we would then have $0 = g(p) = 1$.

Therefore, there cannot exist a $p$ in $\beta X \setminus X$, and since X is a subset of $\beta X$, it follows that $\beta X = X$. Hence $X$ is compact, a contradiction.

Therefore, BX cannot have been metrizable.

Indeed, BX is metrizable iff X is a compact metrizable space.

azarel
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