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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 26, 2023 at 17:27
  • $\begingroup$ @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 $\omega_2$. $\endgroup$
    – PatrickR
    Commented Dec 26, 2023 at 20:07
  • $\begingroup$ @StevenClontz See my partial answer. Maybe you can find a way to generalize it. $\endgroup$
    – PatrickR
    Commented Dec 26, 2023 at 23:34

3 Answers 3

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Generalizing @PatrickR's answer.

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 $\kappa=\lambda$.

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  • $\begingroup$ 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 $|C|\le\kappa$. $\endgroup$
    – PatrickR
    Commented Dec 27, 2023 at 4:08
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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.

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  • $\begingroup$ 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. $\endgroup$
    – PatrickR
    Commented Dec 26, 2023 at 20:17
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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 $a<p$, the set $(\leftarrow,a)$ is finite. And the collection of sets $(\leftarrow,a)$ with $a<p$ 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.

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