In a general topological space $(X,\tau)$ we define an accumulation point $x_0$ of a set $A$ to be a point such that any open neighbourhood about $x_0$ intersects $A$.

Now it is certainly true that if a sequence $x_n\in A$ tends to some limit $x \in X$, $x$ must be an accumulation point of $A$ since $x_n$ lies in any open neighbourhood of $x$ for all $n$ sufficiently large, and so lies in the intersection of this open neighbourhood and $A$.

What I would like to know is: Are there any (preferably elementary) examples of a topological space with a subset $A$ that has an accumulation point which is not the limit of any sequence in $A$. I would also appreciate information on any conditions on a space which imply that any accumulation point of $A$ is the limit of some sequence in $A$.

For example, if $X$ is first countable (e.g. any metric space) then it is easy to show that any accumulation point of $A$ must be the limit of some sequence of points in $A$. Intuitively this is because for a given point $x$, we can find a nested sequence of open sets that "get smaller" and can eventually be contained in any open neighbourhood of $x$, so these nested open sets "contract around $x$, allowing us to find such a sequence.


4 Answers 4


Here are two examples:

  1. The closed ordinal space $\omega_1 + 1 = [ 0 , \omega_1 ]$ consisting of all countable ordinals as well as the least uncountable ordinal $\omega_1$. It is easy to see that $\omega_1 \in \overline{[ 0 , \omega_1 )}$, but since any sequence in $[0 , \omega_1 )$ is countable, it has an upper bound $\beta$ which is a countable ordinal, and $( \beta , \omega_1 ]$ is an open neighbourhood of $\omega_1$ disjoint from the sequence.

  2. The Stone–Čech compactification $\beta \mathbb{N}$ of the discrete space $\mathbb{N}$. This space has the property that there are no non-trivial (i.e., not eventually constant) convergent sequences. (For quite a bit of basic information about this space see Dan Ma's Topology Blog.) We have that $\mathbb{N}$ is a dense subset of $\beta \mathbb{N}$, but no sequence in $\mathbb{N}$ can converge outside of $\mathbb{N}$.

(Consider also this previous answer of mine, and possibly also this question and its answers.)

For the interested:

One may be lead to believe that the problem is that sequences are "too short", and that things would change if only we were allowed to use "longer" sequences. This is almost true.

Definition. A topological space $X$ is called radial if for any $A \subseteq X$ and any $x \in \overline{A}$ there is a transfinite sequence $\langle x_\xi \rangle_{\xi < \alpha}$ (where $\alpha$ is an ordinal) in $A$ which converges to $x$. (The definition of convergence in this sense is the natural extension of sequential convergence.)

It is not too difficult to show that $\omega_1 +1$ is radial (in fact, we only need to allow sequences of length $\omega_1$ to capture this property).

However $\beta \mathbb{N}$ is not even radial!

  • $\begingroup$ Thank you very much for the (very thorough) answer! $\endgroup$ Jun 2, 2013 at 14:36
  • $\begingroup$ Hi Arthur, I give another example, which you may be interested in. $\endgroup$
    – Paul
    Jun 3, 2013 at 0:38

Another example: An uncountable space $X$ with countable complement (co-countable) topology.

Proof: Pick any point $x\in X$. It is the accumulation point of $A=X\setminus \{x\}$, however it is not the limit point of $A$. Since for any countable subset $S$ of $A$, $U=X\setminus S$ is an open set of $x$, which satisfies that $U \cap S=\emptyset$.

  • 2
    $\begingroup$ Help clarify the proof: any sequence in $A$ can be regarded as a countable subset $S$ of $A$. The proof actually says for any countable set $S$ of $A$, there exits an open set $U$ containing $x$ that does not contain any points in $S$. So no sequence in $A$ converges to $x$. $\endgroup$
    – Hua
    Sep 16, 2016 at 8:21

Here is a compact Hausdorff example you might consider more friendly.

Let $X$ be the "cube" $[0,1]^{[0,1]}$. That is, the space of all functions $[0,1] \to [0,1]$ in the topology of pointwise convergence. Let $A = \{ f \in X : f \text{ is continuous, and } \int_0^1 f(t) \ dt \geq 1/2\}$. Then, the zero function is an accumulation point of $A$ (for every finite set $x_1,\ldots,x_n \in [0,1]$, there is an $f \in A$ which vanishes at all of them). But, there can be no sequence $f_n \in A$ converging pointwise to zero (hint: use the Lebesgue dominated convergence theorem).

One can also cut $[0,1]^{[0,1]}$ down to the subspace of polynomial functions with rational coefficients to get an example where the space $X$

  • has countably many points
  • is a Tychonoff space (being a subspace of $[0,1]^{[0,1]}$)
  • is zero-dimensional in the sense that, whenever $f,g \in X$ are distinct, there is a parition of $X$ into open sets $U,V$ so that $f \in U, g \in V$.

I took this example from a little "mock paper" I wrote for a course once. See Example 4 in http://arxiv.org/abs/1210.1008.

  • $\begingroup$ @JSempliner: OK, I've removed my comment about pointwise convergence, as it was apparently unnecessary. Anyway, I don't understand what you don't like about the example. Do you agree with the following claim? Given any $\epsilon > 0$ and any finite set $\{ x_1, \ldots, x_n\} \subset [0,1]$, there is a polynomial $p(x)$ with rational coefficients such that (1) $0 \leq p(x)\leq 1$ when $x \in [0,1]$, (2) $\int_0^1 p(x) \ dx \geq 1/2$, and (3) $p(x_i) < \epsilon$ for $i=1,\ldots,n$. $\endgroup$
    – Mike F
    Jan 22, 2014 at 2:30
  • $\begingroup$ No sorry to give that impression! I was unaware that there were countable spaces that were not Frechet-Urysohn, so it was quite enlightening! It just clashed with my intuition late last night so I wanted to ask about it. $\endgroup$
    – Sempliner
    Jan 22, 2014 at 2:34
  • $\begingroup$ @JSempliner: I don't mind, I was glad to be reminded of this example. I quite like it. $\endgroup$
    – Mike F
    Jan 22, 2014 at 2:38

Now it is certainly true that if a sequence $x_n∈A$ tends to some limit $x∈X$, $x$ must be an accumulation point of $A$

Unless the sequence is constant, right?

For example, if X is first countable (e.g. any metric space)

Metric spaces don't need countable space?

  • $\begingroup$ No, my definition of accumulation point allows for (eventually) constant sequences, which would provide a counter-example to the same statement made for limit points. I'm also not sure what you mean by the question "Metric spaces don't need countable space?". Are you saying that metric spaces aren't necessarily countable? I agree with that, but that's not implied by being first countable (which all metric spaces are), so I'm not sure what you're saying... $\endgroup$ Sep 14, 2014 at 9:01
  • $\begingroup$ @TomOldfield An eventually constant sequence is not evidence for an accumulation point $\endgroup$ Sep 14, 2014 at 10:31
  • $\begingroup$ I don't understand what you mean. From the definition I gave, any element of $A$ is an accumulation point. So even if the sequence is constant, the limit is still an accumulation point. $\endgroup$ Sep 14, 2014 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.