Is a point where any sequence that converges to it contains it the same thing as an isolated point?

Question

Suppose I have a topological space $(X, \tau^X)$. Based on the Wikipedia definition of an isolated point, $p \in X$ is an isolated point if and only if $\{p\} \in \tau^X$.

In one specific example I'm looking at, I have one point that seems very different from all the rest.

The first thing that I noticed about this point $p$ is that every sequence $s$ that converges to $p$ is eventually constantly $p$. I heard the phrase "isolated point" before, so I looked it up on Wikipedia, but I didn't see a theorem connecting the open-set-flavored definition to sequence/net/filter-flavored definition, which made me curious.

• I'm wondering whether the sequence characterization is equivalent in general to being an isolated point.
• Assuming the sequence characterization is inequivalent, is there a more general kind of pseudosequence like a net that can capture the notion of being an isolated point.

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What follows is motivation. The question above is self-contained without the movation.

I asked this question earlier today and tried to come up with a linear order without endpoints with no fixed-point-free order-automorphisms.

My first attempt was the order $ \mathbb{R} + \{0\} + \mathbb{R} $ which was unsuccessful (since it is equivalent to $\mathbb{R}$ as Noah Schweber points out here).

My next and current attempt is the order $\mathbb{R} + \{ 0, 1, 2 \} + \mathbb{R}$, which I'll call $\alpha$. I'm pretty sure an order-automorphism of this ordering must fix $1$.

If I endow $\alpha$ with the order topology, then $1$ is an isolated point because the open interval $(0, 2)$ is $\{1\}$.

$1$ also has the property that any sequence that approaches it, intuitively, will eventually need to "hop over" the chasm between $0$ and $1$ or between $1$ and $2$.

The visual metaphor of "eventually needing to jump" is pretty natural, so it seems like it must correspond to something.

Answer 1

If $p$ is an isolated point in $X$, then every sequence converging to $p$ contains $p$. Stronger, every sequence converging to $p$ is eventually equal to $p$. This is immediate from the definition: Suppose $(x_n)_{n\in \mathbb{N}}$ converges to $p$. Since $\{p\}$ is an open neighborhood of $p$, there exists $N\in \mathbb{N}$ such that $x_n\in \{p\}$ for all $n\geq N$, so $x_n = p$ for all $n\geq N$.

The converse is false. For a counterexample, take the ordinal $\omega_1+1$ ($\omega_1$ is the least uncountable ordinal) with the order topology. The point $p = \omega_1$ is not isolated, since it does not an immediate predecessor. But for any sequence $(\alpha_n)_{n\in \mathbb{N}}$, which does not contain $\omega_1$, $\sup\{\alpha_n\mid n\in \mathbb{N}\}$ is an ordinal $\beta<\omega_1$ (since $\omega_1$ is regular), so the sequence does not meet the interval $(\beta,\infty)$, and hence it does not converge to $\omega_1$. Taking the contrapositive, any sequence converging to $\omega_1$ contains $\omega_1$.

Replacing sequences with nets, the analogue of your statement is true, with a straightforward proof. The following are equivalent:

1. $p$ is an isolated point.
2. For every net $(x_d)_{d\in D}$ converging to $p$, there exists $a\in D$ such that $x_b = p$ for all $b\geq a$.
3. Every net $(x_d)_{d\in D}$ converging to $p$ contains $p$.

$1\implies 2$: Just as for sequences, suppose $(x_d)_{d\in D}$ is a net converging to $p$. Since $\{p\}$ is an open neighborhood of $p$, there exists $a\in D$ such that $x_b\in \{p\}$ for all $b\geq a$, so $x_b = p$ for all $b\geq a$.

$2\implies 3$: Trivial.

$3\implies 1$: Suppose $p$ is not an isolated point. We construct a net converging to $p$ which does not contain $p$. Let $D$ be the set of all open neighborhoods of $p$, ordered by reverse inclusion, so $U\leq V$ if and only if $V\subseteq U$. This is a directed set, since for any $U,V\in D$, $U\cap V$ is an open neighborhood of $p$, and we have $U\leq (U\cap V)$ and $V\leq (U\cap V)$. Now for any $U\in D$, since $p$ is not isolated, $U\neq \{p\}$, so we can pick some $x_U\in U$ with $x_U\neq p$. The net $(x_U)_{U\in D}$ does not contain $p$, but it converges to $p$. Indeed, for an open neighborhood $U$ of $p$, if $U\leq V\in D$, then $x_v\in V\subseteq U$, so $x_V\in U$ for all $V\geq U$.

Answer 2

As @AlexKruckman pointed out, my initial attempt (in the spoiler) was a complete flunk! However, after some time I figured out and cleared the road towards these more elementary examples (more elementary than the ordinals):

1. An uncountable set $X$ and the family of sets $Y$ such that $A\setminus Y$ is at most countable.

2. The plane $\mathbb{R}^2$ and the family of sets of the form $\mathbb{R}^2$ minus countably many straight lines through $(0,0)$ plus the point $(0,0)$. We can also drop in all the regular open sets of $\mathbb{R}^2$ and generate the topology out of it; that still keeps the point $(0,0)$ special.

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This is how I was thinking:
Your statement:

If every sequence converging to the point contains the point, then the limit is an isolated point.

In the case of first-countable topologies (where every point has a countable local basis), and assuming countable choice, your statement is true. This is because in these cases for every non-isolated point we can construct a sequence that converges to it and doesn't contain it. So if there isn't such a sequence for some point, then we can't construct it and consequentially it can't be non-isolated.

This is how we construct it: Enumerate the basis around a non-isolated point $p$: $A_1,A_2\ldots$ Construct another sequence of open sets where $n$-th set is $B_n=\bigcap_{k=1}^n A_k$. The sequence $B$ is non-increasing. Let's clear it from duplicates and we will end up with a decreasing "sequence" (because it's maybe finite) $C$. If $C$ is finite, then the last element from $C$ is the minimal neighborhood of $p$ (meaning it is contained in all other neighborhoods) and it isn't empty or $\{p\}$, so there is another point $q\neq p$ in that minimal neighborhood and we can take a constant sequence which is equal to $q$ always. We are left to cover the case when $C$ is a proper sequence (infinite). From each of the sets $D_n=C_n\setminus C_{n+1}$ we can take one point $d_n\in D_n,d_n\neq p$ (here we are using the countable choice) and we claim that this sequence $d$ is the one we were looking for. Every neighborhood of $p$ has a subneighborhood of form $A_k$, which further has a subneighborhood of form $B_k$, which is equivalent to some element of the sequence $C$, say $C_m$, which means that for $h\ge m$, $d_h$ is in that initially chosen neighborhood. Hence $d_n$ was the sequence we were looking for.

This shows that if we want to find a counter-example to your statement then we need a point whose every basis is uncountable. From the other side, your statement would be false if every countable set which doesn't contain a non-isolated point $p$ does not intersect some neighborhood of $p$, which is of course true if that countable set is a closed set itself. And this is true if the space is $T_1$ and intersection of any countable family of open sets is an open set itself. Now this condition does exclude a lot of nice spaces (which is maybe why it isn't easy to find an example), but it is nice in two senses: it is not a local condition, and it is a slight generalization of topology: in topology you can intersect at most finitely many open sets and here you can intersect countably many open sets, which is the very first bigger cardinal than all finite ones. Then the examples I gave seemed not too far from these conditions to me.

The flunk:

I think you can take the topology $\tau$ on the set of all sequences of real numbers where $\tau$ is generated by sets that are products of finitely many open intervals (we limit the value on finitely many indices) and infinitely many all reals (on the rest of indices we don't limit). If I remember well, this is what product topology is. What I described will be the base of the topology it generates. Now you can make a sequence of sequences where $n$-th sequence starts with zeros and then at $n$-th index switches to $1$ and stays there. For every neighborhood of constant zero sequence, the sequence of sequences I described will eventually fall into that neighborhood and never leave, meaning it converges to zero. But no sequence is isolated here so your statement should be false.