Let me start with a precise statement of the question. For a subset $A\subseteq \mathbb{N}$ and a series of positive real numbers $\sum_{n=0}^\infty a_n$, I'll use the notation $\sum_A a_n$ as a shorthand for $\sum_{n\in A} a_n$.

Is there a bijection $f:\mathbb{N}\rightarrow \mathbb{N}$ with the property that for every $A\subseteq \mathbb{N}$, $\sum_A \frac{1}{n}$ converges iff $\sum_{f(A)} \frac{1}{\sqrt{n}}$ converges?


Fix a series of positive terms $\sum_{n=0}^\infty a_n$. Given a subset $A\subseteq \mathbb{N}$, call it $a_n$-small if $\sum_A a_n$ converges. The following proposition is easy to prove:

Proposition: The set $\{A\subseteq \mathbb{N}: A\text{ is } a_n\text{-small}\}\cup \{\mathbb{N}\}$ is a topology of closed sets on $\mathbb{N}$.

I will use the notation $(\mathbb{N},a_n)$ to refer to this topology. Then one can ask how topological properties of $(\mathbb{N},a_n)$ are related to series properties of $\sum a_n$. For example, I can show:

Proposition: The following are equivalent.

  1. $\sum a_n$ converges.

  2. $(\mathbb{N},a_n)$ is discrete.

  3. $(\mathbb{N},a_n)$ is disconnected.

  4. $(\mathbb{N},a_n)$ is Hausdorff

And there are other nice things. For example, $(\mathbb{N},a_n)$ is compact iff $(\mathbb{N},a_n)$ is cofinite iff $\liminf a_n > 0$.

With this language, my question can be reformulated as...

Are $\left(\mathbb{N}, \frac{1}{n}\right)$ and $\left(\mathbb{N}, \frac{1}{\sqrt{n}}\right)$ homeomorphic?

I have made very little progress on this. Of course, the identity function $i:(\mathbb{N}, \frac{1}{\sqrt{n}})\rightarrow (\mathbb{N}, \frac{1}{n})$ is a continuous bijection, but the inverse map is not continuous. Also, if there is such a bijection, there there is such a bijection with agrees with $i$ on any preassigned finite set.

Edit I thought I add in a slightly suprising (to me, at least) example when things work out to be homeomorphic.

Begin with a convergent series $\sum c_n$. and divergent series $\sum a_n$ with $\lim a_n = 0$. Create a new series $b_n$ using all the terms of $c_n$ and $a_n$ (in whatever order you wish). Then $(\mathbb{N}, a_n)$ and $(\mathbb{N}, b_n)$ are homeomorphic. The idea is that since $\lim a_n = 0$, there is a convergent infinite subseries $\sum_A a_n$. Then we use a bijection which with $A$ and $A\cup \{\text{indices of }c_n\}$ to "squeeze" the $c_n$ in without changing the topology. Of course, I am glossing over many details, but I can include them if desired.

  • $\begingroup$ Note that $\left(\mathbb N,\frac{1}{\sqrt{n}}\right)$ has a homeomorphic image of $\left(\mathbb N,\frac{1}{n}\right)$ as a subspace. Not sure how that helps. It definitely seems like they shouldn't be homeomorphic... $\endgroup$ Commented Sep 14, 2017 at 15:41
  • $\begingroup$ These topologies are based on filters, in that the set of open sets are a "filter" in $\mathcal P(\mathbb N)$. $\endgroup$ Commented Sep 14, 2017 at 15:44
  • $\begingroup$ I agree they shouldn't be homeomorphic. I learned about filters many years ago, maybe its time to dust off the old topology book... $\endgroup$ Commented Sep 14, 2017 at 15:46
  • $\begingroup$ You might define $A$ to be closed if $\sum_A\frac{1}{n^p}$ is convergent for some $p\in(0,1)$. Then I'd even be surprised if this topology was homeomorphic to $(\mathbb N,1/n)$. $\endgroup$ Commented Sep 14, 2017 at 16:00
  • $\begingroup$ Your observation at the end says that if $\sum a_n=\infty$ and $a_n\to 0$ then for any open $U\subseteq (\mathbb N,a_n)$ (other than $U=\emptyset$) you have $U$ is homeomorphic to $\mathbb N$. $\endgroup$ Commented Sep 14, 2017 at 16:36

2 Answers 2


In what follows, I assume that by $\mathbb{N}$ you mean $\mathbb{Z}_{>0}$, this way I don't have to deal with $\frac 1 0$. Of course, the answer is the same for $\mathbb{Z}_{\geq 0}$.

No, such a bijection doesn't exist.

Consider the sequence $(b_n)_{n=1}^\infty$ given by $b_1 = 1$, $b_n = n [\log_2 n]^2$ for $n>1$, where $[x]$ denotes the integer part of $x$. Then the following 2 statements are true:

(1) $\sum_{n=1}^\infty \frac{1}{b_n}$ converges.

(2) Denote $s_k := \sum_{i=1}^k b_i$. Then $\sum_{n=1}^\infty \frac{1}{\sqrt{s_n}}$ diverges.

(1) follows from the fact that for $k\geq 1$, $$ \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{b_n} < \frac{2^k}{b_{2^k}} = \frac{1}{k^2}, $$ and $\sum \frac{1}{k^2}$ converges. To prove (2), note that $s_{2^{k+1}-1} < k^2 \sum_{i=1}^{2^{k+1}-1} i < k^2 2^{2(k+1)}$, so $$ \sum_{n=2^k}^{2^{k+1}-1} \frac{1}{\sqrt{s_n}} > \frac{2^k}{\sqrt{s_{2^{k+1}-1}}} > \frac{1}{2k}, $$ and $\sum \frac{1}{2k}$ diverges.

Now let $f$ be a bijection from $\mathbb{N}$ to $\mathbb{N}$. Set $u_n = \max\{f^{-1}(i)\mid s_{n-1}< i \leq s_n\}$ and $A = \{u_n\mid n\in \mathbb{Z}_{>0}\}$ (assuming $s_0 = 0$). Note that $u_m\neq u_n$ for $m\neq n$ since $f(u_m)\neq f(u_n)$. Also, $u_n\geq b_n$ since $u_n$ is a maximum of $b_n$ positive integers. Then $\sum_A \frac{1}{n} = \sum_{n=1}^\infty \frac{1}{u_n} \leq \sum_{n=1}^\infty \frac{1}{b_n}$, which converges. On the other hand, $f(u_n) \leq s_n$, so $\sum_{f(A)} \frac{1}{\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{\sqrt{f(u_n)}} \geq \sum_{n=1}^\infty \frac{1}{\sqrt{s_n}}$, which diverges.

Edit: Actually, a much more general statement can be proven:

Let $(a_n)_{n\in\mathbb{N}}$ and $(a'_n)_{n\in\mathbb{N}}$ be two sequences of positive numbers. Assume that $\lim\limits_{n\to\infty} a'_n = \lim\limits_{n\to\infty} \frac{a'_n}{a_n} = 0$ and that $\sum_{n\in \mathbb{N}} a_n$ diverges. Then $\left(\mathbb{N}, a_n\right)$ and $\left(\mathbb{N}, a'_n\right)$ are not homeomorphic, i.e., there is no bijection $f: \mathbb{N}\to \mathbb{N}$ such that for every $A\subseteq\mathbb{N}$, $\sum_A a_n$ converges iff $\sum_{f(A)} a'_n$ converges.

Proof: First of all, it's enough to prove the statement for the cases where the sequence $a'_n$ is non-increasing. Since $a'_n>0$ and $a'_n\to 0$, the sequence $a'_n$ can be rearranged in a non-increasing order, i.e., there is a bijection $p:\mathbb{N}\to \mathbb{N}$ such that $a'_{p(n+1)}\leq a'_{p(n)}$ for all $n\in \mathbb{N}$. The sequences $a_{p(n)}$ and $a'_{p(n)}$ satisfy the conditions of the statement, and $f$ is a homeomorphism between $\left(\mathbb{N}, a_{p(n)}\right)$ and $\left(\mathbb{N}, a'_{p(n)}\right)$ iff $p^{-1}\circ f\circ p$ is a homeomorphism between $\left(\mathbb{N}, a_n\right)$ and $\left(\mathbb{N}, a'_n\right)$. So in what follows I assume that $a'_n$ is non-increasing.

Let us denote partial sums of $\sum a_n$ and $\sum a'_n$ by $S_n$ and $S'_n$, i.e., $S_n:= \sum_{k\leq n} a_n$ and $S'_n:= \sum_{k\leq n} a'_n$.

Lemma 1: $\lim\limits_{n\to\infty} \frac{S'_n}{S_n} = 0$.

Proof of Lemma 1: For any $\epsilon>0$, there is $N\in \mathbb{N}$ such that $\frac{a'_n}{a_n}<\frac{\epsilon}2$ for all $n>N$. Then, for $n>N$, $$ \frac{S'_n}{S_n} = \frac{S'_N + \sum\limits_{N<k\leq n} a'_k}{S_n} < \frac{S'_N + \frac{\epsilon}{2}\sum\limits_{N<k\leq n} a_k}{S_n} = \frac{S'_N + \frac{\epsilon}{2}(S_n-S_N)}{S_n}, $$ which is $<\epsilon$ if $S_n> \frac{2}{\epsilon}S'_N - S_N$. Since $\sum a_n$ diverges, $S_n\to\infty$, so $\frac{S'_n}{S_n}<\epsilon$ for all large enough $n$. End of proof of Lemma 1

Now let $f: \mathbb{N}\to \mathbb{N}$ be a bijection. For $\epsilon>0$, denote $M^f_\epsilon := \{n\in\mathbb{N}\mid a'_{f(n)}<\epsilon a_n\}$.

Lemma 2: For all $\epsilon>0$, $\sum_{M^f_\epsilon} a_n$ diverges.

Proof of Lemma 2: Assume that $\sum_{M^f_\epsilon} a_n$ converges, and denote its value by $\mu$.

Since $f$ is a bijection and $a'_n$ is non-increasing, $\sum_{k\leq n} a'_{f(k)} \leq \sum_{k\leq n} a'_{k} = S'_n$, so $$ \frac{S'_n}{S_n} \geq \frac{\sum\limits_{k\leq n} a'_{f(k)}}{S_n} \geq \frac{\sum\limits_{k\leq n,~ k\not\in M^f_\epsilon} a'_{f(k)}}{S_n} \geq \frac{\epsilon\sum\limits_{k\leq n,~ k\not\in M^f_\epsilon} a_{k}}{S_n} \geq \frac{\epsilon(S_n - \mu)}{S_n}. $$ The last expression is $\geq \epsilon/2$ if $S_n>2\mu$; since $S_n\to\infty$, it contradicts Lemma 1. End of proof of Lemma 2

Following your terminology, I will call a subset $X\subseteq \mathbb{N}$ $a$-large if $\sum_X a_n$ diverges. So $M^f_\epsilon$ is $a$-large.

Finally, let us construct a counterexaple set $A$ as a union of pairwise disjoint finite sets $A_m$, $m\in\mathbb{N}$. We construct $A_m$ inductively as follows: consider the set $$ M^f_{2^{-m}} \backslash \left(\{n\in\mathbb{N}\mid a'_{f(n)}\geq 2^{-m}\}\cup\bigcup\limits_{k<m} A_k\right). $$ This set is $a$-large, since it's a difference of $a$-large set $M^f_{2^{-m}}$ and a finite set. (The set $\{n\in\mathbb{N}\mid a'_{f(n)}\geq 2^{-m}\}$ is finite since $a'_n\to 0$ and f is a bijection, and $\cup_{k<m} A_k$ is finite since all $A_k$ are finite.) Thus, it contains a finite subset $X$ such that $\sum_X a_n>1$; let $A_m$ be a minimal such subset w.r.t. inclusion and let $x$ be some element of $A_m$. Then $$ \sum_{f(A_m)} a'_n = \sum\limits_{k\in A_m} a'_{f(k)} = a'_{f(x)} + \sum\limits_{k\in A_m\backslash \{x\}} a'_{f(k)} < 2^{-m} + 2^{-m}\sum\limits_{k\in A_m\backslash \{x\}} a_k \leq 2^{-m+1}. $$ Since $A_m$ are by construction pairwise disjoint, for $A = \cup_{m\in\mathbb{N}} A_m$, $\sum_A a_n = \sum_{m\in\mathbb{N}}\sum_{A_m} a_n$ diverges (by construction, $\sum_{A_m} a_n>1$), and $\sum_{f(A)} a_n = \sum_{m\in\mathbb{N}}\sum_{f(A_m)} a_n < \sum_{m\in\mathbb{N}} 2^{-m+1}$, which converges. End of proof

  • $\begingroup$ Very nice! Minor comment: I think the bound on $s_{2^{k+1}-1}$ is $k^2 2^{3(k+1)}$ (because $\sum_{i=1}^n i^2$ is a cubic in $n$), but it doesn't affect the rest of the argument at all. Do you have any insight as to how you thought of the definition of $b_n$, $s_n$, and $u_n$? I feel like I get your proof on a line by line level, but not in any kind of big picture way. $\endgroup$ Commented Sep 14, 2017 at 22:07
  • $\begingroup$ Superb! Just a thought, your proof works for f surjective it seems to me. $\endgroup$
    – orangeskid
    Commented Sep 15, 2017 at 1:50
  • $\begingroup$ @orangeskid: A weak form of Injectivity is used in the definition of $u_n$: if $f^{-1}(i)$ is infinite for any particular $i$, then $u_n$ is not always well defined. I agree that the argument works so long as each $f^{-1}(i)$ is a finite set. $\endgroup$ Commented Sep 15, 2017 at 3:26
  • $\begingroup$ @JasonDeVito Actually, the bound is correct, but the sum must be of $i$, not of $i^2$ (I've fixed it now). The bound $k^2 2^{3(k+1)}$ wouldn't be enough. My logic went approximately like this: it's pretty clear that if we want to find a way to construct a counter-example $A$ for a given $f$, then we should look for $A$ such that $\sum_A 1/n$ converges and $\sum_{f(A)} 1/\sqrt{n}$ diverges, not the other way around. So if we construct $A$ out of elements $u_i$, then we need to have some kind of lower bound on $u_i$, but at the same time to have a upper bound on $f(u_i)$. $\endgroup$
    – Litho
    Commented Sep 15, 2017 at 7:29
  • 1
    $\begingroup$ @JasonDeVito Please see the update to my answer. It proves a much more general statement now. $\endgroup$
    – Litho
    Commented Sep 18, 2017 at 10:39

I claim that there is no such bijection for any distinct exponents not exceeding $1$.

Lemma If positive vanishing sequences $(a_n),(b_n)$ are such that $\sum_n a_n = \infty$, $b_n = o(a_n)$, $n\to\infty$, then there exists $A\subset \mathbb N$ such that $\sum_A a_n=\infty$, $\sum_A b_n<\infty$.

Proof Let $N_k$ be such that $b_n<a_n/k$, $n\ge N_k$. Since $(a_n)$ vanishes and $\sum_n a_n = \infty$, there exist some $n_1'>n_1> N_1$ such that $\sum_{n=n_1}^{n_1'} a_n \in (1,2)$. Inductively, for each $k\ge 2$ there exist $n_k'>n_k > N_k\vee n_{k-1}'$ such that $\sum_{n=n_k}^{n_k'} a_n \in (1/k,2/k)$. Obviously, $A = \bigcup_{k\ge 1} \{n_k,n_k+1,\dots,n_k'\}$ is as required.

Now assume that for exponents $a<b\le 1$ there exists a bijection $f\colon \mathbb N\to \mathbb N$ such that $\sum_A n^{-b}$ converges iff $\sum_{f(A)} n^{-a}$ converges.

Choose some $r\in (1,b/a)$. Clearly, $\sum_{n: f(n)>n^{r}} \frac{1}{f(n)} <\infty$ and $\sum_{n} \frac{1}{f(n)} =\infty$, so $\sum_{n: f(n)\le n^{r}} \frac{1}{f(n)} =\infty$. Therefore, $\sum_{n: f(n)\le n^{r}} \frac{1}{f(n)^a} =\infty$. If $f(n)\le n^r$, then $f(n)^a\le n^{ra} = o(n^b)$, $n\to\infty$. Setting $a_n = 1/f(n)^{a}$ and$^*$ $b_n = 1/n^b$ and applying lemma, we get that $\sum_A f(n)^{-a}=\infty$ and $\sum_A n^{-b} <\infty$, a contradiction.

$^*$ apologies for the pun.

  • 1
    $\begingroup$ I started writing this as the last edit of @Litho's post was not there yet; now it's no longer needed, but I think it is still worth to keep it. $\endgroup$
    – zhoraster
    Commented Sep 18, 2017 at 13:51
  • $\begingroup$ Thanks for posting! I agree it's worth keeping. Does the definition of $A$ in the proof of the lemma need a union? Otherwise it seems as though $A$ is a finite set... (Gotta go teach for a bit; I will return to this later.) $\endgroup$ Commented Sep 18, 2017 at 13:58
  • $\begingroup$ @JasonDeVito, sure it does (otherwise $A$ is even empty), thanks! $\endgroup$
    – zhoraster
    Commented Sep 18, 2017 at 14:03
  • $\begingroup$ I've now had time to read it - looks great to me. Thanks again! $\endgroup$ Commented Sep 18, 2017 at 15:55
  • $\begingroup$ It turns out ( a year later), that I am writing up this stuff (with an undergraduate) for submission to an undergrad journal. While we used Litho's result for the general homeomorphism question, we were able to modify your approach to give the following result: For $0\leq p\leq 1, p< q$, any continuous function $(\mathbb{N}, 1/n^p)\rightarrow (\mathbb{N}, 1/n^q)$ is constant. We would like to thank you in the paper. If you are ok with that, what name would you prefer that we use? $\endgroup$ Commented Sep 9, 2018 at 4:58

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