# Can every locally compact Hausdorff space be recognized as a subspace of a cube that has an open underlying set?

In this question cubes are topological spaces of the form $$[0,1]^J$$ with product topology and $$[0,1]$$ with usual topology. Further a space is a Tychonoff space if and only if it is a completely regular space. Also note that cubes are compact Hausdorff spaces.

In the wide look that two spaces that are homeomorphic are actually the same I was eager to classify certain spaces as subspaces of a cube, and made the following observations:

• Tychonoff spaces are exactly the subspaces of cubes.

• Compact Hausdorff spaces are exactly the subspaces of cubes that have a closed subset of the cube as underlying set.

• Locally compact Hausdorff spaces are exactly the subspaces of cubes that have the intersection of a closed and an open subset of the cube as underlying set.

The first observation is a corollary of the Imbedding theorem for Tychonoff spaces (Munkres Th. 34.2).

The second observation follows from the fact that compact Hausdorff spaces are Tychonoff spaces. They are compact in the cube which is Hausdorff, so their underlying set is a closed set.

The third observation is important in my question. It is based on the statement that a space is homeomorphic to an open subspace of a compact Hausdorff space if and only if $$X$$ is locally compact Hausdorff (Munkres Cor. 29.4).

First question (a check of someone else never harms):

Are these observations correct?

Second question (the main):

Can locally compact Hausdorff spaces also be classified (up to homeomorphisms) as exactly the subspaces of cubes that have an open subset of the cube as underlying set?

So asked is actually whether I can replace my third observation (intersection of open and closed set) with this (more comfortable) one.

If the answer is "no" then of course I would like to see a counterexample.

• Your observations are correct. For the last question: obviously "no" because evey compact space is locally compact. For a more sophisticated example: this will fail if you take a disjoint union of a locally compact space and a point. – freakish Dec 26 '18 at 11:32
• @freakish If I understand well then your argument is that compact (hence also locally compact) sets exist then do not have a clopen homeomorphic image in a cube. Do I understand you correctly? – drhab Dec 26 '18 at 11:45
• Yes, a simpliest example is a point. – freakish Dec 26 '18 at 12:02
• Thank you. Concise and convincing counterexample. If you make this an answer then I will accept it. – drhab Dec 26 '18 at 12:13

No. Compact spaces are locally compact. And so the simpliest counterexample is a singleton $$\{*\}$$. Also if you take any locally compact space $$X$$ and the disjoint union $$Z:=X\sqcup Y$$ with a compact space $$Y$$ then $$Z$$ cannot be an open subset of a cube even though it is locally compact.
Your observations are correct because of the following theorem: if $$X$$ is compact Hausdorff and $$Y \subseteq X$$ is locally compact then $$Y = C \cap O$$ where $$C\subseteq Y$$ is closed and $$O \subseteq Y$$ is open. Applied to $$X = [0,1]^J$$, and $$Y$$ any Tychonoff space (seen as a subspace of $$X$$).
Of course compact Hausdorff spaces cannot be seen as open subsets of Tychonoff cubes (they would be clopen, contradicting the connectedness of those cubes). Moreover open subsets of e.g. $$[0,1]^\mathbb{N}$$ are infinite dimensional and not all locally compact sets are. You cannot do better in general then "open subset of a closed subset of a Tychonoff cube".