Let $X$ be a topological space. Let $A \subset X$. The point $x$ is a limit point of $A$ iff every neighborhood of $x$ contains a point $a$ of $A$ not equal to $x$.

I am thinking about the following question: If $x$ is a limit point of $A$ does it hold that there exists a sequence $x_n $ in $A$ converging to $x$?

Please can you tell me if my thoughts are correct: If $x$ has infinitely many neighborhoods then there are infinitely many different points $x_n \in A$ and for larger $n$ they are nearer $x$ therefore $x_n \to x$ and it is true. If $x$ only has finitely many neighborhoods then it is also true: take the smallest neighborhood and the point $x_n = a \in A$ in that neighborhood and then the constant sequence $a$ converges to $x$ (even though $x\neq a$).

  • $\begingroup$ I think that you are looking for $x_n\neq x$, in which case you would need infinitely many neighborhoods. You need to be more explicit in how you're choosing $x_n$. And rework words like "nearer" to have a topological meaning. Hint: You'll want to use intersections at some point. $\endgroup$ – T.J. Gaffney Feb 18 '14 at 9:23
  • $\begingroup$ It's not true. Take the set $\Omega$ of ordinals $\le\ \omega_1$ in the order topology. Then $\omega_1$ is a limit point but not a sequential limit point of $\Omega\setminus\{\omega_1\}$. $\endgroup$ – David Mitra Feb 18 '14 at 9:30
  • $\begingroup$ A couple of (somewhat) related questions: one, two. $\endgroup$ – user642796 Feb 18 '14 at 9:43

The property that you're describing does not hold in every topological space. If the limit points can be characterized by limits of convergent sequences, then one calls $X$ a sequential space (Edit: as pointed out in the comment, there are some subtleties involved in the exact definition, see here).

Every first-countable space is sequential. Most "real-world" topological spaces are first-countable.

But one can also make up simple examples of nonsequential spaces. Let $X$ be any uncountable set and define a topology on it by calling every subset of $X$ whose complement is countable open. Additionally, call the empty set open. This is also called the cocountable topology. That space answers your question in the negative.

The problem with sequences is that they by nature countable, because they are indexed by natural numbers. You can however generalize the concept of sequences to nets (Moore-Smith sequences), where you can use any set as index set instead. In the language of nets it is really true that every limit point is the limit of a convergent net.

Edit: Let us prove that the cocountable topology is not sequential.

Claim: Let $X$ be an uncountable set equipped with the cocountable topology and $x_n\rightarrow x$ a convergent sequence. Then there exists $N$ such that $x_n=x$ for all $n\ge N$.

Proof: Define $Y=\{x_n\,:\,n\in\mathbb{N},\,x_n\not=x\}\subset X$. Because $Y$ is countable, its complement $U=X\backslash Y$ is open. Also, $x\in U$ because $x\not\in Y$. So $U$ is a neighborhood of $x$. By definition of convergence, there is an $N$ such that $x_n\in U$ for all $n\ge N$. But that means $x_n\not\in Y$. That is, $x_n=x$ for $n\ge N$. $\square$

That means, every convergent sequence is eventually constant. So if we consider an arbitrary set $A\subset X$, then we do not gain any new points by considering limits of sequences in $A$. Every set in $X$ is sequentially closed.

  • $\begingroup$ Actually, what you are describing are called Fréchet-Urysohn spaces. Sequential spaces are those such that whenever a subset is not closed, there is a sequence of points from that set converging outside the set. It is easy to show that first-countable implies Fréchet-Urysohn implies sequential, but none of these implications reverse. $\endgroup$ – user642796 Feb 18 '14 at 9:50
  • $\begingroup$ Thank you. So if $X = \mathbb R$ with the cocountable topology I am trying to find $A$ and $x$. If $A$ is countable it does not have any limit points because $A^c$ is open. So let $A$ for example be $\mathbb R \setminus \mathbb Q$ and then $x $ and $q \in \mathbb Q$. But it seems to be the case that every nbhood of $x$ does contain points in $A$. Can you help me more please? $\endgroup$ – blue Feb 18 '14 at 10:42
  • $\begingroup$ @blue: Yes, that is true. Any neighborhood of $x\in\mathbb{Q}$ contains all of $\mathbb{R}$ except countably many numbers, therefore it must intersect with $A$. But: You can not find a sequence in $A$ that converges to $x$. In fact, all convergent sequences in the cocountable topology must eventually be constant (I will add this above). So a sequence in $A$ can impossibly converge to a point outside of $A$. $\endgroup$ – J.R. Feb 18 '14 at 10:52
  • $\begingroup$ @blue: I should not have used the same letter, I'm sorry. The $A$ in the proof above is not the $A$ we talked about here. I will change it. $\endgroup$ – J.R. Feb 18 '14 at 11:27
  • $\begingroup$ Oh I'm sorry, how stupid of me, I will delete my comment. Very good answer! Thank you! $\endgroup$ – blue Feb 18 '14 at 11:29

It's not true.

A few examples:

  • Take the set $\Omega$ of ordinals $\le\omega_1$ in the order topology. Then $\omega_1$ is a limit point but not a sequential limit point of $\Omega\setminus\{\omega_1\}$.
  • Take $=\Bbb R^{\Bbb R}$ with the product topology and set $$ E=\{f\in \Bbb R^{\Bbb R} \mid f(x)=0\ \text{or}\ 1\ \text{and}\ f(x)=0\ \text{only finitely often}\ \}. $$ Then the zero function is a limit point of $E$ but not a sequential limit point of $E$.
  • In $\ell_2$ with the weak topology, take $E=\{e_n+ne_m\mid m>n\}$. Then the zero vector is a limit point of $E$ but not a sequential limit point of $E$.

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