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.

Crossposted with mathoverflow.

  • $\begingroup$ What is "realcompact" versus "compact"? As in, any $\Bbb R$-indexed open cover has a finite subcover? $\endgroup$
    – FShrike
    Jun 14, 2023 at 16:08
  • $\begingroup$ @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". $\endgroup$
    – Jakobian
    Jun 14, 2023 at 16:11
  • $\begingroup$ 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.) $\endgroup$
    – PatrickR
    Jun 15, 2023 at 17:14
  • $\begingroup$ @PatrickR I've added the definition in the edit. $\endgroup$
    – Jakobian
    Jun 15, 2023 at 18:48
  • $\begingroup$ "Zero set" as in, subsets $A$ of the space $X$ for which there is some continuous $f:X\to\Bbb R$ satisfying $f^{-1}\{0\}=A$? $\endgroup$
    – FShrike
    Jun 15, 2023 at 20:17

1 Answer 1


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. $\square$


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