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A text book I'm reading says at one point the following:

"In metric spaces are sequences the ones which chacterize closed sets and continuous functions".

What is exactly the meaning of that phrase?, is that every closed set and every continuous function can be described by sequences.

I know a couple of things about metric spaces and sequences that I believe are related:

$1-$ In a closed sets sequences inside a set converge to a point inside the set

$2-$ A function $f(x)$ is continuous iff $\{f(x_n)\}\to f(x)$ for every sequence convergent to $x$.

But I don't understand what it means by 'characterize', are sequences in metric spaces enough to define what closed sets and continuous functions are or there is something more behind that statement?

Cure
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    For metric spaces convergence of sequences (i.e. which sequences are convergent and what are their limits) fully describes the induced topology. Hence, if we know which sequences are convergent and to what limits, we know which sets are closed (sets that contain it's limit points) and which functions are continuous (functions such that $x_n\to x$ implies $f(x_n)\to f(x)$). It does not hold for general topological spaces. Keyword is first-countability. – Marcin ĊoĊ› May 09 '14 at 06:52
  • can you please provide a reference of the textbook? – johnny09 Apr 21 '19 at 17:45

1 Answers1

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Basically what is going on is that metric spaces are sequential:

Definition. A topological space $X$ is sequential if given $A \subseteq X$, $A$ is closed iff every limit of a (convergent) sequence consisting of points of $A$ is also an element of $A$.1

(Worded slightly differently, given any topological space, we can define the sequential closure, $\mathrm{cl}_{\text{seq}}(A)$, of some $A \subseteq X$ to be the smallest set $B \supseteq A$ which is closed under limits of (convergent) sequences. Then a topological space $X$ is sequenial iff $\overline{A} = \mathrm{cl}_{\text{seq}}(A)$ for each $A \subseteq X$.)

A nice fact about sequential spaces (completely analogous to what you have stated for metric spaces) is the following:

Fact. If $X$ is a sequential space, and $Y$ an arbitrary topological space, then a function $f : X \to Y$ is continuous iff given any (convergent) sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ and any limit $x$ of $\langle x_n \rangle_{n \in \mathbb{N}}$, we have that $f(x)$ is a limit of the sequence $\langle f(x_n) \rangle_{n \in \mathbb{N}}$ in $Y$.

So if you have a sequential space $X$, then just knowing the families of the form $$L(\langle x_n \rangle_{n}) := \{ x \in X : \langle x_n \rangle_{n} \rightarrow x \}$$ is enough to determine

  • the closed sets (these are exactly the subsets $A$ such that $L(\langle x_n \rangle_{n}) \subseteq A$ for every sequence $\langle x_n \rangle_{n}$ of points of $A$); and
  • which function with domain $X$ are continuous (these are exactly the functions $f$ such that $\langle f(x_n) \rangle_{n} \rightarrow f(x)$ where $x \in L(\langle x_n \rangle_{n})$).

In this way, the convergent sequences (and their limits) in a sequential space $X$ completely characterise the closed sets (and therefore also the underlying topology) of $X$ and the continuous functions from $X$.

We also have the following nice implications: $$\text{metric} \Rightarrow \text{first-countable} \Rightarrow \text{sequential}$$ (and, as a bonus, neither implication reverses).

1I am not assuming the spaces are Hausdorff, so sequences can have multiple limits. Of course, metric spaces are Hausdorff.

user642796
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