The definition I have for sequential compactness goes like this:
A subspace $ K \subseteq X $, $X$ a metric space, is sequentially compact if all sequences $(x_n)$, $x_n \in K$ have a convergent subsequence $(x_{n_i})$, $x_{n_i} \rightarrow x \in K$.
My interpretation of this is the following statement:
Let $ K \subseteq X $, $X$ a metric space. Then:
all sequences $(x_n)$, $x_n \in K$ have a convergent subsequence $(x_{n_i})$, $x_{n_i} \rightarrow x \in K$ $\implies$ $K$ is compact.
But I was wondering if/when this is true in the other direction? I thought that if $K$ is compact and it contains some sequence, then that sequence must have a convergent subsequence? But perhaps there is no guarantee that $K$ contains a sequence? But then if $K$ is non-empty, it must have points in it which form some sort of sequence?
I'd be grateful if someone could clarify this for me. Many thanks.
Best Answer
Definitions in math are biconditional statements. Confusingly, they are often not stated that way. However, it is the case that
$$\text{All sequences $(x_n)$ with elements in $K$ have a convergent subsequence $(x_{n_i})$ $\Leftrightarrow$ $K$ is compact}$$
The case of the empty set is interesting. As an analogy, suppose I said "all the irrational integers are zero". That statement is nonsense, clearly, but it's also true. Since there are no irrational integers, it doesn't matter what statement I make about them. Such a statement is said to be "vacuously true".
So if $K$ is empty, it is true that every sequence in $K$ contains a convergent subsequence, but it's vacuously true since $K$ does not contain any sequences.