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 $h:{\mathbb{N}}^{\ast }\to [0,1]$, and write $a=h(\mathrm{\infty })$.

The open interval $(a-1\mathrm{/}3,a+1\mathrm{/}3)\cap [0,1]$ contains $a$, so continuity gives a neighbourhood of $\mathrm{\infty }$ 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 $1\mathrm{/}3$. 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 ${\mathbb{N}}^{\ast }$ a Hausdorff compactification.

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

The same cofinite-neighbourhood argument as follows directly from The one-point (Alexandroff) compactification $X^{\ast }=X\cup \{\mathrm{\infty }\}$, whose open sets are the open sets of $X$ together with the complements in $X^{\ast }$ of the closed compact subsets of $X$: continuity at $\mathrm{\infty }$ would make $p$ eventually lie in an interval of radius $1\mathrm{/}3$, 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.