$x \in \overline{A}$ with respect to the topology induced by a metric iff there exists a sequence of points in $A$ that converge to $x$.

Question

I'm looking at this theorem in my textbook, and I was wondering: is this not true for any general topological space?

i.e. Does the following hold?

Let $X$ be a topological space and $A \subset X$. Then $x \in \overline{A}$ if and only if there exists a sequence of points in $A$ that converge to $x$.

Why is the title version of the theorem restricted to the metric case?

Here $\overline{A}$ denotes the closure of $A$ in $X$.

Answer 1

No, it doesn't hold.

Consider the Cocountable topology on an uncountable set; for example, on $\mathbb{R}$.
Here the open sets are $\varnothing$ and those sets whose complement is countable.

On this topological space every convergent sequence is eventually constant, as you can see in an answer to this question; in particular, limits of sequences are unique.

Fix $r \in \mathbb{R}$, and let $A = \mathbb{R} \setminus \{r\}$.
Then $A$ is not countable, whence $\{r\}$ is not open, and so $A$ is not closed. Therefore, $\bar{A} = \mathbb{R}$.

Now if $(x_n) \subseteq A$ is a convergent sequence, then $x_n \to a \in A$ because $x_n = a$, for some $a \in \mathbb{R}$, for $n \geq n_0 \in \mathbb{N}$, and so $(x_n)$ doesn't converge to $r$.

Answer 2

As you've been pointed out, your question has to do with first-countability of the topological space in question.

Definition: Topological space $X$ is first-countable if for any $x\in X$ there exists a countable basis of neighborhoods for $x$, that is, a family $\mathcal U_x$, such that for any neighborhood $U$ of $x$ there exists a neighborhood $V\in \mathcal U_x$ of $x$, so that $V\subset U$.

Proposition: Let X be a topological space and $A \subset X$. Let $$ L(A):=\{x\in X\;|\;\exists\text{ sequence }(a_n)_{n\in\mathbb N}\text{ in }A\text{ with }\lim_{n\rightarrow\infty}a_n=x\}, $$ i.e. $L(A)$ is the set of all limits of all sequences with terms in $A$. Then

1. $L(A)\subset\overline A$ holds for any topological space $X$,
2. $\overline A\subset L(A)$ holds if X is first-countable.

Proof. We'll be using the characterization of closure of $A$ that states $$\overline A=\{x\in X\;|\;\text{any nbh. of }x\text{ intersects }A\}.$$

1. Let $x\in L(A)$, i.e. $\exists (a_n)_{n\in\mathbb N}$ such that $\forall n\in\mathbb N : a_n\in A$ and $\lim_{n\rightarrow\infty}a_n=x$. If $V$ is a nbh. of $x$, all terms except maybe finitely many must be inside $V$, which means $V$ intersects $A$, and since $V$ was arbitrary, $x\in\overline A$.
2. Let $x\in\overline A$. Because $X$ is first-countable, there exists (by definition) a countable basis of neighborhoods $\mathcal U_x =\{V_n\;|\;n\in\mathbb N\}$ for point $x$. Since $x\in\overline A$, $V_n\cap A\neq \emptyset$, so let $a_n \in V_n\cap A$ for all $n\in\mathbb N$. We've constructed a sequence $(a_n)_{n\in\mathbb N}$; we need to check $\lim_{n\rightarrow\infty}a_n=x$. Since for every nbh. $W$ of $x$ there exists $n_0\in\mathbb N$ such that $V_{n_0}\subset W$ (as this is what it means for $\mathcal U_x$ to be a basis of neighborhoods), $\forall n\geq n_0: a_n\in V_n\subset V_{n_0}\subset W$. Thus $x\in L(A)$.

The idea with the second part is to take a point from each neighborhood-basic set. Also, it's easy to see that every metrizable space is first-countable, as $\mathcal U_x=\{B(x,\frac{1}{n})\;|\;n\in\mathbb N\}$ is a countable basis of neighborhoods for any point $x\in X$.

Answer 3

The class of topological spaces is very broad class and includes spaces in which the closure operator cannot be defined in terms of convergent countable sequences. Here is an example.

Let $X$ be the set of all functions from $\Bbb N$ to $\Bbb R.$ For brevity let $X^+ = \{g\in X: \forall n\in \Bbb N\;(g(n)>0)\}.$

For $f\in X$ and $g\in X^+ $ let $B(f,g)=\{h\in X: \forall n\in \Bbb N\;(\;|h(n)-f(n)|<g(n)\;)\}.$ Then $\mathcal B=\{B(f,g):f\in X\land g\in X^+\}$ is a base (basis) for a topology $T$ on $X.$

(Note: $T$ is equal to the Box-Product topology on $X,$ which is usually defined in a slightly different manner.)

Take $f_0\in X$ where $f_0(n)=0$ for all $n\in \Bbb N.$ It is easily shown that $f_0\in \overline {X^+}:$ For if $U$ is a nbhd of $f_0$ then $U\supset B(f_0,g)$ for some $g\in X^+,$ so let $h(n)=g(n)/2$ for all $n\in \Bbb N.$ Then $h\in U\cap X^+.$

But if $(g_n)_{n\in \Bbb N}$ is a sequence of members of $X^+,$ then for $n\in \Bbb N$ let $g(n)=g_n(n)/2.$ Then $B(f_0, g)$ is a nbhd of $f_0$ which does not contain $any$ member of $\{g_n:n\in \Bbb N\}.$

FOOTNOTE. To show that $\mathcal B$ is the base for a topology: For a set $B$ of subsets of a set $Y$ to be a base for a topology on $Y$ it is necessary and sufficient that (i)...$\cup B=Y$ (i.e. every point of $Y$ belongs to to least one member of $B$), and (ii)... If $b_1,b_2\in Y$ and $x\in b_1\cap b_2$ then there exists $b_3 \in B$ such that $x\in b_3\subset b_1\cap b_2.$

Now $\mathcal B,$ defined above, obviously satisfies $\cup \mathcal B=X.$

If $b_i=B(f_i,g_i)\in \mathcal B$ for $i\in \{1,2\}$ and $h\in b_1\cap b_2$ then for each $n\in \Bbb N$ take $g_3(n)>0$ where $g_3(n)$ is small enough that $$(-g_3(n)+h(n), g_3(n)+h(n))\subset (-g_i(n)+f_i(n), g_i(n)+f_i(n)) \; \text {for } i\in \{1,2\}.$$ Let $b_3=B(h, g_3).$ Then $b_3\in \mathcal B$ and $h\in b_3\subset b_1\cap b_2.$