Existence of convergent sequences in vector analysis and topology

Question

Many theorems in vector analysis and topology use the concept of converging sequences to e.g. display that a set is closed.

E.g.

A set F is closed in some metric space M if and only if $\{x_k\}_{i=1}^{∞}$ ⊂ F and $\{x_k\}_{i=1}^{∞} \rightarrow x$ implies that $x \in F$.

Why do we know that a convergent sequence can always be selected?
Can all such sequences be listed, that is, given some set, can one display all the (converging) sequences it contains?

Answer 1

As soon as $M$ is non-empty, i.e. as soon as there is some $a\in M$, there are sequences in $M$, e.g. the constant sequence $x_k:=a$. So you can pick a sequence $(x_k)\subseteq M$ just like you are picking an element of a set (in fact any such sequence is an element of the set of sequences in $M$, which in turn is the set of maps $\mathbb{N}\longrightarrow M$).

As for your second question, what do you mean by listed? If the question is if there are only countably many sequences (so they can be "listed"), then this depends on the size of $M$: if $M$ is infinite, then it has uncountably many infinite subsets (this follows from a result of Cantor that power sets of sets are of larger cardinality) so the sequences cannot be listed in the above sense. But if $M$ is finite, then this is possible because then there are only countably many functions $\mathbb{N}\longrightarrow M$.

Answer 2

Notice that it's not relevant whether such a sequence exists or not. The definition of the statement $F$ is closed is the following implication: $$ \Big( \{x_i\}_{i=1}^\infty \subset F \text{ and } x_i\to x \Big) \implies \Big( x\in F \Big). $$ This doesn't assert the existence of any sequences; instead it says IF the hypothesis is true (i.e., if there is a sequence in $F$ that converges to some point $x$), THEN the conclusion must be true ($x$ is in $F$).

It might well happen that there do not exist any sequences in $F$. (Can you think of a set $F$ for which this is the case?) In that case, the hypothesis of this implication is always false, so the implication itself is "vacuously true," and therefore $F$ is closed.