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This question arises after the realisation that countable sets and uncountable sets does not necessary differ in the way where the former has gaps while the latter has no gaps. The best topological notion that capture the property of gaps is total disconnectedness. Quoting Wikipedia:

A totally disconnected set is a set $S$ where the only connected subsets are the singletons and the empty set.

So for example,

$\Bbb{Z},\Bbb{N}$ are totally disconnected because they are discrete.

$\Bbb{R}$ is connected "everywhere" (under the open interval topology) since any intervals cannot be wrote as unions of open intervals without overlapping.

whereas $\Bbb{Q}$ is totally disconnected as between any two rationals, we can pick an irrational $r$ and construct open intervals $(-\infty,r)\cup(r,\infty)$ to separate them.

Via the proof of this question, the cantor set $\mathcal{C}$ is also totally disconnected as it consists of only endpoints and limit points of said endpoints thus it is totally disconnected,

and via the proof here, the p adics $\Bbb{Z}_p$ are also totally disconnected

All ordinals, and in general, well ordered sets, are totally disconnected. Take any ordinal $\alpha$ as an example, as shown here, we can take the partitions $\{[0,\beta +1)\cup(\beta,\infty),\beta \in \alpha\}$ thus the only connected components are the singletons.

So we basically have the following table:

enter image description here

Therefore:

Regardless of whether we accept or reject the axiom of choice, (generalied) continuum hypothesis. Using ZF, do non well ordered countably infinite sets $X$ which has no gaps exists (i.e. can we have a "countable continuum", analogous to how $\Bbb{R}$ formed a continuum), if yes what are the examples?

If such set cannot exists, what is the theorem that suggests that?

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    $\begingroup$ I assume you want a countable connected space. Am I right? $\endgroup$ – Hanul Jeon Jul 26 '17 at 15:27
  • $\begingroup$ Notice that the Cantor set and the space of $p$-adic integers are homeomorphic, so they are in fact same (as topological spaces). $\endgroup$ – Hanul Jeon Jul 26 '17 at 15:29
  • $\begingroup$ more accurately, analogous to reals, I want a countable connected space such that every point is part of some connected set and the whole set cannot be written in terms of disjoint unions of open sets. That is something that can possibly be the best candidate for a countable 'continuum' (where continuum I mean it is intuitively continuous with no gaps, not whether it is totally ordered) $\endgroup$ – Secret Jul 26 '17 at 15:29
  • $\begingroup$ I think you asks there is a countable locally connected space. $\endgroup$ – Hanul Jeon Jul 26 '17 at 15:33
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    $\begingroup$ @Secret: "the whole set cannot be written in terms of disjoint union of (more than one, nonempty) open sets" is exactly what "connected" means -- so you don't need to specify that twice. $\endgroup$ – Henning Makholm Jul 26 '17 at 15:46
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First of all, it is meaningless to talk about a set being "connected" or "totally disconnected". These words only have meaning when you are talking about a topological space, not just a set.

As for a countably infinite connected topological space, there are plenty of such spaces. The easiest example is to take any countably infinite set $X$ and give it the indiscrete topology, where the only open sets are $X$ and $\emptyset$. Intuitively, you can think of this as a "space" where all the points are right on top of each other, so you can't tell any of them apart (but there are nevertheless infinitely many of them).

More surprisingly, there exist countably infinite connected Hausdorff spaces. See https://mathoverflow.net/questions/46986/countable-connected-hausdorff-space for some examples.

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  • $\begingroup$ Perhaps add a result in the other direction: A countably infinite connected normal space must have at least the cardinality of the continuum, because Urysohn's Lemma gives you a non-constant continuous map to the reals. (I'm using "normal" in the sense that includes $T_1$, so "$T_1$ and $T_4$.) $\endgroup$ – Andreas Blass Jul 26 '17 at 15:47
  • $\begingroup$ This also holds for countable $T_3$ (which includes $T_1$) spaces as these are $T_4$ by Lindelöfness). So better than Hausdorff we cannot really do. $\endgroup$ – Henno Brandsma Jul 26 '17 at 17:05
  • $\begingroup$ here are all the spaces from Counterexamples in Topology (5 of them): (relatively) prime integer topology, irrational slope topology, etc. $\endgroup$ – Henno Brandsma Jul 26 '17 at 17:10
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A simple example, which is a $T_1$ space, but not $T_2:$ Let $X=\mathbb N$ and let $T$ be the co-finite topology on $X$ : Any non-empty $S\subset X$ is open iff $X$ \ $S$ is finite.

It is connected, as any two non -empty open sets have infinite intersection. And if $Y$ is any infinite subspace of $X$ then $Y$ is homeomorphic to $X$, so $Y$ is connected. And if $p\in V\subset X,$ where $V$ is a nbhd of $p$, then $V$ is connected.

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