The Tychonoff Embedding and the Stone–Čech Compactification: Examples and Counterexamples

Every two distinct points of a discrete space have disjoint singleton neighbourhoods, so $D$ is Hausdorff; it is compact by [L1].

The Stone–Čech compactification of a compact Hausdorff space adds no points applied to step 1.1 gives the claim.

The map $g$ is continuous by [L1]. By [L2], the interval $[0,1]$ is compact Hausdorff, so the extension and uniqueness clause of The Stone–Čech compactification by its compact-Hausdorff extension property gives a unique continuous $\bar g:\beta\mathbb N\to[0,1]$ with $\bar g\circ i=g$.

This is the asserted extension.

Suppose $p$ extends continuously to $h:\mathbb N^*\to[0,1]$, and write $a=h(\infty)$.

The open interval $(a-1/3,a+1/3)\cap[0,1]$ contains $a$, so continuity gives a neighbourhood of $\infty$ on which $h$ has values in that interval. By [L1] and [L2], this neighbourhood contains every natural except finitely many.

Both an even and an odd natural lie outside every finite subset of $\mathbb N$. Their $p$-values are $0$ and $1$, which cannot both belong to an interval of radius $1/3$. This contradicts step 1.2.

Hence $p$ has no continuous extension; the asserted Stone–Čech property is false.

Discrete $\mathbb N$ is locally compact, Hausdorff, and noncompact, so [L1] makes $\mathbb N^*$ a Hausdorff compactification.

Suppose it had the Stone–Čech property. Its parity map would then extend continuously to $\mathbb N^*$.

The same cofinite-neighbourhood argument as follows directly from The one-point (Alexandroff) compactification $X^{*} = X \cup \{\infty\}$, whose open sets are the open sets of $X$ together with the complements in $X^{*}$ of the closed compact subsets of $X$: continuity at $\infty$ would make $p$ eventually lie in an interval of radius $1/3$, while arbitrarily large even and odd naturals have values $0$ and $1$. This is impossible.

Therefore this Hausdorff compactification is not Stone–Čech, refuting the displayed statement.