Let $(X,\leq)$ be a poset.

For every $A\subseteq X$, $a \in A$ is called isolated iff there are only finite number of elements in $A$ that can be compared to it.

A continuous (probably not the best name) set is one with only a finite number of isolated points.

Assume now that every $a \in A$ is isolated.

  1. Can we define a new partial order on $X$ that subsumes the original one and under the new relation $A$ is continuous?
  2. Now let $\leq '$ be a partial order on $A$ (that agrees with $\leq |\underset{A}{}$) under which $A$ is continuous. Is there a way to define a partial order on X that agrees with both $\leq$ and $\leq '$?

I'm particulary interested in (2) where A is countable.

Some motivation: If both answers are 'yes' then we could extend properties of continuous sets to isolated ones (surprisingly this could lead to some interesting local-global theorems that I 'proved' - I'm missing this construction) by talking about this property for every extension described above and take the common denominator of the properties (running on the extensions), if there is one.

Followup question on some reverse notion of this extension:

A classification of the types of poset extensions that raise local-global principle is equivelent to the axiom of choice?


1 Answer 1


Unless I am missing something obvious, every partial order can be extended to a total order. And in a total order, $A$ is either finite or will be continuous.

Here is the proof. $X$ is a set. Let $O_X$ be the set of possible partial orders on $X$. If $o_1, o_2 \in O_X$ we say that $o_1 \leq o_2$ iff every comparison that can be made in $o_1$ can be made in $o_2$ and gives the same result. This relation is a partial order on $O_X$. Now if $C$ is any chain in $O_X$ then you can define a partial order $o_C$ where two things are comparable iff they are comparable for any order in $C$. $o_C$ will be an upper bound for $C$.

Therefore by Zorn's lemma, $O_X$ contains maximal elements. In fact for any partial order $o \in O_X$ the subset of $O_X$ that is comparable to $o$ must contain a maximal element.

To finish all that you have to do is prove that a maximal element of $O_X$ must be a total order. Which means proving that if $o$ is an order such that there are two elements $x$ and $y$ which are not comparable, then $o$ is not maximal.

However I believe that if $o$ is such an order and $x$ is not comparable to $y$, then you can construct a partial order $o'$ that is larger than $o$ which agrees with $o$ when things are comparable to it, and whenever $x' \leq x$ and $y \leq y'$ adds the relation $x' \leq y'$. Do the mechanical work of verifying that $o'$ is well-defined and a partial order and you're done.

If you grant the axiom of choice, every partial order on $X$ can be extended to a total order on $X$.

  • $\begingroup$ Thank you! I knew I was mising something obvious. One property I was talking about was a specific subset, so the common denominator I was talking about was an intersection of extensions -- the main theorem I was going for: which I know isn't empty --(but maybe there are no extensions and that's where this construction came into play). $\endgroup$ May 18, 2016 at 21:11

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