The smallest compactification

Solution 1:

Theorem 3.5.12 in Engelking: If in the family $\mathscr{C}(X)$ of all compactifications of a non-compact Tikhonov space $X$ there exists an element $cX$ which is smallest with respect to the order $\le$, then $X$ is locally compact and $cX$ is equivalent to the Alexandroff compactification $\omega X$ of $X$.

If $cX$ is a compactification of $X$, I’m going to identify $X$ with $c[X]$.

Proof. Suppose that $X$ has a smallest compactification $cX$, and suppose that there are distinct points $x$ and $y$ in the remainder $cX\setminus X$. Let $Y=cX\setminus\{x,y\}$. $Y$ is an open subspace of $cX$, so it’s locally compact and has a one-point compactification $\omega Y$. But $\omega Y$ is a compactification $c_1 X$, and by hypothesis $cX\le c_1 X$, so there is a continuous surjection $f:c_1 X\to cX$ such that $f\upharpoonright X=\mathrm{id}_X$. $X$ is dense in $Y\subseteq\omega X=c_1 X$, and $\mathrm{id}_Y\upharpoonright X=\mathrm{id}_X=f\upharpoonright X$, so $\mathrm{id}_Y=f\upharpoonright Y$. In other words, $f$ is a continuous function from $c_1 X$, a compactification of $Y$, onto $cX$, another compactification of $Y$, that fixes $Y$ pointwise. But this is clearly impossible, since $|c_1 X\setminus Y|=1$ and $|cX\setminus Y|=2$. Thus, $cX\setminus X$ is a singleton, $X$ is locally compact, and it’s straightforward to check that $cX=\omega X$. $\dashv$

Solution 2:

According to Wikipedia, the answer is no. More precisely, a noncompact Tychonoff space has a smallest compactification if and only if it has a one-point compactification (if and only if it is locally compact). So it suffices to find a noncompact Tychonoff space which is not locally compact; $\mathbb{Q}$ is probably the most familiar example.