2
$\begingroup$

Given a topological space $(X, \mathcal{T})$, and a subset $A$ of $X$, define $p$ to be a strict limit point of $A$ if there exists a sequence $(x_n)_n$ in $A \setminus \{p\}$ such that $x_n \rightarrow p$. (Arbogast, Methods of Applied Mathematics page 9 - book here)

Is easy to show that if $p$ is a strict limit point of $A$ then $p$ is an accumulation point of $A$.

The converse is also true if the space is first-countable (every point has a countable base of neighbourhoods).

Is the converse also true when the space is not first-countable?

$\endgroup$
2
$\begingroup$

A relatively simple counterexample: let $X = \mathbb{R}$ with the topology $\mathcal{T} = \{\emptyset\} \cup \{X \setminus A: A \text { at most countable }\}$, the so-called co-countable topology. Check that this is indeed a topology. It's indeed not first countable.

Then set $A = \mathbb{R}\setminus \{0\}$ and $0$ is an accumulation point of $A$. But if $x_n$ is any sequence of points from $A$, then $X \setminus \{x_n: n \in \mathbb{N}\}$ is open, contains $0$ and no point of the sequence, $0$ is not a strict limit point of $A$.

If you know ordinals, $\omega_1 + 1$ is also an example with $A = \omega_1$, and this has nicer separation axioms (and is also sequentially compact). This is Giulio's example, essentially.

The class can be broadened to so called sequential spaces, instead of the more strict first countable. But some condition is needed.

$\endgroup$
  • $\begingroup$ thanks! I was trying with co-finite topology and don't succeed $\endgroup$ – nadapez Apr 27 '16 at 17:53
0
$\begingroup$

No, but the only counterexample occurring to my mind is easy if and only if you know the basics of the theory of ordinals. Well, since there exists a proof in the first-countable case, counterexamples need some sort of uncountable construction, and hence you'll need something like the theory of ordinals.

Take a well ordered set $\alpha$ whose cofinality is strictly greater than $\aleph_0$. This means that every function $f:\aleph_0\to\alpha$ respecting the order is limited, i.e. there exists $y\in\alpha$ such that, for every $x\in\aleph_0$, $y>f(x)$. Now take $\beta=\alpha\cup\{*\}$, where $*$ is an element greater than any element of $\alpha$. Put on $\beta$ the following topology: a subset $U\subseteq \beta$ is open if $x<y$ and $x\in U$ imply $y\in U$. Except one case: we impose the subset $\{*\}\subseteq\beta$ to be not open. It is easy to check that this defines a topology, and that $\{*\}$ is an accumulation point. Still, since the cofinality of $\alpha$ is strictly greater than $\aleph_0$, there cannot be any sequence in $\alpha$ converging to $*$, hence $*$ is not a limit point of $\beta$.

$\endgroup$

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.