[Math] Suppose every convergent Sequence has a unique limit point in space $X$, then $X$ is Hausdorff

general-topology

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

Best Answer

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.

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