Construct a sequence to show $A$ is closed (topology)

Question

Here is the question:

Let $(X,\mathcal{T})$ be a topological space and $A \subset X$. Prove that $A$ is closed if and only if every sequence $(x_n)_{n\in\mathbb{N}}$ of points from $A$ and $x \in X$ such that $(x_n)_{n\in\mathbb{N}}$ converges to $x$, we have $x \in A$.

I am able to show the forward direction, but I am having lots of trouble with the backward direction. Here is what I have so far:

Suppose that for every sequence of points $(x_n)_{n\in\mathbb{N}}$ from $A$ and $x \in X$ such that $(x_n)_{n\in\mathbb{N}}$ converges to $x$, we have $x \in A$.

We assume for a contradiction that $A$ is not closed. If $A$ is not closed, there exist limit points of $A$ that are not in $A$. Let $l$ be such a limit point. We shall construct a sequence of points in $A$ that converges to $l \notin A$ (and hence giving us a contradiction).

Because $l$ is a limit point of $A$ (and because $l \notin A$), we have that $l$ is a cluster point of $A$. Therefore, \begin{equation} (U \backslash \{l\}) \cap A \ne \emptyset. \end{equation} Hence, there exists some $z \in (U \backslash \{l\}) \cap A$. But then, $z \in A$. We consider the sequence given by, \begin{equation} z_n = z, \text{ for all } n \in \mathbb{N}. \end{equation}

It is true that we can do this for any $U \in \mathcal{T}$. But the $(z_n)$ we have constructed does not converge to $l$.

Am I heading in the wrong direction? Any help is appreciated.

---

Here are the definitions I am using:

$(x_n)_{n\in\mathbb{N}}$ converges to $x$ if for any $U \in \mathcal{T}$ such that $x \in U$, we have that $\exists N_U \in \mathbb{N}$ such that $\forall n > N_U$, we have $x_n \in U$

$x$ is a limit point of $A$ if $\forall U \in \mathcal{T}$ such that $x \in U$, we have $U \cap A \ne \emptyset$.

$x$ is a cluster point of $A$ if $\forall U \in \mathcal{T}$ such that $x \in U$, we have $(U\backslash\{x\}) \cap A \ne \emptyset$.

Answer 1

Let us focus on the main question of this post: The characterization of closedness via sequences:

Let $(X,\mathcal{T})$ be a topological space and $A\subset X$. Prove that $A$ is closed if and only if every sequence $(x_n)_{n\in\mathbb{N}}$ of points from $A$ and $x\in X$ such that $(x_n)_{n\in\mathbb{N}}$ converges to $x$, we have $x\in A$.

1. Is this true?

This characterization is not true in general. The implication $\Rightarrow$ always holds, but in general we have $\not\Leftarrow$. As an example, consider $X$ an uncountable set and $\mathcal{T}$ the co-countable topology. A property of this space (that was also discussed in this post) is that any sequence $(x_n)_{n\in\mathbb{N}}$ is convergent in $(X,\mathcal{T})$ if and only if it is eventually constant.

Consider any $A\subseteq X$ countable. Hence, $A$ is not closed. However, take any sequence $(x_n)_{n\in\mathbb{N}}\subseteq A$ convergent to some $x$. By the previous comment, $(x_n)_{n\in\mathbb{N}}$ is eventually constant, so there exists some $n_0\in\mathbb{R}$ such that for all $n\geq n_0$, $x=x_n\in A$.

2. When the other implication holds?

The previous point proved that, generally, sequences cannot be regarded as an ideal tool to define topologies, since it does not convey the whole topological information of the space. The spaces that can be described using sequences (i.e., the other implication is satisfied) are known as sequential spaces. This is one of the weakest axioms of countability. For instance, every first countable space is sequential, so are metric spaces.

3. Why sequences cannot describe topologies in general?

Short answer: because sequences are defined by $\mathbb{N}$, which is a countable set. We need countability axioms because not always topologies can be described using objects with countability properties. A way to solve this problem is: how can I generalized the notion of sequences such that does not depend on $\mathbb{N}$ (hence, on being countable) and can characterize topologies? This is the concept of a net.

A net is just a map $\Lambda\to X$, where $\Lambda$ is a directed set, i.e., $\Lambda$ has a reflexive, transitive relation $\preceq$ such that $\forall \lambda_1,\lambda_2\in\Lambda$, $\exists\lambda_3\in\Lambda$ s.t. $\lambda_1,\lambda_2\preceq\lambda_3$ . In this context, $\Lambda$ can be of any size, so restricting to countable sets is not our case. Defining the notion of converge of a net by parallelism with converge of a set, we reach that nets can characterize topologies and, thus, the following property holds:

Let $(X,\mathcal{T})$ be a topological space and $A\subset X$. $A$ is closed if and only if every net $(x_\lambda)_{\lambda\in\Lambda}$ of points from $A$ and $x\in X$ such that $(x_\lambda)_{\lambda\in\Lambda}$ converges to $x$, we have $x\in A$.

4. Where your proof fails?

The process you proposed of choosing elements $z_n\in X$ in the intersection is, in fact, how this proposition is proven when $X$ is considered to be first countable. In your case, you have chosen a $z_n$ for each $U\in\mathcal{T}$. However, how many open sets $U$ can you intersect with? Probably uncountably many times. But these are more elements than the sequence requires (as many as $\mathbb{N}$, i.e., countable). When we suppose that $X$ is first countable, we use the following property:

If $(X,\mathcal{T})$ is first countable, then for all $x\in X$, there exists a countable neighborhood basis $\mathcal{B}(x)=\{B_n\}_{n\in\mathbb{N}}$ such that $B_1\supset B_2\supset B_3\supset\ldots$.

Instead of intersection with open sets $U$, we could intersect with $B_n$. Since there are countably many $B_n$, we can find countably many $z_n$, obtaining a sequence. And since $B_n$ are in a chain of inclusion, this will guarantee the convergence of $(z_n)_{n\in\mathbb{N}}$.

I hope this answer solves your inquiry.