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Consider the set of all subsets of the naturals, $2^\Bbb{N}$. We call a subset $A \subseteq \Bbb{N}$ small if $\sum_{a \in A} \frac{1}{a} < \infty$, and large otherwise. The set of small subsets of $\Bbb{N}$, $$X := \{ A \in 2^\Bbb{N}: A \text{ is small} \},$$ is closed under arbitrary intersection and finite union. This suggests that we can define a topology on $\Bbb{N}$ as follows: Call an element $U \in 2^{\Bbb{N}}$ co-small if $U^c := \Bbb{N} \setminus U$ is small. Then the co-small topology on $\Bbb{N}$ is the topology where the nontrivial open sets (i.e. besides $\Bbb{N}$ and the empty set) are the co-small sets. This is a topology because the set of co-small sets is closed under arbitrary union and finite intersection. Under this topology, $\Bbb{N}$ is $T_1$ (for any two distinct points $a, b$ there is a neighborhood of $a$ disjoint from $b$ and vice versa) but not Hausdorff (since any two co-small sets have co-small intersection, any two neighborhoods of distinct points $a, b$ will overlap). The only compact sets in $\Bbb{N}$ under this topology are the finite sets; however, $\Bbb{N}$ is not discrete in this topology (since one-point sets cannot be co-small).

Questions:

  1. Is there a formal name for this topology, and is it studied in the literature at all? Is $\Bbb{N}$ in the co-small topology homeomorphic to another, better known or understood space?

  2. What are the continuous functions from $\Bbb{N}$ to itself under the co-small topology, aside from trivial examples like the constant function or the identity function? (The only continuous maps from $\Bbb{N}$ in the co-small topology to $\Bbb{R}$ in the usual topology are the constant functions.)

  3. We can also think of $\Bbb{N}$ as a discrete measure space with the obvious measure $$\mu(A) := \sum_{a \in A} \frac{1}{a}.$$ Every open set has infinite measure, and sets with finite measure are closed. Also, every element of $2^\Bbb{N}$ is measurable, from which it trivially follows that every function $f: \Bbb{N} \to \Bbb{R}$ or $\Bbb{C}$ is measurable. The measure also scales in a nice way: we have $\mu(kA) = \frac{\mu(A)}{k},$ where $kA := \{ ka: a \in A \}$. Can this measure be applied to any interesting problems in number theory or combinatorics? Is it used to prove ergodicity of any maps?

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  • $\begingroup$ A subset of the naturals $2^{\mathbb N}$ doesn’t make sense. Can you revisit this sentence? $\endgroup$ Aug 28, 2020 at 20:24
  • $\begingroup$ You mean the first sentence? Are you asking me to use $2^{\aleph_0}$ instead? $2^\Bbb{N}$ means the set of all functions $f: \Bbb{N} \to \{ 0, 1 \}$. These functions are in 1-1 correspondence with subsets of $\Bbb{N}$ via $A := f^{-1}(1)$ (i.e. $f = 1_{A}$ is the indicator function of $A$). $\endgroup$ Aug 28, 2020 at 20:26
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    $\begingroup$ Slightly off-topic, but I have a feeling, and maybe it's noteworthy, that for many sets $A \subset \mathbb N$ it will be quite a problem to decide whether it is closed in this topology or not. Cf. mathoverflow.net/a/65978. This might severely limit the use of this topology (unless there is some miracle which would make it possible to conclude "the other way around"). $\endgroup$ Aug 29, 2020 at 0:58
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    $\begingroup$ This is my favorite example of a countable space which is not first countable. Or rather, the Hausdorff space you get by declaring the neighborhoods of $1$ to be the "co-small" sets containing $1$ and making all other points isolated; i.e., the open sets are the "co-small" sets containing $1$ and all sets not containing $1$. I don't know if either of these topologies has a name. $\endgroup$
    – bof
    Aug 29, 2020 at 1:02
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    $\begingroup$ @RiversMcForge Oh no, it makes sense and is technically correct. It's really just a minor point and more a matter of taste than anything. I prefer (and I think so do most mathematicians) to speak either in terms of subsets of $X$ or elements $2^X$ and to use both at the same time only when absolutely necessary. It tends to make it easier to read. For example, someone not reading closely might think you were trying to topologize $2^\mathbb{N}$ at first. $\endgroup$ Sep 7, 2020 at 20:46

1 Answer 1

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Concerning the continuous functions from $\mathbb N$ to itself in the co-small topology:

A map $f\colon \mathbb N\to \mathbb N$ is continuous in the co-small topology if and only if it is either constant or maps large sets to large sets.

I will use the following observations without further notice:

  1. Subsets of small sets are small. Super-sets of large sets are large.
  2. $\overline X=\mathbb N$ for all large $X$.
  3. Co-small sets are large.

Let’s dive into the proof of my claim above. Of course, $f\colon \mathbb N\to \mathbb N$ is continuous if and only if $f(\overline{X})\subset \overline{f(X)}$ for all $X\subset \mathbb N$. If $X$ is closed, this is trivially the case. If $X$ is not closed, i. e., neither small nor all of $\mathbb N$, then $\overline X= \mathbb N$ and so the condition for $f$ to be continuous is that for all large $X$ we have to have $f(\mathbb N)\subset \overline{f(X)}$. There are two essentially different cases to distinguish, namely, whether $f( \mathbb N)$ is small or not.

If $f(\mathbb N)$ is large, then for $f$ to be continuous, $\overline{f(X)}$ must contain a large set, hence, has to be large itself. But then $f(X)$ must have been large to begin with. Conversely, if $f$ maps large sets to large sets, then clearly $f(\mathbb N)\subset \mathbb N=\overline{f(X)}$, as desired.

If $f( \mathbb N)$ is small, then so is each $f(X)$; thus, $f(X)=f(\mathbb N)$ for every large $X$, for $f(X)\subset f(\mathbb N)\subset \overline{f(X)}=f(X)$. I claim that that’s possible only if $f$ is constant.

In fact, let $m\in f(\mathbb N)$ be arbitrary. Since $\{m\}$ is small, it’s closed. Therefore, $f^{-1}m$ is closed, hence either small or all of $\mathbb N$. The latter case means that $f$ is constant. It remains to rule out the former case: If $f^{-1}m$ were small, then $X=\mathbb N-f^{-1}m$ would be co-small, hence large, satisfying $f(X)=f(\mathbb N)-m\subsetneq f(\mathbb N)$, a contradiction. This completes the proof.

I have no idea concerning the 1. and 3. question.

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  • $\begingroup$ I'm actually looking for a more specific characterization of continuous functions on $\Bbb{N}$ than what is (more or less) a restatement of the definition of "continuity". Specifically, I want conditions on how fast or how slow a continuous function $f: \Bbb{N} \to \Bbb{N}$ under this topology must grow. I was already aware that if $f: \Bbb{N} \to \Bbb{N}$ is continuous and the image of some large set is small, then $\Bbb{N}$ must be constant, but I appreciate you typing this result up for newcomers. $\endgroup$ Sep 7, 2020 at 20:07
  • $\begingroup$ Rather than edit my original question, I clarified the type of answer I was looking for here, and credited you for a partial answer: math.stackexchange.com/questions/3821767/… $\endgroup$ Sep 11, 2020 at 0:34

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