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For the duration of this question, let a "surface" be any connected Hausdorff topological space that is locally homeomorphic to R2. Note that we make no assumption about a countable base to its topology.

For instance, if L denotes the long ray 𝜔1 × [0, 1) with the lexicographic order topology (𝜔1 is the first uncountable ordinal) and O denotes its endpoint, then let the long plane P be L × S1 with {O} × S1 identified to a point.

Define a maximal surface as one that is not a proper subspace of any other surface.

Edit: Because I naïvely thought the long plane et al. were contractible, it is necessary to define a similar notion that holds for these non-metrizable surfaces.

Definition: Call any surface jordan if every simple closed curve bounds a topological 2-disk D2 and the complement of this 2-disk is noncompact.

Then the long plane P is a jordan maximal surface.

Questions:

1. Are all jordan maximal surfaces homeomorphic to P?

2. Is it true that every jordan surface N is a subspace of a jordan maximal one?

(I suspect there is a proof with Zorn's Lemma, but the details are delicate and have so far escaped me.)

3. If 2. is true, is a maximal surface containing a given jordan surface N unique up to homeomorphism?

4. In particular: Consider the Prüfer manifold M.* If M is a subspace of a jordan maximal surface, what is its description, and is it unique up to homeomorphism?


* One way to define M: For each c ∈ ℝ, take a disjoint copy Hc of the open upper half-plane, and one more copy called H. Now for each c ∈ ℝ, identify H with the subspace of Hc obtained by mapping each point p ∈ H to the point q ∈ Hc corresponding to pushing p directly away from c by exactly one unit. (I.e., q = p + (p-c)/|p-c|, if we view an upper half-plane as complex.) The image of H will be Hc minus a semidisk of radius 1 about c. M is the resulting identification space. (M is clearly not maximal.)

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    $\begingroup$ Long line (and hence the long plane) is not contractibe in the usual sense, see this post. $\endgroup$
    – Zerox
    Commented Jul 9, 2023 at 6:29
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    $\begingroup$ For the long ray, did you mean $\omega_1\times[0,1)$? If you use $\omega_1\times[0,1]$ as you said, it won't be connected and won't be locally like $\mathbb{R}$ etc. $\endgroup$ Commented Jul 10, 2023 at 23:27
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    $\begingroup$ Thanks, JDH, for catching my typo. $\endgroup$ Commented Jul 11, 2023 at 0:53

2 Answers 2

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The answer to 1 and 3 should be negative. A space is called $\omega$-bounded if the closure of any countable set is compact. $\omega$-bounded surfaces are well studied (see for instance Bagpipe theorem and the reference), and an $\omega$-bounded surface $X$ can be written as an increasing union of separable open subsets $\bigcup_{\alpha<\omega_1}X_\alpha$ where the closure of $X_\alpha$ is contained in $X_{\alpha+1}$.

There are many non-homeomorphic $\omega$-bounded surfaces $X$ that can be written as $\bigcup_{\alpha<\omega_1}X_\alpha$ where each $X_\alpha\simeq\mathbb{R}^2$; they are essentially the "long pipes" in the Wikipedia article. Let $L$ be the closed long ray and consider $P=\{(x,y)\in L\times L:y\leq x\}$. The $x$-axis $L\times\{0\}$ and the diagonal $\{(x,y)\in P:x=y\}$ behave very differently. one can show that a continuous map $f:L\rightarrow P$ is either null-homotopic, or homotopic to the inclusion map of $x$-axis, or homotopic to the diagonal map $x\mapsto (x,x)$; in the second case it is either eventually disjoint from the $x$-axis or eventually contained in the $x$-axis, but in the third case it can enter and exit the diagonal unboundedly many times. This can be used to show that the long plane is not homeomorphic to the square of long line. Gluing the diagonal and $x$-axis of $P$ together gives yet another surface, and there are many variation.

Any surface $X$ with the above property is Jordan maximal. It is Jordan because the image of any continuous map from $S^1$ must fall into one of the $X_\alpha$, and is maximal because the image of an $\omega$-bounded space is $\omega$-bounded, and therefore closed (at least when the target space is Hausdorff and first countable, I believe), so the image of an embedding of $\omega$-bounded surface into any surface is clopen, hence equal to that surface.

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These type of questions have been studied in detail by P. Nyikos, for example in his article Non-metrizable manifolds in the Handbook of Set Theoretic Topology, where he shows that there are $2^{\aleph_1}$ non-homeomorphic “long planes”, that is, simply connected $\omega$-bounded surfaces. (He actually shows that there is such a family of non-homeomorphic longpipes, but it is easy to sew a disc in a longpipe to obtain a longplane. See the article for definitions.) Hence, as noted in a previous answer by new account, there are many non-homeomorphic surfaces that are maximal in OP's sense.

Nyikos also showed (Thm 6.1) in

On first countable, countably compact spaces III: The problem of obtaining separable noncompact examples, Open Problems in Topology, J. van Mill and G.M. Reed (Editors), 1990, pp. 130–161.

that a $n$-manifold $M$ is properly contained in another $n$-manifold iff there is a closed copy in $M$ of the closed unit $n$-ball with a deleted point in its boundary. As a corollary (Cor. 6.2), a pseudocompact manifold is maximal. (Recall that a space is pseudocompact iff any real valued map on it is bounded.) Nyikos' Example 6.7 in the same paper is a maximal surface whose properties depend on the model of set theory: under the continuum hypothesis, it can be made countably compact, and under $\mathfrak{p}>\omega_1$, it cannot be pseudocompact. Hence, the relationship between pseudocompactness and maximality can sometimes be a subtle matter.

I don't know the answer to Question 2 but I strongly suspect that it is negative.

The answer to question 3 is clearly no, as also noted in the same previous answer by new account, but a small variation of it seems more interesting:

Question 3': If $N$ is a Jordan surface (in OP's sense) properly contained in two maximal Jordan surfaces $M_1$, $M_2$ and dense in them, are $M_1$ and $M_2$ homeorphic ?

The answer to this question is also negative: Nyikos' example 6.7 is the union of a copy of the longray and the $2$-sphere with a deleted point. The long ray is in a sense “pushed inside” the hole left by the deleted point. One can delete more points and push copies of the longray in each scar. This gives non-homeomorphic maximal surfaces which all have a dense subset homeomorphic to open $2$-disk. (There are certainly simpler counter-examples to Question 3'.)

As to Question 4, I admit that I do not really understand your definition of Prüfer Surface (maybe because I am on holidays and somewhat slow at thinking), but I guess that it must be the classical example of attaching uncountably many copies of $\mathbb{R}$ to the half plane, yielding a surface whose boundary is these uncountable copies of $\mathbb{R}$. If that is the case, I believe that another example in the same paper by Nyikos (ex 6.3) can be adapted to obtain a pseudocompact (hence maximal) surface containing the Prüfer surface, however the construction works only when $\mathfrak{b}=\omega_1$. (Nyikos' example is described in more details in D. Gauld's book “Non-Metrisable Manifolds”, ex. 1.29 on page 15).

Note: $\mathfrak{b}$ and $\mathfrak{p}$ are classical small uncountable cardinals defined for instance in Nyikos' above cited article.

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  • $\begingroup$ Thank you for pointing out that my definition of the Prüfer manifold needed fixing. $\endgroup$ Commented Jul 13, 2023 at 19:11

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