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A linearly-ordered topological space or LOTS is one whose topology admits a basis generated by open intervals of a total ordering of its points.

A well-based space is one which admits a local basis of each point that is totally ordered by set inclusion.

It appeared to me that the former implied the latter, and I attempted to prove it via the following: for any point $p$ in a LOTS take (possibly transfinite) sequences $a:\Gamma\rightarrow[-\infty,p)$ and $b:\Gamma\rightarrow(p,\infty]$ for some ordinal $\Gamma$ which are monotonic and surjective with $a_0=-\infty$ and $b_0=\infty$. Then for any open interval $(c,d)$ containing $p$ there is an ordinal $\beta$ such that $c\le a_\beta<p<b_\beta\le d$, so $\{(a_\alpha,b_\alpha)\mid\alpha\in\Gamma\}$ is a local basis of $p$.

However, I got a response that a LOTS can be not well-based if there exists a point with different cofinalities on the left and the right, giving the example of $\omega_1+1+\omega*$ with the order topology (where $+$ denotes order concatenation and $\omega*$ the reverse order of $\omega$). This doesn't jibe with my understanding though, since there should exist exist monotonic surjections from $\omega_1$ to $\omega$ to make my argument work. Is this response wrong, or is there some shortcoming of my argument?

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  • $\begingroup$ Where your argument is wrong that there doesn't have to exist ordinal $\Gamma$ which would be the same for both sides. $\endgroup$
    – Jakobian
    Commented Mar 21 at 15:40

3 Answers 3

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Your argument is wrong because such ordinal $\Gamma$ and monotone surjections need not exist. For example, there is no (not necessarily strictly) increasing surjection $f:\omega_1\to \omega$, for otherwise by picking $\alpha_n\in f^{-1}(n)$ for each $n$, we'd obtain a countable cofinal subset of $\omega_1$. But cofinality of $\omega_1$ is uncountable.

Let $X = \omega_1+\{n' : n\in\mathbb{N}_0\}$ and let $x = 0'$ be the point greater than all of points of $\omega_1$, where I decided to use $\{n' : n\in\mathbb{N}\}$ instead of $1+\omega*$ for the clarity of argument. Here $n' < k'$ iff $n > k$.

Suppose that $U_t$, $t\in T$ is a linearly ordered base of $x$ where $T$ is a totally ordered set. We can write $T = \bigcup_{n=1}^\infty T_n$ where $T_n = \{t\in T : \min\{k : k'\in U_t\} = n\}$. Since we know that $k'\notin U_t$, $t\in T_n$ for $n < k$, we see that for $s\in T_k$ we must have $U_s\supsetneq U_t$ and so $t < s$ since the map $a\mapsto U_a$ is increasing. So $T_n$ decompose $T$ into disjoint sets such that, with notation $A < B$ iff $\forall_{a\in A, b\in B}\ a < b$, we have $T_n < T_k$ when $n < k$.

If we pick a sequence $t_k\in T_{n_k}$ (some of the $T_n$ can be empty, but $T_n\neq \emptyset$ for infinitely many $n$), this then means that $U_{t_k}$ is a countable local base at $x$. But this would then mean that $U_{t_k}\cap (\omega_1\cup \{0'\})$ is a countable local base at $x$. In other words, $\omega_1\cup \{0'\}\cong \omega_1+1$ would be first countable. But it isn't, since it has no countable local base at the point $\omega_1$ - again, for the reasons of uncountable cofinality.

You can see how this argument generalizes to show that if $X$ is an ordered space, then for $x\in X$ to have a totally ordered local base at $x$, we need to have that if the cofinality of $\{y\in X : y < x\}$ and coinitiality $\{y\in X : y > x\}$ is infinite, then they have to agree.

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    $\begingroup$ minor tweak for the last paragraph: "(or one of sets is empty, or the point $x$ has an immediate predecessor or has an immediate successor)" $\endgroup$
    – PatrickR
    Commented Mar 21 at 17:28
  • $\begingroup$ @PatrickR Thanks $\endgroup$
    – Jakobian
    Commented Mar 21 at 17:38
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There is no monotonic surjection $s:\kappa\to\lambda$ when $cf(\kappa)>\lambda$. To see this, we construct a cofinal map $f_s:\lambda\to\kappa$ by $f(\alpha)=\min\{\beta<\kappa:s(\beta)=\alpha\}$.

We then see that $f$ is cofinal: given $\beta<\kappa$, consider $s(\beta)+1<\lambda$. There is some minimal $\gamma<\kappa$ such that $s(\gamma)=s(\beta)+1$, so $f(s(\gamma))=\gamma$. Since $s$ is monotonic and $s(\gamma)>s(\beta)$, we have $\gamma>\beta$.

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    $\begingroup$ This answer is incomplete. $\endgroup$
    – Jakobian
    Commented Mar 21 at 15:40
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(This proves the same thing as Jakobian's answer, but presented in the more general context of partially ordered sets.)

Some generalities about posets

Let $(P, \le)$ be a partially ordered set. A subset $A\subseteq P$ is a chain if any two of its elements are comparable. A subset $A\subseteq P$ is cofinal if every element of $P$ is less than or equal to some element of $A$.

If $P$ is totally ordered, there is a well-ordered cofinal subset of some cardinality $\kappa$. The smallest such cardinal is the cofinality of $P$ and is a regular cardinal (or $1$ is the poset has a maximum element) and there is a strictly increasing (transfinite) sequence $(x_\alpha)_{\alpha<\kappa}$ cofinal in $P$.

In a general (non-totally ordered) poset $P$ there may not exist a cofinal chain. If $P$ has a cofinal chain, one says that $P$ has true cofinality (Jech, Set Theory, p. 461). In that case, as shown in In a poset with a cofinal chain, does every cofinal subset admit a cofinal chain?, every cofinal subset $Q\subseteq P$ (considered as a poset in its own right) also has a cofinal chain. And conversely, every cofinal chain in $Q$ is a cofinal chain in $P$. So to determine if $P$ has true cofinality, it is enough to restrict attention to a suitably chosen cofinal subset.

(Note: if $P$ has true cofinality, all cofinal chains in $P$ have the same cofinality. See here.)

Applying this to topology

Let $X$ be a topological space and $x$ be a point of $X$.

Consider the poset $\mathcal N$ consisting of all nbhds of $x$, ordered by reverse inclusion ($\supseteq$).

  • A local base of nbhds of $x$ is precisely a cofinal subset of $\mathcal N$.
  • $X$ being well-based at $x$ is equivalent to $\mathcal N$ having a cofinal chain; or equivalently, the existence in $(\mathcal N,\supseteq)$ of a cofinal strictly increasing sequence $(x_\alpha)_{\alpha<\mu}$ indexed by some regular cardinal $\mu$ (unless $x$ itself has a smallest nbhd).

Specializing to the case of LOTS, suppose $(X,\le)$ is a totally ordered set with the corresponding order topology. And let $x\in X$.

If $x$ has no element to one side or has an immediate predecessor or an immediate successor, then it is easy to check $X$ is well-based at $x$ by considering intervals of the form $(a,x]$ or $[x,b)$.

Otherwise, $x$ has infinite cofinality $\kappa$ on its left (i.e., the cofinality of $(\leftarrow,x)$) and infinite (downward) cofinality $\lambda$ on its right (i.e., the cofinality of $(x,\rightarrow)$ ordered by $\ge$). Then there is an increasing sequence $(a_\alpha)_{\alpha<\kappa}$ converging to $x$ from the left and a decreasing sequence $(b_\beta)_{\beta<\lambda}$ converging to $x$ from the right. And the collection of intervals $(a_\alpha,b_\beta)$ forms a local base at $x$. The corresponding poset ordered by reverse inclusion is isomorphic to the product poset $\kappa\times\lambda$ (ordered by $\langle\alpha_1,\beta_1\rangle \le \langle\alpha_2,\beta_2\rangle$ iff $\alpha_1\le\alpha_2$ and $\beta_1\le\beta_2$).

Proposition: Let $\kappa$ and $\lambda$ be infinite regular cardinals. The product poset $P=\kappa\times\lambda$ has true cofinality iff $\kappa=\lambda$.

Proof: If $\kappa=\lambda$, the diagonal $\{\langle\alpha,\alpha\rangle : \alpha<\kappa\}$ is a cofinal chain.

The rest of the proof uses the same argument used to show that the Dieudonné plank is not radial (see here) or that the Tychonoff plank is not radial (see here).

Assume $\kappa$ and $\lambda$ are not equal, say $\kappa>\lambda$. Suppose there is a strictly increasing cofinal sequence $(x_\alpha)_{\alpha<\mu}$ cofinal in $P$ and indexed by an infinite regular cardinal $\mu$. Projecting to the first component gives the sequence $(\pi_1(x_\alpha))_\alpha$ cofinal in $\kappa$. Therefore $\kappa=\text{cof}(\kappa)\le\mu$. On the other hand, $P$ has cardinality $\kappa$, hence $\mu\le\kappa$. So $\mu=\kappa$.

Now projecting to the second component give the sequence $(\pi_2(x_\alpha))_\alpha$ cofinal in $\lambda$. So for each $\beta<\lambda$ there is some $\alpha<\mu$ with $\pi_2(x_\alpha)\ge\beta$ and since the sequence of $x_\alpha$ is increasing in $P$, the set $\{\alpha<\mu:\pi_2(x_\alpha)<\beta\}$ has cardinality less than $\lambda$. And since every $\alpha$ belongs to one of these sets and $\lambda<\mu$, $\mu$ is a union of less than $\mu$ sets, each of cardinality less than $\mu$. That is, $\mu$ is not a regular cardinal, which contradicts the regularity of $\kappa$.


And the final result for LOTS:

Theorem: Let $(X,\le)$ be a totally ordered set with its order topology and let $x\in X$. The topology on $X$ is not well-based at $x$ iff $x$ has infinite and unequal cofinalities on its left and right sides.

Stated equivalently, $X$ is well-based at $x$ iff $x$ has no element to its left or to its right, or $x$ has an immediate predecessor or immediate successor, or $x$ has infinite and equal cofinalities on both sides.

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