I'm trying to understand the following definition of pre-compactness (from Wikipedia):
A relatively compact subspace (or relatively compact subset, or precompact) $Y$ of a topological space $X$ is a subset whose closure is compact. Since closed subsets of a compact space are compact, every subset of a compact space is relatively compact. In the case of a metric topology, or more generally when sequences may be used to test for compactness, the criterion for relative compactness becomes that any sequence in $Y$ has a subsequence convergent in $X$.
I'm thinking of $X$ as a functional space, e.g., the space of all right continuous functions with left limits and of $Y$ as a sequence of functions in $X$ indexed by $n$, say $\{f^{n}(t)\}$.
In this case, according to the definition, the criterion for relative compactness is that ${\it any}$ sequence in $Y$ has a convergent subsequence. But what does that mean in the above case where $X$ itself is a sequence of functions? Is it simply that the sequence $\{f^{n}(t)\}$ must have a convergent subsequence? Could you give an example where the ${\it any}$ part makes sense?
As the space of all right continuous functions with left limits, I believe $X$ is not necessarily a metric space, since it is not generally bounded. Therefore, the equivalence between sequential compactness and compactness does not necessarily hold for $Y$.
If you add the condition that the functions in $X$ be bounded, then then space can be expressed as a metric space, equipped with the $sup$ norm
By construction, subset $Y$ itself is a sequence of functions where $Y$ is defined as $\{f^{n}(t)\}$ and so you are correct in that $\{f^{n}(t)\}$ must have a convergent subsequence.
Note $\{f^{n}(t)\}$ itself need not converge as it is the closure of $Y$ that is compact.