I'd like to see if one can capture the notion of a space being "complete" or a "continuum" in a way that doesn't require the use of metrics. Specifically, I want to see if the intuitive notion of a "continuum" (a "continuous" space with no holes) can be pinned down precisely by only using a topology on the underlying set. After some thought, I arrived at the following definition:
Suppose $(X,\tau)$ is a connected topological space ($\tau$ is our collection of open sets) with the property that for any nonempty open set $S\in\tau$, there is an infinite sequence of non-empty open sets $\left\{S_i\right\}_{i=1}^\infty$ with $S_1\subset S$ and $S_{i+1}\subset S_i$ for every $i\in\mathbb N$. Simply put, for any open set $S\neq\emptyset$, we can nest infinitely many non-empty open sets inside $S$, giving us the infinite chain $S\supset S_1\supset S_2\supset\cdots$ of inclusions.
We call $(X,\tau)$ a continuum if any infinite sequence of non-empty open sets $\left\{S_i\right\}_{i=1}^\infty$ that can be nested in this way has a nonempty intersection: $$\bigcap_{i=1}^\infty S_i\neq \emptyset$$
This definition is modeled on the nested interval property possessed by the real numbers. I've tested this definition on the examples below, and it seems to accurately pick out which spaces are continuums and which ones aren't.
- The real numbers $\mathbb R$ equipped with the standard topology (open sets are unions of open intervals) are a continuum under this definition because the reals have the nested interval property. For similar reasons, $\mathbb R^n$ with its standard topology is also a continuum.
- The rational numbers $\mathbb Q$ equipped with the order topology (open intervals $(a,b):=\{x\in\mathbb Q:a<x<b\}$ form the base of the topology) are not a continuum because the chain of open intervals $$(3,4)\supset (3.1,3.2)\supset (3.14,3.15)\supset (3.141,3.142)\supset\cdots$$ has an empty intersection. In the reals, this intersection would be $\pi$.
- The set $(0,1)\cup (1,2)\subset \mathbb R$ with the usual subspace topology is disconnected. In line with the definition, we rule it out as a candidate for a continuum.
Some other spaces are more difficult to reason about. For instance, I think it's reasonable to call $(0,1]\subset\mathbb R$ a continuum (I'm open to objections), but it fails to satisfy the definition in the subspace topology because the then-open sets $(0,1), (0,1/2), (0,1/3),\dots$ have an empty intersection.
Another example of a problem space is $\mathbb Z$. It is intuitively not a continuum, but because $\mathbb Z$ doesn't seem to have a "most-reasonable" topology, we can't immediately rule it out as a continuum under this definition. I know two topologies on $\mathbb Z$ that rule it out as a possible continuum: the discrete topology where all subsets are open sets (because we can't nest a chain in $\{1,2,3\}$, for example), and the subspace topology when we identify $\mathbb Z$ as a subset of $\mathbb R$ (again, can't nest a chain in $\{1, 2, 3\}$.
Given all these observations, does this definition, perhaps with a few modifications, seem like a candidate for pinning down what we mean by a continuum? Are there any examples of spaces that are intuitively continuums, but nevertheless fail to satisfy this definition or any variant of it? Any and all input is greatly appreciated.