# Can a noncompact metric space have a maximal metrizable Hausdorff compactification?

We know the Stone-Čech compactification $$(h, \beta X)$$ of a Tychonoff space $$X$$ is its largest (in particular, a maximal) Hausdorff compactification, in the sense that if $$(k,\gamma X$$) is any other Hausdorff compactification, then there is a continuous map $$f:\beta X \rightarrow \gamma X$$ such that $$fh=k$$.

What about if I restrict to considering just all metrizable Hausdorff compactifications? To fix ideas, say $$X$$ is a noncompact metric space. (In particular $$\beta X$$ will not be metrizable.) Is there a maximal metrizable Hausdorff compactification of $$X$$?

I am guessing no: if we're given any metrizable Hausdorff compactification of noncompact metric space $$X$$, we can make another metrizable Hausdorff compactification that is larger (in the sense of the first paragraph above).

No. Let $$X$$ be a noncompact metric space and let $$Y$$ be a metrizable compactification of $$X$$. Pick a point $$y\in Y$$ and a sequence $$(x_n)$$ of distinct points in $$X$$ converging to $$y$$. Let $$A$$ consist of the points $$x_n$$ for all even $$n$$ and let $$B$$ consist of the points $$x_n$$ for all odd $$n$$. Then $$A$$ and $$B$$ are closed and disjoint in $$X$$, so there exists a continuous $$f:X\to[0,1]$$ such that $$f$$ maps $$A$$ to $$0$$ and $$B$$ to $$1$$. This map $$f$$ does not extend continuously to $$Y$$.
We can now embed $$X$$ in $$Y\times [0,1]$$ by using the given embedding $$X\to Y$$ on the first coordinate and $$f$$ on the second coordinate. The closure of the image of $$X$$ in $$Y\times[0,1]$$ is then a metrizable compactification of $$X$$, which is strictly larger than $$Y$$ since $$f$$ extends continuously to it.
For another perspective on this, note that the algebra $$C_b(X)$$ of bounded continuous functions on $$X$$ can canonically be identified with $$C(\beta X)$$, and an arbitrary compactification of $$X$$ just corresponds to a closed subalgebra of $$C_b(X)$$ which separates points from closed sets on $$X$$. Moreover, a compactification is metrizable iff the corresponding subalgebra is separable. "Larger" compactifications in your sense just corresponds to the inclusion order on subalgebras.
So, given any metrizable compactification corresponding to a separable subalgebra $$C(Y)\subset C_b(X)$$, you can get a larger separable subalgebra by just taking some $$f\in C_b(X)\setminus C(Y)$$ and taking the closed subalgebra generated by $$f$$ and $$C(Y)$$.
• May I ask why the closure of the image of $X$ in $Y×[0,1]$ is a compactification of $X$? I've gone through the other details, but in particular why is the inverse function of the embedding continuous? – SSF Dec 8 '18 at 2:20
• The inverse function is the composition of the projection $Y\times [0,1]\to Y$ and the inverse of the embedding $X\to Y$. – Eric Wofsey Dec 8 '18 at 2:24