Neighbourhoods via sequences

Question

I'm studying general topology and a question has come to my mind.

By definition, a sequence of points in a topological space converges to a point if its values are eventually in every neighbourhood of the point. Rephrasing, the neighbourhoods of a point satisfy the condition that the values of every sequence converging to the point are eventually in them.

I wonder if that is a characterising property of the neighbourhoods, viz. given a point and a subset of a topological space, if every sequence converging to the point eventually belongs to the subset, is it true that the subset is a neighbourhood of the point?

If not (this feels having something to do with countability properties), I'd like to get to know about sufficient conditions on the spaces under which this happens and how to prove it.

Disclaimer For a subset, to be a neighbourhood of a point I mean containing an open subset which contains the point.

Take a topological space $X$ and a non-isolated point $x\in X$ such that $x$ is not the limit of any sequence in $X\setminus {x}$. Then the point $x$ and the subset ${x}$ serve as a counterexample. One such space would be $X=\omega_1+=[0,\omega_1]$, where $\omega_1$ is the first uncountable ordinal, and $x=\omega_1$. As you said, this is the uncountability at play. — user469053, Apr 10 '24 at 20:28
There is more than uncountability involved, although for ordinary $\omega$-length sequences it's certainly relevant. See is a net stronger than a transfinite sequence for characterizing topology? and my comments to Intuition behind nets. — Dave L. Renfro, Apr 10 '24 at 20:57
If you replace "sequences" with "nets", though, then the statement does become true. — Daniel Schepler, Apr 11 '24 at 00:03

Steven Clontz · Answer 1 · 2024-04-12T01:47:43.467

Given a point and a subset of a topological space, if every sequence converging to the point eventually belongs to the subset, is it true that the subset is a neighbourhood of the point?

This is not true. For example, consider the class of sequentially discrete spaces, where sequences only converge if they are eventually constant. If your statement held, then each point must have its singleton as a neighborhood, that is, the space must be discrete.

However, there exist sequentially discrete spaces that are not discrete: https://topology.pi-base.org/spaces?q=Sequentially+discrete%2B%7EDiscrete

If not (this feels having something to do with countability properties), I'd like to get to know about sufficient conditions on the spaces under which this happens and how to prove it.

The property where the topology is indeed determined by sequences is known as sequential. Sufficient properties for sequential include Frechet Urysohn and more generally First countable.

The question clearly states "If not (this feels having something to do with countability properties), I'd like to get to know about sufficient conditions on the spaces under which this happens and how to prove it." Something like this should be added to the answer so that it can be considered complete. — Smiley1000, Apr 11 '24 at 08:41

score 0 · Accepted Answer · answered Apr 11 '24 at 08:38

Let $X$ be a topological space that is first countable, meaning that each point $x$ has a countable neighbourhood basis $(U_{x,n})_{n \in \mathbb{N}}$. Without loss of generality, assume that $U_{x,1} \supseteq U_{x,2} \supseteq U_{x,3} \supseteq \cdots$.

Let $x \in A \subseteq X$. Assume that for all sequences $(x_n)_{n \in \mathbb{N}}$ in $X$ with $x_n \to x$, there exists $N \in \mathbb{N}$ with $\forall n \geq N: x_n \in A$. We will show that there exists an open set $U$ with $x \in U \subseteq A$, meaning that $A$ is a neighborhood of $x$.

Striving for a contradiction, assume that for every open set $U$, we have either $x \not\in U$ or $U \not\subseteq A$. In particular, for every $n \in \mathbb{N}$, we have either $x \not\in U_{x,n}$ or $U_{x,n} \not\subseteq A$. By assumption, $x \in U_{x,n}$, so we must have $U_{x,n} \not\subseteq A$, hence there must exist some point which we call $x_n$ with $x_n \in U_{x,n}$ but $x_n \not\in A$. Take the so-constructed sequence $(x_n)_{n \in \mathbb{N}}$.

We will now show that $x_n \to x$, but clearly $\forall n \in \mathbb{N}: x_n \not\in A$, and this gives a contradiction to the assumption. So take any open set $V$ with $x \in V$. By first countability, there exists $N \in \mathbb{N}$ with $x \in U_{x,N} \subseteq V$. Hence for all $n \geq N$, we have that $x_n \in U_{x,n} \subseteq U_{x,N} \subseteq V$. Therefore, $x_n \to x$. This contradiction finishes the proof.

Neighbourhoods via sequences

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