What is a counterexample for If $x\in\overline{A},$ $x$ is not necessarily a limit of a sequence in $A$?

Question

Let $(X,\tau)$ be a topological space, $\emptyset\neq A\subseteq X$ and $(x_n)$ be a sequence in $A$. I could easily show that any limit of $(x_n)$ belongs to $\overline{A}$. My note says that if $x\in\overline{A},$ $x$ is not necessarily a limit of a sequence in $A$. But I cannot find a counterexample. Now in my attempt to find a counterexample, which I failed, I could conjecture that if the topological space is metrizable then, if $x\in\overline{A}$ then $x$ is a limit of a sequence in $A$. Is my conjecture true? And how would you find a counterexample to the original problem? Please help. Thanks.

Answer 1

It's true in a metrisable space, and more generally a first countable space.

The standard counterexample is $X=\omega_1+1$, the first uncountable ordinal plus $1$, under the order topology. This is the set of all countable ordinals, together with $\omega_1$. Let $A=X-\{\omega_1\}$. Then $\overline A=X$. Each countable sequence in $A$ is bounded by a countable ordinal, so can't converge to $\omega_1$

Answer 2

Another example is: let $X := [0, 1]^\omega$ with the box topology, and let $A = (0, 1]^\omega$. Then it's fairly easy to see $(0, 0, \ldots) \in \overline{A}$. But suppose $(x_{11}, x_{12}, \ldots), (x_{21}, x_{22}, \ldots), \ldots$ is a sequence in $A$. Then $[0, \frac{1}{2} x_{11}) \times [0, \frac{1}{2} x_{22}) \times \cdots$ is an open neighborhood of $(0, 0, \ldots)$; however, no member of the sequence is in this neighborhood. Therefore, no sequence in $A$ converges to $(0, 0, \ldots)$.

Answer 3

I think the simplest example is an uncountable set like $X= \mathbb{R}$ in the co-countable topology. This is the topology $\{\emptyset\} \cup \{A \subseteq X: X \setminus A \text{ at most countable }\}$, so the closed sets are all finite and all countable sets and $X$ itself.

This implies that any uncountable set $D \subseteq X$ is dense (the only closed set that contains it is $X$), but a sequence $(x_n)$ converges to $x$ iff $\exists N: \forall n \ge N: x_n =x$, i.e. only the eventually constant sequences converge. This implies that any sequence from $D$ only can converge to a point of $D$, so all points of $X\setminus D$ are in the closure but cannot be reached by a convergent sequence.

To show the sequence statement: suppose $x_n \to x$. Define $C = \{x_n : x_n \neq x\}$ Then $C$ is at most countable and does not contain $x$ so $X\setminus C$ is an open neighbourhood of $x$ in the co-countable topology. The convergence then shows us there is some $N$ such that $\forall n \ge N: x_n \in X\setminus C$. But $x_n \in X\setminus C$ iff $x_n = x$. So the sequence is eventually constant with value $x$.

So $\mathbb{R}$ in the co-countable topology is an example where all $x \in \mathbb{Q}$ are in the closure of $D = \mathbb{R} \setminus \mathbb{Q}$ but not a sequential limit from it.