0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks a lot for your answer. Yes, I'm concerned with a specific set of functions that are bounded on compact sets. I'm still not sure however what the ${\it any}$ subsequence means in the definition? Would that make sense when $Y$ is a certain family of functions and not necessarily a sequence of functions? In that case what would be those subsequences? Could you give an example? $\endgroup$ – Submartingale Sep 15 '16 at 3:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.