If every convergent sequence has a unique limit point in space $X$, then is $X$ Hausdorff?

Question

My attempt

Let $a, b \in X$ are two distinct points . we will show that there exist two open sets $G_1$ and $G_2$ such that $a \in G_1$ and $b \in G_2$ and $G_1 \cap G_2 = \phi$ or there exists $n \in \mathbb N$ such that $S_{1/n}(a) \cap S_{1/n}(b) = \phi$.

Let if possible for each $n \in \mathbb N$ $S_{1/n}(a) \cap S_{1/n}(b) \neq \phi$ , then there exist $x_n \in S_{1/n}(a) \cap S_{1/n}(b)$ for each $n \in \mathbb N$ . So there exist a sequence $\{x_n \} \in X$ suct that converges to two distinct point $a$ and $b$, which is a contradiction.

I would be thankful , if someone check my Solution

An open ball of a particular radius is defined only when your space has a distance measure, such as in metric spaces. — Batominovski, Jul 22 '15 at 02:31
I gave a wrong counterexample earlier. Now, equip $\mathbb{R}$ with the cocountable topology (i.e. $X\subseteq\mathbb{R}$ is open iff $\mathbb{R}\setminus X$ is countable). Any convergent sequence $\left(x_n\right)_{n\in\mathbb{N}}$ in $\mathbb{R}$ which converges to $x\in\mathbb{R}$ cannot converge to another point $y\in\mathbb{R}$ because $x_n=x$ for all sufficiently large $n\in\mathbb{N}$. — Batominovski, Jul 22 '15 at 02:36
Why? My space is non-Hausdorff and any of its convergent sequences has a unique limit point. — Batominovski, Jul 22 '15 at 02:40
@Struggler: Don't try to prove that it's Hausdorff. It isn't Hausdorff. — Cameron Buie, Jul 22 '15 at 03:07

score 6 · Accepted Answer · edited Apr 13 '17 at 12:21

The main problem with your proof is that you are trying to prove something that is false. The following are standard, though not very common, definitions:

A topological space $X$ is called a KC-space is every compact subset of $X$ is closed.
A topological space $X$ is called a US-space if all convergent sequences have a unique limit point.

Now, it isn't too hard to show that the implications T₂ ⇒ KC ⇒ US ⇒ T₁ hold. However none of the arrows reverse.

The example provided by Batominovski in a comment to the question

[E]quip $\mathbb R$ with the cocountable topology (i.e. $X \subseteq\mathbb R$ is open iff $\mathbb R \setminus X$ is countable). Any convergent sequence $\left(x_n\right)_{n\in\mathbb{N}}$ in $\mathbb R$ which converges to $x\in\mathbb R$ cannot converge to another point $y\in\mathbb R$ because $x_n=x$ for all sufficiently large $n\in\mathbb{N}$.

is actually an example of a KC-space which is not Hausdorff. (It is a KC-space because the only compact subsets are finite, which are closed by T₁-ness.)

For an example of a US-space which is not a KC-space, consider the space $X = [0,1] \cup \{ \infty \}$ where $[0,1]$ is an open subspace (with the usual topology), and the open neighborhoods of $\infty$ are of the form $\{ \infty \} \cup U$ where $U$ is a dense open subset of $[0,1]$.

This is a US-space essentially because a sequence only converges to $\infty$ if eventually all its points are $\infty$ (and $[0,1]$ is an open set not containing $\infty$).

In more detail, if $(x_n)_n$ is a sequence in $[0,1]$ then by the compactness of $[0,1]$ some subsequence $(x_{n_k})_k$ must converge to a point $x \in [0,1]$. Now it isn't hard to show that $F := \{ x_{n_k} : k \geq 1 \} \cup \{ x \}$ is a closed nowhere dense subset of $[0,1]$, and so $U := \{ \infty \} \cup ( [0,1] \setminus F)$ is an open neighborhood of $\infty$. But then $U$ cannot contain a tail of the sequence $(x_n)_n$, meaning the sequence cannot converge to $\infty$.
It not a KC-space because $[0,1]$ is a compact subset which is not closed.

The major false move in your attempted proof is the implicit assumption of first-countability, where you are looking at the neighborhoods $S_{1/n} (a)$ and $S_{1/n}(b)$ ($n \in \mathbb N$) for $a,b$, respectively.

But actually, your proof basically gives a correct argument that every first-countable US-space is Hausdorff:

If $X$ is a first-countable space which is not Hausdorff, then there are distinct points $a,b \in X$ which cannot be separated by disjoint open neighborhoods. By first-countability, $a,b$ have descending open neighborhood bases $\{ U_n \}_{n \in \mathbb N}$, $\{ V_n \}_{n \in \mathbb N}$, respectively. By choice of $a,b$, for each $n \in \mathbb N$ we have that $U_n \cap V_n \neq \emptyset$, so pick $x_n \in U_n \cap V_n$. Then $( x_n )_n$ can be shown to be a sequence which converges to both $a$ and $b$. Thus $X$ is not a US-space.

If every convergent sequence has a unique limit point in space $X$, then is $X$ Hausdorff?

1 Answers1

Linked