# Mysior plane is not realcompact

Let $$X = \mathbb{R}^2$$ with $$(x, y)\in X$$ for $$y\neq 0$$ isolated and $$(x, 0)$$ having neighbourhood basis of the form $$U_n(x) = \{(x, y) : y\in (-1/n, 1/n)\}\cup \{(x+y+1, y) : 0 < y < 1/n\}\cup \{(x+\sqrt{2}+y, -y) : 0 < y < 1/n\}.$$ The space $$X$$ is called Mysior plane, and it's an example of a space which is a union of two closed subspaces, $$X_+ = \mathbb{R}\times [0, \infty)$$ and $$X_- = \mathbb{R}\times (-\infty, 0]$$, which are realcompact, but isn't itself realcompact.

Since the map $$f:X_+\sqcup X_-\to X$$ is perfect, this shows realcompactness is not invariant under perfect mappings.

The space $$X$$ is Tychonoff: Say $$(x, y), (w, t)\in X$$. If $$y\neq 0$$ or $$t\neq 0$$ this is quite obvious since one of the points is isolated. If $$y = t = 0$$, and $$x < w$$, then by splitting into three cases $$x < w \leq x+1, x+1 < w \leq x+\sqrt{2}$$ and $$w\geq x+\sqrt{2}$$ one can easily see that we have disjoint neighbourhoods. Thus $$X$$ is Hausdorff. If $$(x, y)\notin F$$ and $$F$$ is closed, of course the only interesting case is when $$y = 0$$ for otherwise $$\{(x, y)\}$$ is clopen. In that case we can find $$n$$ with $$U_n(x)\subseteq F^c$$. By defining $$f(w, t) = 1$$ for $$(w, t)\notin U_n(x)$$ and $$f(w, t) = nt$$ for $$(w, t)\in U_n(x)$$, we see that $$f$$ is continuous and so $$X$$ is Tychonoff.

To see that $$X_+$$ and $$X_-$$ are realcompact, lets focus on $$X_+$$. If $$\mathcal{F}$$ is a real z-ultrafilter on $$X_+$$, then since $$\mathbb{R}\times [0, a]$$ for $$a > 0$$ is clopen, we can see that if $$\mathbb{R}\times (a, \infty)\in\mathcal{F}$$, then $$\mathcal{F}\restriction_{\mathbb{R}\times (a, \infty)}$$ is a real z-ultrafilter on the realcompact space $$\mathbb{R}\times (a, \infty)$$, and as such is fixed, so that $$\mathcal{F}$$ is fixed. Otherwise, if for all $$a > 0$$, $$\mathbb{R}\times [0, a]\in \mathcal{F}$$, then $$\mathbb{R}\times \{0\}\in\mathcal{F}$$ since $$\mathcal{F}$$ is closed under countable intersections. One can observe that $$([a, b]+\mathbb{Z})\times \{0\}$$ for $$a < b$$ are zero sets of $$X_+$$ by construction appropriate functions (similar to the one in the proof that $$X$$ is Tychonoff). Thus $$([0, 1/2]+\mathbb{Z})\times \{0\}\in \mathcal{F}$$ or $$([1/2, 1]+\mathbb{Z})\times\{0\}\in \mathcal{F}$$, for example the former case holds. Take $$a = \sup \{x\in [0, 1/2] : ([x, 1/2]+\mathbb{Z})\times\{0\}\in \mathcal{F}\}$$ and $$b = \inf\{x\in [a, 1/2] : ([a, x]+\mathbb{Z})\times\{0\}\in\mathcal{F}\}$$. Once again one would show that $$[a, b]+\mathbb{Z}\in \mathcal{F}$$ from closure under countable intersections, and if $$a\neq b$$ we could take $$a < c < b$$ and by considering $$([a, c]+\mathbb{Z})\times\{0\}$$ and $$([c, b]+\mathbb{Z})\times\{0\}$$ obtain a contradiction with choice of $$a, b$$. Thus $$(a+\mathbb{Z})\times\{0\}\in\mathcal{F}$$ for some $$a\in \mathbb{R}$$. However, since $$(a+\mathbb{Z})\times\{0\} = \bigcup_{n\in \mathbb{Z}}\{(a+n, 0)\}$$ is a countable union of zero sets, and since $$\mathcal{F}$$ is a real z-ultrafilter, we must have $$\{(a+n, 0)\}\in\mathcal{F}$$ for some $$n$$, that is $$\mathcal{F}$$ is a fixed z-ultrafilter. This proves $$X_+$$ is realcompact, the proof for $$X_-$$ is the same.

Could anyone help me on how to show that $$X$$ is not realcompact?

Edit: A z-ultrafilter is an ultrafilter on the lattice of zero sets (as to distinguish them from ultrafilters on the lattice of sets), and a real z-ultrafilter is a z-ultrafilter closed under countable intersections.

• What is "realcompact" versus "compact"? As in, any $\Bbb R$-indexed open cover has a finite subcover? Jun 14, 2023 at 16:08
• @FShrike It means that any real z-ultrafilter is fixed. Sorry but I can't express it in any simpler terms. For reference see Gillman and Jerison "Rings of continuous functions". Jun 14, 2023 at 16:11
• What is a z-ultrafilter, and what is a "real z-ultrafilter"? (It would be helpful to add all that to the text of the question.) Jun 15, 2023 at 17:14
• @PatrickR I've added the definition in the edit. Jun 15, 2023 at 18:48
• "Zero set" as in, subsets $A$ of the space $X$ for which there is some continuous $f:X\to\Bbb R$ satisfying $f^{-1}\{0\}=A$? Jun 15, 2023 at 20:17

This is a more detailed version of this answer, but written how I'd write it. I'd love if you were to credit the author there!

In the following proofs we'll write $$Z(f) = \{x\in X : f(x) = 0\} = Z\subseteq\mathbb{R}\times\{0\}$$ of $$X$$, $$f\in C(X)$$ and $$W = \{x\in\mathbb{R} : (x, 0)\in Z\}$$ for simplicity.

For any $$A\subseteq \mathbb{R}$$ define $$D(A) = \{x\in\mathbb{R} : \text{there is no neighbourhood }U\text{ of }x\text{ such that }U\cap A\text{ is meager}\}.$$

Proposition 1. $$D(A) = \emptyset$$ iff $$A$$ is meager.

Proof: If $$A$$ is meager then obviously $$D(A) = \emptyset$$. Conversely if $$D(A) = \emptyset$$ then taking cover of $$\mathbb{R}$$ by open sets $$U_x$$, $$x\in U_x$$, such that $$U_x\cap A$$ is meager, and taking a countable subcover $$U_1, U_2, ...$$ (note that $$\mathbb{R}$$ is Lindelöf) we see that $$\bigcup_{i=1}^\infty (U_i\cap A) = A$$ is meager as a countable union of meager sets. $$\square$$

Proposition 2. $$D(A\setminus D(A)) = \emptyset$$ for all $$A\subseteq \mathbb{R}$$, that is $$A\setminus D(A)$$ is always meager.

Proof: For all $$x\in A\setminus D(A)$$ consider $$U_x$$, $$x\in U_x$$ such that $$U_x\cap A$$ is meager. Since $$\bigcup_{x\in A\setminus D(A)} U_x$$ is again Lindelöf, we can find $$U_1, U_2, ...$$ such that $$U_i = U_{x_i}$$ for some $$x_i\in A\setminus D(A)$$ and $$\bigcup_{i=1}^\infty U_i = \bigcup_{x\in A\setminus D(A)} U_x$$, thus showing that $$A\setminus D(A)\subseteq \bigcup_{x\in A\setminus D(A)} (U_x\cap A) = \bigcup_{i=1}^\infty (U_i\cap A)$$ is meager. $$\square$$

Proposition 3. $$D(A)$$ is closed.

Proof: If $$U\cap A$$ is meager, then of course $$U\subseteq \mathbb{R}\setminus D(A)$$. $$\square$$

Theorem 1. For a zero set $$Z\subseteq \mathbb{R}\times\{0\}$$ of $$X$$, its projection onto $$\mathbb{R}$$, namely $$W$$, is either meager or comeager in $$\mathbb{R}$$ with Euclidean topology.

Proof: Case 1. If for all intervals $$(a, b)\subseteq \mathbb{R}$$ we can find a subinterval $$(p, q)\subseteq (a, b)$$ such that $$(p, q)\cap W$$ is meager, then $$D(W)$$ has empty interior, thus is nowhere dense. It follows that $$W = D(W)\cup (W\setminus D(W))$$ is meager (and in fact $$D(W) = \emptyset$$).

Case 2. There is interval $$(a, b)\subseteq \mathbb{R}$$ such that for any subinterval $$(p, q)\subseteq (a, b)$$, $$(p, q)\cap W$$ is not meager.

Fix $$\varepsilon > 0$$. For every $$x\in W$$ take $$n_x$$ such that $$f[U_{n_x}(x)]\subseteq [-\varepsilon, \varepsilon]$$. If $$(p, q)\subseteq (a, b)$$ then $$F_n = \{x\in (p, q)\cap Z : n_x = n\}$$ can't be nowhere dense for all $$n$$ since $$\bigcup_{n=1}^\infty F_n = (p, q)\cap Z$$ which by assumption is not meager. So can pick $$n$$ for which $$F_n$$ is not nowhere dense, and further an interval $$(r, s)\subseteq \overline{F_n}$$. If $$x\in (r, s)$$, we can pick an increasing $$(t_k)_{k\in\mathbb{N}}$$ and decreasing $$(s_k)_{k\in\mathbb{N}}$$ sequence of elements of $$F_n$$, both converging to $$x$$. Since for any $$m\in\mathbb{N}$$ and large enough $$k\in\mathbb{N}$$, $$U_m(x+1)$$ and $$U_m(x+\sqrt{2})$$ intersect $$U_n(t_k)$$, $$U_m(x-1)$$ and $$U_m(x-\sqrt{2})$$ intersect $$U_n(s_k)$$, we must have $$|f(x\pm 1, 0)|\leq \varepsilon$$ and $$|f(x\pm\sqrt{2}, 0)|\leq \varepsilon$$. This shows that the sets $$A_1^\pm(\varepsilon)=\{x\in (a, b) : |f(x\pm 1, 0)|\leq \varepsilon\}, A_2^\pm(\varepsilon) = \{x\in (a, b) : |f(x\pm\sqrt{2}, 0)|\leq \varepsilon\}$$ contain dense open sets in $$(a, b)$$ by taking union of all intervals $$(r, s)$$ i.e. their complement is nowhere dense. Thus $$A = \bigcap_{n=1}^\infty (A_1^+(1/n)\cap A_1^-(1/n)\cap A_2^+(1/n)\cap A_2^-(1/n))$$ is such that $$f(x\pm 1, 0) = f(x\pm \sqrt{2}) = 0$$ for $$x\in A$$ and $$(a, b)\setminus A$$ is meager. So $$(a\pm 1, b\pm 1)\setminus W$$ and $$(a\pm \sqrt{2}, b\pm \sqrt{2})\setminus W$$ must be meager by translating. Of course, all of those intervals again satisfy the assumption of case 2 since if $$(a, b)\setminus W$$ is meager, and $$(p, q)\cap W$$ were meager for some subinterval $$(p, q)\subseteq (a, b)$$, then $$(p, q)$$ would also be meager, which it isn't. So by induction, $$(a+c, b+c)\setminus W$$ is meager for all $$c\in \{n+\sqrt{2}m : n, m\in\mathbb{Z}\}$$. Of course $$\bigcup_{c\in \{n+\sqrt{2}m : n, m\in\mathbb{Z}\}}(a+c, b+c) = \mathbb{R}$$ since $$\{n+\sqrt{2}m : n, m\in\mathbb{Z}\}$$ is dense in $$\mathbb{R}$$, so that $$\mathbb{R}\setminus W$$ is meager as a countable union of meager sets. Thus $$W$$ is comeager. $$\square$$

Theorem 2. $$X$$ is not realcompact.

Proof: For simplicity I'll refer to subsets of $$\mathbb{R}\times \{0\}$$ as meager or comeager when they are so in Euclidean topology when projected onto $$\mathbb{R}$$. Since countable intersection of comeager zero sets is a comeager zero set, we can speak of the $$z$$-filter $$\mathcal{F}$$ generated by the comeager zero sets of $$\mathbb{R}\times \{0\}$$, and it's closed under countable intersections. To show that $$\mathcal{F}$$ is a $$z$$-ultrafilter, we can restrict our attention to the zero sets in $$\mathbb{R}\times \{0\}\in \mathcal{F}$$. If $$Z = Z(f)\subseteq \mathbb{R}\times \{0\}$$ is a meager zero set, and if $$\{x\in\mathbb{R}\times\{0\} : |f(x)|\leq \frac{1}{n}\}$$ were comeager for all $$n$$, then $$Z$$ would be comeager, which is impossible. So there is $$n$$ such that $$\{x\in\mathbb{R}\times\{0\} : |f(x)|\leq \frac{1}{n}\}$$ is meager, thus $$\{x\in\mathbb{R}\times\{0\} : |f(x)|\geq \frac{1}{n}\}$$ is a comeager zero set disjoint from $$Z$$. Thus $$\mathcal{F}$$ is a real $$z$$-ultrafilter. Any $$x\in \mathbb{R}\times \{0\}$$ is a meager zero set, so we can find comeager zero set $$Z_x$$ disjoint from it (in fact $$(\mathbb{R}\times\{0\})\setminus \{x\}$$ is a comeager zero set), $$\bigcap \mathcal{F} \subseteq \bigcap_x Z_x = \emptyset$$ so that $$\mathcal{F}$$ is free. This shows that $$X$$ is not realcompact. $$\square$$