In non-standard analysis, we have the following equivalent definition for a (standard) topological space X which is a Hausdorff space to be compact:
X is compact iff every non-standard pseudo-element of X* has a standard approximation in X.
Here is a clear advantage of using
NSA. The "standard"
(pun intended) definition of compactness has
two quantifiers: For
all open coverings of X, there
exists a
finite subcovering. Worse, we're quantifying over
sets of
subsets of X! The "non-standard" (but equivalent) definition uses only
one quantifier, and only on pseudo-elements of X.
In a compact space, every pseudo-element is close to some element. Compare with the example given for N in standard approximation, where no pseudo-element is infinitesimally close to a standard element.
⇒:
Suppose X is a compact topological space. We must prove that every element of X* has a standard approximation in X. That is, for any non-standard y in X*, we must find x∈X such that for any (standard!) open neighborhood x∈U⊆X of x in X, y is a pseudo-element of U ("y∈U*").
Suppose some y had no standard approximation. Then for any standard x∈X there is some open neighborhood Ux⊆X of which y is not a pseudo-element. Obviously, the set of all Ux's contains all x's, thus is a covering of X. As X is compact, it has a finite sub-covering Ux1,...,Uxn. But then the proposition "∀t.t∈X→t∈Ux1 v ... v t∈Uxn" is true of the standard world (it just says "Ux1,...,Uxn is a covering of X"), but not of the non-standard world -- a contradiction, as all propositions are true of the standard world iff they are true of the non-standard world! Thus, every y has a standard approximation. Note the use of finiteness: if X had only infinite sub-coverings with Ux's, then we would be unable to form our proposition -- all propositions have finite length!
⇐:
Suppose now every y has a standard approximation. Given any open covering {Uα}α∈I of X, we must find a finite sub-covering. Suppose no such finite sub-covering existed. Then for any choice of α1,...,αn there would be some xα1,...,αn not in any of Uα1,...,Uαn. Form all the propositions "∃x.x∈X&x∉Uα". Then any finite subset of these propositions has some standard x∈X satistfying it, thus some (non-standard) y∈X* satisfies all these propositions, simultaneously. But this y cannot have a standard approximation -- every standard x is in some Uα, and y is in none of them.
Thus, if every non-standard y has a standard approximation in X, then X must be compact.