I'm trying to extract more useful information from the following definition:

A point $x^*$ is an accumulation point of the sequence $\{x_n\}$ if, for every open set containing x , there are infinitely many indices such that the corresponding elements of the sequence belong to the open set.

I would like to say that this is equivalent to saying the following:

$x^*$ is an accumulation point if there exists a subsequence $x_{a_n}$ such that $lim_{n\rightarrow\infty} x_{a_n}=x^*$.

Is this true?

  • 2
    $\begingroup$ It is indeed true, provided that you’re working in a space in which sequences determine the topology; in particular, it’s true in metric spaces. $\endgroup$ – Brian M. Scott Sep 7 '12 at 18:10
  • $\begingroup$ As I'm working in $R^n$ space, I think that should work! :) $\endgroup$ – Paul Sep 7 '12 at 18:20
  • $\begingroup$ I figured that you probably were. :-) $\endgroup$ – Brian M. Scott Sep 7 '12 at 18:26
  • $\begingroup$ Actually, because your definition of accumulation point is slightly different from most books, the statement is an "if and only if". $\endgroup$ – Matt Sep 7 '12 at 19:19
  • $\begingroup$ Ah, so it does work both ways! That's awesome! :) $\endgroup$ – Paul Sep 7 '12 at 21:47

Before we proceed with your question, I'd like to give an interesting example to clarify a point. We say that a set $Y$ is cofinite in a superset $X$ of $Y$ if and only if $X\smallsetminus Y$ is finite. Observe that a union of cofinite sets is cofinite by Demorgan's laws (since an intersection of finite sets is finite), and that a finite intersection of cofinite sets is likewise cofinite (since a finite union of finite sets is finite). Consequently, we can describe a cofinite topology on any set $X$, by saying that $Y\subseteq X$ is "open" if and only if $Y$ is cofinite or $Y$ is empty. Here's where it gets interesting: If $X$ is an infinite set with the cofinite topology, then a sequence of points converges to every point of $X$ if and only if it has infinitely many distinct terms--for example, considering $\Bbb R$ in the cofinite topology, the sequences $1,\frac12,\frac13,\frac14,...$ and $1,0,\frac12,0,\frac13,0,...$ converge to every point of $\Bbb R$!!

In these (and other) problematic spaces, we don't have uniqueness of sequence limits, so saying $\lim\limits_{n\to\infty}x_n$ is not meaningful in general, even if it is known that $\{x_n\}$ converges. To avoid this, it is sufficient (but not necessary) that a space be Hausdorff.

Let's say (to distinguish between the two definitions) that $x^*$ is a cluster point of $\{x_n\}$ if it is a limit point of some subsequence of $\{x_n\}$. Note that I said a limit point--the discussion above demonstrates why that's relevant. This generalizes your proposed alternate definition for "accumulation point" to general spaces.

Now suppose $x^*$ is a cluster point of the sequence $\{x_n\}$. Take any open set $U$ containing $x^*$. Since $x^*$ is a cluster point of $\{x_n\}$, then there is some $\{x_{n_k}\}$ such that $x_{n_k}\to x^*$. Then for sufficiently large $k$, each $x_{n_k}\in U$ by definition of sequence convergence, which implies that there exist infinitely many $x_{n_k}$ (and so infinitely-many $x_n$) in $U$. Thus, every cluster point is an accumulation point.

As pointed out above, the converse need not hold in general, so they are related, but not generally equivalent concepts. Again, as pointed out above, first-countability is sufficient for the equivalence of the two concepts (and may be necessary, too, but I'm not sure about that).


This is not true in general. If a space is first countable, i.e. has a countable basis at each point, then this result is true. Metric spaces are first countable.

To prove this, suppose $X$ is a first countable space and $A \subset X$. (In your case, $A = \{x_n : n \in \mathbb{N}$ where $(x_n)$ is a sequence.) Let $(U_n)_{n \in \mathbb{N}}$ denote a countable basis at the point $x^*$, where $x^*$ is an accumulation point of $A$. By definition, of accumulation point, there exists a $x_0 \in A$ and $x_0 \in U_0$. $U_0 \cap U_1$ is an open set. Again since $x^*$ is an accumulation point, there exists a $x_1 \in A$ and $x_1 \in U_0 \cap U_1$. Continue this to create a sequence $(x_n)_{n \in \mathbb{N}}$. Then since $U_n$ is a countable basis, you can prove that $\lim_{n \rightarrow \infty} x_n = x^*$. This is because given any open set $U$, there exists a $n$ such that $U_n \subset U$ such that $x^* \in U_n$ (definition countable basis). By construction, for all $m > n$, $x_m \in U_n \subset U$.

In metric spaces, the proof is bit nicer. You can take your countable basis to be $U_n = B_{\frac{1}{n}}(x^*)$. So using the definition of accumulation point, just choose $x_n$ such that $x_n \in B_\frac{1}{n}(x^*)$.


If you are interested in a counterexample, here is one:

For simplicity I assume that the sequence $s$ is just $(0,1,2,\dots)$ and $a\notin{\bf N}$ is some fresh element, we'd now like to construct a topology $\tau$ on $\{a\}\cup{\bf N}$ such that $(i)$ $a$ is limit point of $s$ and $(ii)$ no subsequence of $s$ converges to $a$. I assume that $\tau$ is generated by $\{n\}$ for $n\in{\bf N}$ and $\{a\}\cup X$ for $X\in{\cal X}$ with some ${\cal X}\subseteq{\cal P}({\bf N})$ to be figured out yet. If ${\cal X}$ is a filter then the neighborhoods of $a$ are exactly the sets $\{a\}\cup X$ for $X\in{\cal X}$, which I will also assume.

Condition $(i)$ demands that each $X\in{\cal X}$ is infinite. Assume condition $(ii)$ were not satisfied, then some subsequence $s'$ of $s$ would have a tail living in $X$ for every $X\in{\cal X}$, i.e. if we call $X'$ the result of adding to $X$ the finitely many elements of $s'$ not in $X$ then $s'$ lives completely in $A:=\cap_{X\in{\cal X}} X'$.

Take ${\cal X}$ as some free ultrafilter, where free just means empty intersection. Then all sets in ${\cal X}$ are infinite. Since $A$ is infinite, we can just remove countably many elements but also leave countably many in place, the ultrafilter property ensures that this new set or its complement is contained in ${\cal X}$, in any case some set $B$ which lacks countably many elements of $A$. But our construction tells us that $A$ should be contained in $B'$, a contradiction hence condition $(ii)$ is also satisfied.


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