# LOTS and radial properties for generalized Fort/Fortissimo spaces

This question explores the linearly orderable and radial properties of Fort and Fortissimo spaces and their generalizations.

A Fort space is a set $$X$$ with a distinguished point, call it $$\infty$$, and a topology defined by having every point $$x\ne\infty$$ isolated and the nbhds of $$\infty$$ being the cofinite subsets of $$X$$ containing $$\infty$$.

A Fortissimo space is a set $$X$$ with a distinguished point, call it $$\infty$$, and a topology defined by having every point $$x\ne\infty$$ isolated and the nbhds of $$\infty$$ being the cocountable subsets of $$X$$ containing $$\infty$$.

More generally, let $$\kappa\le\lambda$$ be two infinite cardinals. We can define a generalized Fort space $$F_{\lambda,\kappa}$$ as follows. Take a set $$X$$ of cardinality $$\lambda$$ with a distinguished point $$\infty\in X$$ and define a topology by having every point $$x\ne\infty$$ isolated and the nbhds of $$\infty$$ being the subsets $$V\subseteq X$$ with $$\infty\in V$$ and $$|X\setminus V|<\kappa$$. The closed sets in $$X$$ are the sets containing $$\infty$$ together with the sets of cardinality less than $$\kappa$$.

If someone has a reference with an official name for these spaces in the literature, I'd be interested.

Clearly up to homeomorphism these spaces are completely determined by the cardinals $$\lambda$$ and $$\kappa$$. Also, Fort space = $$F_{\lambda,\omega}$$ and Fortissimo space = $$F_{\lambda,\omega_1}$$.

Note: In the same way that a Fort space can be seen as the one-point compactification of a discrete space and a Fortissimo space can be seen as the one-point Lindelofication of a discrete space, the $$F_{\lambda,\kappa}$$ can be viewed as some one-point ...-fication of a discrete space of size $$\lambda$$ (where the three dots stand for some generalized compactness property, already described by Alexandroff & Urysohn: every open cover of the space has a subcover of size less than $$\kappa$$).

A space $$X$$ is linearly orderable (or LOTS) if has the order topology induced by some total order on the underlying set.

A space $$X$$ is radial if for every set $$A\subseteq X$$ and every point $$p\in\overline A\setminus A$$ there is a transfinite sequence $$(x_\alpha)_{\alpha<\lambda}$$ with each $$x_\alpha\in A$$ and converging to $$p$$. (Here $$\lambda$$ is a limit ordinal and can always be taken to be a regular cardinal; and saying $$(x_\alpha)$$ converges to $$p$$ means every nbhd of $$p$$ contains a tail of the sequence.)

It is easy to see that every LOTS is radial.

Proposition 1: A generalized Fort spaces $$F_{\lambda,\kappa}$$ is radial iff $$\kappa$$ is a regular cardinal.

See below for a proof. In fact, if $$\kappa$$ is a singular cardinal, $$F_{\lambda,\kappa}$$ is not even pseudoradial as it contains a radially closed set that is not closed.

Question: Which of the generalized Fort spaces $$F_{\lambda,\kappa}$$ are linearly orderable (LOTS)?

From Proposition 1, if $$\kappa$$ is a singular cardinal, $$F_{\lambda,\kappa}$$ is not a LOTS. And I can show the following partial result:

Proposition 2: If $$\kappa$$ is a regular cardinal, the space $$F_{\kappa,\kappa}$$ is linearly orderable.

See further below for a proof. This includes $$F_{\omega_1,\omega_1}$$ = Fortissimo on $$\omega_1$$. What about other cases with $$\kappa$$ regular?

To highlight the answer, SUMMARY for LOTS, based on Prop. 2 and the answer of Steven Clontz:

Proposition 3: A generalized Fort spaces $$F_{\lambda,\kappa}$$ is linearly orderable iff $$\lambda=\kappa$$ with $$\kappa$$ a regular cardinal.

(Prop. 1) Proof that $$F_{\lambda,\kappa}$$ is radial if $$\kappa$$ is regular:

Assume $$\kappa$$ is regular and write $$X=F_{\lambda,\kappa}$$. Let $$A$$ be a non-closed subset of $$X$$ and $$p\in\overline A\setminus A$$. Necessarily $$p=\infty$$ and $$|A|\ge\kappa$$. Take an injective function $$\kappa\to A$$. This defines a corresponding transfinite sequence $$(x_\alpha)_{\alpha<\kappa}$$ with values in $$A$$ and all values distinct. I claim that it converges to $$\infty$$. Let $$V$$ by a nbhd of $$\infty$$ in $$X$$. Then $$|\{\alpha\in\kappa:x_\alpha\notin V\}|\le |X\setminus V|<\kappa$$ and since $$\kappa$$ is regular, the set of such $$\alpha$$ is bounded in $$\kappa$$. So $$V$$ contains a tail of the sequence.

(Prop. 1) Proof that $$F_{\lambda,\kappa}$$ is not radial if $$\kappa$$ is singular:

Assume $$\kappa$$ is singular and write $$X=F_{\lambda,\kappa}$$. Take a set $$A\subseteq X\setminus\{\infty\}$$ of cardinality $$\kappa$$. We have $$\overline A=A\cup\{\infty\}$$; in particular $$A$$ is not closed. I claim $$A$$ is radially closed, which will imply $$X$$ is not radial (and not pseudoradial). Suppose by contradiction that there is a transfinite sequence $$(x_\alpha)_{\alpha<\mu}$$ in $$A$$ that converges to $$\infty$$ (the only point in $$\overline A\setminus A$$). Since $$X$$ is $$T_1$$, $$\mu$$ must be a limit ordinal and one can assume wlog that $$\mu$$ is an infinite regular cardinal not exceeding $$|A|=\kappa$$ (see for example the discussion at the end of https://math.stackexchange.com/a/4785111). But $$\kappa$$ is not regular, so $$\mu<\kappa$$, which implies that $$\{x_\alpha:\alpha<\mu\}$$ is closed in $$X$$. So the transfinite sequence cannot converge to $$\infty$$.

(Prop. 2) Proof that $$F_{\kappa,\kappa}$$ is linearly orderable if $$\kappa$$ is regular:

Assume $$\kappa$$ is regular. Take a totally ordered set of order type $$\gamma\cdot\kappa+1$$, where $$\gamma$$ is the order type of $$\mathbb Z$$. Specifically, $$X=(\mathbb Z\times\kappa)\cup\{\infty\}$$, where the product $$\mathbb Z\times\kappa$$ has the antilexicographic order, and $$\infty$$ is an element larger than all the others. The set $$X$$ has cardinality $$\kappa$$. Let $$\tau_{\le}$$ be the corresponding order topology on $$X$$ and let $$\tau$$ be the generalized Fort topology on $$X$$ with $$\lambda=\kappa$$.

Every point $$x\ne\infty$$ is isolated for $$\tau_\le$$ (thanks to the $$\mathbb Z$$ factor). Every nbhd of $$\infty$$ for $$\tau_\le$$ has a complement of cardinality less than $$\kappa$$. Conversely, every subset $$A\subseteq\mathbb Z\times\kappa$$ of cardinality less than $$\kappa$$ is bounded in $$\mathbb Z\times\kappa$$ (because $$\kappa$$ is regular). So the complement of such an $$A$$ is a nbhd of $$\infty$$ for $$\tau_{\le}$$. This shows that the nbhds of $$\infty$$ are the same for the two topologies $$\tau_{\le}$$ and $$\tau$$. So the two topologies are the same and $$\tau$$ is linearly orderable.

• Do you have a slick proof that $F_{\omega_1,\omega}$ is not LOTS? I wrote half of one up, but got stuck in the details and ran out of time for now. Commented Dec 26, 2023 at 17:27
• @StevenClontz Yeah, I have one for $F_{\lambda,\omega}$. Will write it up when I am back at my desk in a few hours. I would guess that $F{\lambda,\kappa}$ is not LOTS in general when $\lambda>\kappa$, but not sure about it. For example $F_{\omega_2,\omega_1}$ = Fortissimo on $\omega_2$. Commented Dec 26, 2023 at 20:07
• @StevenClontz See my partial answer. Maybe you can find a way to generalize it. Commented Dec 26, 2023 at 23:34

Let $$\kappa\leq\lambda$$. We will show that if $$F_{\lambda,\kappa}=\lambda\cup\{p\}$$ is a LOTS ordered by $$\preceq$$, then $$\kappa=\lambda$$.

Choose a maximal subset $$C$$ of $$(\leftarrow,p)$$ that's well-ordered by $$\preceq$$. Note that $$C\cong\alpha\leq\kappa$$. If not, choose $$c\in C$$ with $$C\cap(\leftarrow,c)\cong \kappa$$. Then $$|(\leftarrow,c]|\geq\kappa$$ but its complement is a neighborhood of $$p$$, a contradiction.

Then we have $$(\leftarrow,p)=\bigcup_{c\in C}(\leftarrow,c]$$ (if not, $$C$$ wasn't maximal), with each $$|(\leftarrow,c]|<\kappa$$. Since $$|C|\leq\kappa$$, we have $$|(\leftarrow,p)|\leq\kappa$$.

Similarly, we can show $$|(p,\rightarrow)|\leq\kappa$$, and thus $$\lambda=|F_{\lambda,\kappa}|\leq\kappa$$, and conclude $$\kappa=\lambda$$.

• Very nice! Your argument does not depend on $\kappa$ being regular or singular. And you have shown a handy lemma: If $(C,\le)$ is a well-ordered set such that every initial segment $\{x\in C:x\le c\}$ for $c\in C$ has size less than $\kappa$, then $|C|\le\kappa$. Commented Dec 27, 2023 at 4:08

Comment too long for a comment: here's another way to see one direction of Prop 1.

Proof that $$F_{\lambda,\kappa}$$ is not pseudoradial if $$\kappa$$ is singular:

Given $$A\in[X\setminus\{\infty\}]^{\kappa}$$, note that $$U=X\setminus A$$ contains $$\infty$$ and $$|X\setminus U|=|A|\not<\kappa$$. Thus $$U$$ is not a neighborhood of $$\infty$$ and $$A$$ is not closed.

Let $$a_\alpha$$ be a transfinite sequence of length $$\gamma$$ of points from $$A$$. If $$\gamma<\kappa$$, then $$X\setminus\{a_\alpha:\alpha<\gamma\}$$ is open and $$\{a_\alpha:\alpha<\gamma\}$$ is closed and thus does not converge outside $$A$$. If $$\gamma=\kappa$$, choose a cofinal subset $$C\subseteq\gamma$$ with $$|C|<\gamma$$. Then $$X\setminus C$$ is an open neighborhood of $$\infty$$ that misses a cofinal subsequence of $$a_\alpha$$, so $$a_\alpha\not\to\infty$$. In particular, $$a_\alpha$$ either fails to converge, or converges (trivially) to some point of $$\{a_\alpha:\alpha\in C\}\subseteq A$$. Thus $$A$$ is is radially closed.

Since $$A$$ is radially closed, but not closed, $$F_{\lambda,\kappa}$$ is not pseduoradial.

• Yeah, it's very similar to what I have. I had relegated some of the details of the choice of $C$ to math.stackexchange.com/a/4785111, but you choose $C$ explicitly, which makes it more self-contained. Maybe one minor thing. $\gamma$ is some limit ordinal, but it could even be that $\gamma>\kappa$ (as long as $\gamma<\kappa^{+}$). So don't we need to ensure first that we can assume $\gamma\le|A|=\kappa$? That's what the related question was about. Commented Dec 26, 2023 at 20:17

Partial result:

If $$\lambda$$ is uncountable, the Fort space $$F_{\lambda,\omega}$$ is not a LOTS.

Proof: Suppose there is a total order $$\le$$ on $$X=F_{\lambda,\omega}$$ such that the corresponding order topology coincides with the Fort topology. Every nbhd of $$p=\infty$$ is cofinite. So given any $$a, the set $$(\leftarrow,a)$$ is finite. And the collection of sets $$(\leftarrow,a)$$ with $$a forms a chain in the poset of finite subsets of $$(\leftarrow, p)$$ ordered by inclusion. Such a chain is necessarily countable, as there cannot be more than one element of given finite cardinality in such a chain. This shows that there are only countably many elements of $$X$$ less than $$p$$. And similarly for elements larger than $$p$$. Consequently $$\lambda$$ must be countable.

I am not sure how to generalize this. For example, for $$X= F_{\omega_2,\omega_1}$$ = Fortissimo space on $$\omega_2$$.

What is the maximum size of a chain of countable subsets of $$\omega_2$$ ordered by inclusion?

If we could show it cannot be more than $$\omega_1$$, then we could deduce that $$X$$ is not a LOTS.

(Added later) Such a chain can have very large cardinality, as witnessed by subsets $$(\leftarrow,b)\cap\mathbb Q\subseteq\mathbb Q$$ with $$b\in\mathbb R$$. So it does not seem this approach is going to help much.