# Critical topological spaces

This is a follow-up question to Existence of injective neighborhood selection function as separation axiom.

Let $(X, \tau)$ be a topological space. If there is an injective map $f:X\to\tau$ such that $x\in f(x)$ for all $x\in X$, we call $f$ an injective neighborhood selection map.

We say a space is critical if it has an injective neighborhood selection map and every such map $f:X\to\tau$ is "almost surjective", i.e. $\text{im}(f) = \tau\setminus\{\emptyset\}$.

For example, if $\alpha$ is an ordinal, consider $\tau_\alpha = \{\emptyset\} \cup \{\alpha\setminus \beta: \beta \in \alpha\}$, which is the topology of upper sets. Then $(\alpha,\tau_\alpha)$ is critical.

Is every critical space isomorphic to $(\alpha, \tau_\alpha)$ for some ordinal $\alpha$? If not, is there an example of a critical space that is $T_1$ or even Hausdorff?

[EDIT: The second question has become obsolete, see Eric Wofsey's comment below.]

• A $T_1$ space (with more than one point) certainly can't be critical, because you can choose $f$ such that $X\setminus f(x)$ always has only one point. Mar 12 '15 at 9:08
• That's right - thanks Eric! So the second part of the question is obsolete.. But I have no idea about the first part. Mar 12 '15 at 12:05

The answer is no. Consider the space $X=\{0,1,2\}$ with the topology $\tau=\{\emptyset,\{0\},\{1,2\},X\}$. There are precisely two injective neighborhood selectors, both of which are almost surjective in your sense. Namely, we must map $0\mapsto\{0\}$ and then $1$ and $2$ get mapped to $\{1,2\}$ and $X$, in either way.

So this space is critical in your sense, but the topology is not of your form $\tau_\alpha$ for any ordinal $\alpha$. For example, it is not $T_0$, since we cannot separate $1$ and $2$.

• This example also appears in an answer at the other question: mathoverflow.net/a/199567/1946 Mar 12 '15 at 15:23
• But there is also an i.n.sel. which is not almost surjective, namely such that $\ f(1)=f(2)=\{1\ 2\}$ (hence the above is not a counter-example). Dec 15 '18 at 4:08
• @WlodAA That map is not injective. Dec 15 '18 at 12:29
• Joel, thank you -- indeed, I forgot about the injectivity condition, my fault. Dec 16 '18 at 5:56

I claim that every finite $T_{0}$ critical space is isomorphic to $(n,\tau_{n})$.

Suppose that $X$ is a finite $T_{0}$-space. Let $\leq$ be the specialization ordering on $X$. Then $U\subseteq X$ is open if and only if $U$ is upwards closed. Then each finite $T_{0}$-space has an injective selector map, namely $x\mapsto{\uparrow}x$ is a neighborhood selector map.

Now suppose that $X$ is not linearly ordered. Then there are two elements $r,s\in X$ with $r\not\leq s$ and $s\not\leq r$. Then ${\uparrow}r\cup{\uparrow}s$ has no least element, so ${\uparrow}r\cup{\uparrow}s\neq{\uparrow}x$ whenever $x\in X$. Therefore $|X|<|\mathcal{T}\setminus\{\emptyset\}|$, so there is no surjective neighborhood selector map.

Therefore if $X$ is a finite critical $T_{0}$-space, then the specialization ordering on $X$ is a linear ordering and hence isomorphic to $(n,\tau_{n})$.

$\textbf{added 3/14/2015}$

I claim that the specialization ordering on any critical $T_{0}$-space must be well-founded. Suppose to the contrary that $(X,\mathcal{T})$ is a critical $T_{0}$-topological space whose specialization ordering is not well founded. Then there is some injective neighborhood selection map $j:X\rightarrow\mathcal{T}\setminus\{\emptyset\}$. However, since $X$ is not well-founded, there is a descending sequence $x_{0}>x_{1}>x_{2}>x_{3}>...$ Since $j(x_{n+1})$ is an upwards closed set containing $x_{n+1}$, we have $x_{n}\in j(x_{n+1})$ as well.

Now define a mapping $k:X\rightarrow\mathcal{T}\setminus\{\emptyset\}$ by letting $k(x_{n})=j(x_{n+1})$ for all $n\in\omega$ and where $k(x)=j(x)$ whenever $x\in X\setminus\{x_{n}|n\in\omega\}$. Then $k$ is an injective neighborhood selector map which is not surjective onto $\mathcal{T}\setminus\{\emptyset\}$. We conclude that every critical $T_{0}$-topological space has a well-founded specialization ordering.

I however claim that not every $T_{0}$-critical space is isomorphic to $(\alpha,\tau_{\alpha})$. Let $X=(\omega+1)\cup\{0'\}$ where $0'$ takes the place of an extra zero. Partially order $X$ by letting $\omega+1$ have its original partial ordering and where $0'\leq n$ whenever $n\neq 0$ but where $0\not\leq 0',0'\not\leq 0$. Give $X$ the topology $\mathcal{T}$ where a set $U\subseteq X$ is declared to be open if $U$ is upwards closed and where if $\omega\in U$ then $n\in U$ for some $n\in\omega$. Then $\leq$ is the specialization ordering for the topology $\mathcal{T}$.

I first claim that there is an injective neighborhood selector function. Simply let $j(0)=X\setminus\{0'\},j(0')=X\setminus\{0\},j(\omega)=X$ and $j(n)=\uparrow n$ for all $n\in\omega\setminus\{0\}$.

However, I claim that every injective neighborhood selector function is bijective. Suppose that $j:X\rightarrow\mathcal{T}\setminus\{\emptyset\}$ is a neighborhood selector function. Suppose that $j(\omega)\in X,X\setminus\{0\},X\setminus\{0'\}$. Then we have function $j[\{\omega,0,0'\}]\subseteq\{X,X\setminus\{0\},X\setminus\{0'\}\}$. Therefore $j$ maps $\{\omega,0,0'\}$ onto $\{X,X\setminus\{0\},X\setminus\{0'\}\}$. Therefore one can prove by induction that $j(n)=\uparrow n$ for all $n$.

Now assume that $j(\omega)=\uparrow n$ for some $n>0$. Then $j[\{\omega,0,...,n-1,0'\}]\subseteq\{\uparrow 1,...,\uparrow n,X,X\setminus\{0\},X\setminus\{0'\}\}$.

However, since $|\{\omega,0,...,n-1,0'\}|=n+2$ and $|\{\uparrow 1,...,\uparrow n,X,X\setminus\{0\},X\setminus\{0'\}\}|=n+2$, we have $j[\{\omega,0,...,n-1,0'\}]=\{\uparrow 1,...,\uparrow n,X,X\setminus\{0\},X\setminus\{0'\}\}$. By induction, we can conclude $j(m)=\uparrow m$ for each $m\in\omega$ with $m>n$. In this case, the function $j$ will turn out to be surjective.

We conclude that the space $X$ is critical, but $X$ is not isomorphic to any $(\alpha,\tau_{\alpha})$.

• Excellent results -- thanks Joseph! Mar 14 '15 at 17:23