An order that lacks "infinite" transitivity?

Question

The ordering on the real line "$<$" possesses the property that if we have a sequence $(a_n)_{n=1}^{\infty}$ such that $a_n < a_{n+1}$ and if we have a finite limit $a$, then $a_1 < a_2 < a_3 < \dots < a$ and we can conclude that $a_1 < a$ - even though transitivity of "$<$" is a statement of finitely many elements.

Can we have an ordering, not necessarily on the reals, where "infinite" transitivity is not necessarily followed? We'd keep all the other properties: anti-symmetric and transitive; but we could have an $"a"$ such that $a_0$ and $a$ are incomparable but there exists an infinite chain such that $a_0 < a_1 < \dots < a$. Have these orderings been discussed already?

Attempt: An attempt I've thought about is taking the extended reals ($\mathbb{R} \cup{\{\pm \infty\}}$) and defining the normal order $<$ on the finite numbers but every finite number is "greater than" $\infty$, $-\infty > \infty$ , and every finite number is "less than" $-\infty$. So if we had a sequence diverging to $\infty$ , we'd simultaneously arrive at the smallest element while getting larger; which wouldn't be infinite transitive. However, I'm curious about a case where our limit $a$ is still larger than at least one element in our sequence.

Motivation: Went on a wikipedia rabbit hole after reviewing Zorn's lemma.

Answer 1

Taking the most generous interpretation of your question, it seems you are considering a partial order $\leq$ on some topological space $X$, and you are asking if strictly increasing sequences must converge to an upper bound, if the order is "continuous" in that sense.

To put it simply, the answer is absolutely not. Without imposing, a priori, some relationship between $\leq$ and the topology of $X$, there is no reason to believe that such a property would hold, and in a precise sense, it usually won't. Similar to how most functions are not continuous, likewise most total orders will not be "continuous" in this way. For instance, the Well Ordering theorem guarantees the existence of some well order relation on the reals, but not only will this ordering disagree with the usual order relation of the real numbers, it will be "discontinuous" at every point.

Your attempt can be adjusted slightly to give an example which is more intuitive. Define $X:=\mathbb{R}\cup\{-\infty\}$, and define $\leq$ as a total order on $X$ in the expected way, so that $-\infty$ is the global minimum. Moreover, provide $X$ with the topology that results from taking $\mathbb{R}\cup\{-\infty,\infty\}$, and gluing together $-\infty\sim\infty$. In this topology, any strictly increasing and unbounded sequence in $X$ limits to $-\infty$, which I think is what you wanted. The identification of $-\infty\sim\infty$ is necessary, since otherwise the positive infinity "gets in the way" by absorbing all the unbounded increasing sequences, preventing a counter example. We also need to define the ordering after the identification, since otherwise we end up in a situation where $-\infty<-\infty$, which is not allowed.

An equivalent example is to let $X=\{(x,y)\in\mathbb{R}^2 : x^2+y^2=1\}$, which is the unit circle. Give $X$ the usual topology as a topological subspace of $\mathbb{R}^2$. As for the order, for each point $p\in X$, define $\arg(p)$ as the unique value $\theta\in[0,2\pi)$ for which $p=(\cos(\theta),\sin(\theta))$. Now define the ordering $\leq$ of $X$ so that $p\leq q \iff \arg(p)\leq \arg(q)$. In this case, we have the point $(1,0)$ as the global minimum of $X$, and we get our counter example by approaching this point from below the x-axis. This example is actually the same as the previous one, under isomorphism.

Answer 2

Your example is not good. You've simply switched the labels of $\infty$ and $-\infty$. You still treat the "divergence to $\infty$" under the standard order of the real numbers, but then you're using your modified one.

So in reality, you've actually defined two orders which a priori have nothing to do with one another, and then you claim that an upper bound in one order is a lower bound in another. That's not what you've asked, though. You've asked about an order, singular, with some properties.

But what you've asked is impossible. An upper bound is by definition greater than, or equal to, each an every element of the sequence (or subset). The infinite transitivity is not really infinite in any kind of way, except that the set of points below $a$ is infinite, since it includes all the $a_n$s.

Now, orders generate a topology by using cones, that is "all points above" and "all points below" a certain point in the order. For linear orders this is a particularly nice topology. In any case, if you have an increasing sequence, a limit (if such limit exists) is going to be an upper bound in that topology. Indeed, the limits are exactly the minimal upper bounds (and in a linear order, there can be at most one such minimal, and indeed the topology is Hausdorff, and hence nice).