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 math.stackexchange](https://math.stackexchange.com/questions/4718866/mysior-plane-is-not-realcompact).


**Edit 2:** The reason why I reposted my question here, is that I think only small amount of people on math.stackexchange know what realcompactness is. I want to gain exposure to knowledgeable people that have a higher chance to help me and maybe even know the example I'm talking about. The original paper this comes from is "A union of realcompact spaces" by A. Mysior. All books and articles only briefly mention this example (e.g. Engelking and plethora articles about realcompactness and various alternatives). My guess is that the $1$ and $\sqrt{2}$ are there for a reason, and I tried to consider what would happen if e.g. $f\in C(X)$ were $0$ on $[0, 1]\times \{0\}$, because it seems like maybe I could prove that a zero set containing this will contain $[1, 2]\times \{0\}$ and $[\sqrt{2}, 1+\sqrt{2}]\times\{0\}$ etc., but I feel like while $f$ on $U\_n(x)$ for $x\in [0, 1]$ must approach $0$ as $n$ gets larger, they can do so at distinct rates for all $x$, hence my difficulties. Overall, my first intuition would be to figure out the zero sets contained in $\mathbb{R}\times\{0\}$.



Mysior plane is not realcompact
The proof that I can think of applies a Baire-category argument to the normal topology of the real line.
As in your argument for realcompactness you need to look at zero-sets that are subsets of the $x$-axis. Let $Z\_f$ be such a zero-set, and look at $Z=\{x:f(x,0)=0\}$. There are two cases: (1) every nonempty open interval contains a nonempty open interval in which $Z$ is of first category (union of countably many nowhere dense sets), or (2) there is an open interval $(a,b)$ such that for no subinterval $(p,q)$ of $(a,b)$ the intersection $Z\cap(p,q)$ is of first category in $(p,q)$.
In the first case you can show that $Z$ is of first category in $\mathbb{R}$.


Consider the second case; we show that $(a+1,b+1)\setminus Z$ and $(a+\sqrt2,b+\sqrt2)\setminus Z$ are both of first-category.  

Let $\varepsilon>0$. For every $x\in(a,b)\cap Z$ there is an $n\_x$ such that $f[U\_{n\_x}(x)]\subseteq(-\varepsilon,\varepsilon)$.
Let $(p,q)$ be a subinterval of $(a,b)$, by our assumption there is an $n$ such that $F=\{x\in(p,q)\cap Z:n\_x=n\}$ is not nowhere dense, that is there is a further subinterval $(r,s)$ of $(p,q)$ that is contained in the closure (in the normal topology of $\mathbb{R}$) of $F$.
Let $x\in(r,s)$ and take an increasing sequence $\langle t\_k\rangle$ in $F$ that converges to $x$. Now check that $U\_m(x+1)$ interesects $U\_n(t\_k)$ for infinitely many $k$; this implies that $|f(x+1,0)|\le\varepsilon$; similarly you'll find that $|f(x+\sqrt2,0)|\le\varepsilon$.
This shows that the sets $O\_1(\varepsilon)=\{x\in(a,b):|f(x+1, 0)|\le\varepsilon\}$ and $O\_2(\varepsilon)=\{x\in(a,b):|f(x+\sqrt2, 0)|\le\varepsilon\}$ both contain dense open subsets of $(a,b)$.  

Now let $D=\bigcap\_m\bigl(O\_1(2^{-m})\cap O\_2(2^{-m})\bigr)$ then $(a,b)\setminus D$ is of first-category and for all $x\in D$ we have $f(x+1,0)=f(x+\sqrt2,0)=0$.


Using simlar methods you can show that $(a-1,b-1)\setminus Z$ and $(a-\sqrt2,b-\sqrt2)\setminus Z$ are both of first-category.
Now use the fact that $\{a+m+n\sqrt2:m,n\in\mathbb{Z}\}$ is dense in $\mathbb{R}$ to see that, in fact, the complement of $Z$ in $\mathbb{R}$ is of first category.


So a zero-set contained in the $x$-axis either is of first-category in $\mathbb{R}$ or its complement is. The latter form a zero-set ultrafilter with the countable intersection property but with an empty intersection.


