If a set is equipped with a dense, complete order then the corresponding topological space is connected - and hence, so are its continuous images, even in unordered spaces. What happens if we remove the completeness condition on the order, and consider a general densely ordered space X?

I know that f is continuous from X to another ordered space, then the induced order on f(X) need not be dense. But is there some weaker property of densely ordered spaces which is preserved under continuous maps? If so, can such a property be generalized to unordered spaces in the same sort of way that "densely and completely ordered" generalizes to "connected"?

Edit: as discussed below, here is the followup question asked by Noah, as well as another one added by me.

  • $\begingroup$ Very nice question. To start with the countable case, can you characterize the continuous images of the rationals? $\endgroup$ Jun 19 '10 at 2:13

Let me observe that Noah's excellent answer generalizes to solve the full case.

Theorem. Every topological space is a continuous image of a dense linear order.

Proof. Let $\kappa$ be any ordinal number. Let $P$ be the linear order $\mathbb{Q}\times\kappa$, under the lexical order, which is obtained by replacing each ordinal less than $\kappa$ with a copy of the rational order $\mathbb{Q}$. This is a dense linear order, resembling the well-known long line, but with rationals and with $\kappa$ instead of the unit interval and $\omega_1$. You could call it the $\kappa$-long rational line. For each $\alpha\lt\kappa$, let $Q_\alpha=\mathbb{Q}\times\{\alpha\}$ be the copy of $\mathbb{Q}$ surrounding $\alpha$ in $P$. This is an open set, and they partition $P$ just as in Noah's argument.

Now, following Noah, let $X$ be a discrete space with $\kappa$ many points, and let $f:P\to X$ map $Q_\alpha$ to the $\alpha$-th point of $X$. This map is continuous, since the pre-image of every point is open. Thus, we have mapped $P$ continuously to the discrete space of size $\kappa$, and by composition, we can now map $P$ to any space of size $\kappa$. QED

The argument shows that every infinite topological space is a continuous image of a dense linear order of the same size.

Note that this proof uses the Axiom of Choice. But full AC doesn't seem required, since we didn't really need that $X$ was well-orderable, but rather only that it was linearly orderable. That is, if $X$ is a linearly orderable set, then by replacing each point with a copy of the rationals, we get a dense linear order, with a locally constant map back to $X$. So the discrete topology on $X$ is the continuous image of a dense linear order, and therefore any topology on $X$ is the continuous image of a dense linear order. (The assertion that every set admits a linear order is a weak choice principle.)


I will address Joel's easier question, the answer to which makes me guess that not much can be said in general. The rationals can be partitioned into countably many disjoint clopen subsets: for example, the intervals of the form $(n+x,n+1+x)$ where $n$ is an integer and $x$ is some fixed irrational number. We can define a map from $\mathbb{Q}$ onto a countably infinite discrete space $X$ which is constant on each of these clopen sets and therefore locally constant, so continuous.

The continuous images of $X$ are exactly the countable sets (where now I'm including finite in "countable"), so by composition any such set can be a continuous image of $\mathbb{Q}$. Of course any image of $\mathbb{Q}$ must be countable, so the continuous images of $\mathbb{Q}$ are exactly the countable sets.

Perhaps one could recover an interesting question by asking that the maps also be injective?

  • $\begingroup$ Thanks for such a prompt and insight-giving answer. Should I create a new question for the case where the maps are restricted to be injective, or just edit my question above? $\endgroup$ Jun 19 '10 at 18:42
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    $\begingroup$ Robin, since this question and the answers seem to tell a complete story, I would recommend simply asking a new question, providing a link in it back to this one, as well as a link above to the new question. $\endgroup$ Jun 19 '10 at 18:58

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