Let's think about the
Bolzano-Weierstrass theorem in real analysis (every sequence in a compact set has a convergent subsequence). What would it take to generalise this theorem to a topological space $X$?
First, let's write down the statement: we know that
sequences aren't very fundamental to arbitrary topological spaces, so the statement we're looking for would be something like:
every net in a compact set has a convergent subnet. The question is what the definition of "compact" is a general topological space generalising the "closed and bounded" definition for metric spaces (and how to prove the generalised statement from this definition).
Let's try to formulate some equivalent ways of stating that. It is trivial to see that the filter associated with a subnet is a filter refinement -- so we can state it as:
every filter in a compact set has a convergent refinement. This also implies that
every ultrafilter is convergent (as an ultrafilter has no proper refinement), and by the ultrafilter lemma, the implication is an iff. Of course equivalently,
every ultranet is convergent (where an ultranet is obviously a net that for all $S$ is either eventually in $S$ or in $S'$).
You might think of ultranets (nets corresponding to ultrafilters) as all sorts of stuff, like "all convergent nets are ultra", or as generalisations of monotonic sequences -- all these are wrong, and it's a useful exercise to write down counter-examples to this on the real line (hint for the monotonic sequences thing: take the union of a bunch of sets such that a tail set of your sequence is in neither this union nor its complement). You might also think that the "ultrafilter" formulation of the statement is related to the "halving intervals" proof of the Bolzano-Weierstrass theorem, but the filter base generated by this mechanism does not generate an ultrafilter (construct a set that isn't in the filter and its complement isn't either).
It should be clear that if we only wanted
sequences having convergent subsequences, requiring the convergent refinement for
every filter is only necessary needed if every filter has a corresponding sequence, and having a convergent refinement only suffices if every filter has a corresponding sequence --
recall that this occurs precisely in a first-countable space. The statement here is that
in and only in (inn?) a first-countable space, compactness implies sequential compactness.
OK -- we still haven't discovered what this generalised "compactness" is. What is the hypothesis of the generalised Bolzano-Weierstrass theorem?
Let's consider two example spaces where the Bolzano-Weierstrass theorem doesn't apply:
These two are "basically the same" kind of set -- a set with some
limit points removed. Except in the second case, we can actually say "it's a set with some limit points removed", because these limit points exist in the space the set is in, while in the first case, the limit points have been removed from the containing space
itself. You know, let's actually just talk in terms of the subspace topology on the set from now on to avoid having to make a distinction between the two "cases".
Well, can we recover some notion of the limit point after it's been removed? A trick you may be familiar with to find limits is to take intersections -- we could just say that if this is true for all proper filters $\phi$, we have a compact set:
$$\bigcap_{N\in\phi}N\ne\varnothing $$
Right?
Not right. This intersection will actually every often be empty -- it only measures if the net has a
constant subnet (check that this is true). What we need to do is to consider the
closures of each set in the filter, so that the limit points, if any, are actually included -- and since the closures are members of the filter too (being supersets), we can just consider the
intersection of all closed members of the filter.
Note how we cannot just say something about the closure differing from the sequence -- do you see why?
So we can write e.g.
$$\forall \phi\ne \pi(S), \bigcap_{\mathrm{closed}\ N\in\phi}N\ne\varnothing $$
Now while this is a perfectly good condition on its own, we see that it contains some redundant sets -- we're taking an intersection, so this kind of naturally "subsumes" the filter's defining properties of closure under finite intersections and supersets. We can just get rid of these sets and consider a bunch of sets that generate the filter -- this is called a
filter sub-basis. Any set of sets generates a filter, but we specifically want to generate a proper filter -- this requires that every finite intersection of sets in the filter sub-basis is non-empty -- this is called the
finite intersection property (FIP).
So we have our definition of a compact set:
any family of closed sets satisfying the finite intersection property has a nonempty intersection.
Or we could state in terms of the contrapositive:
any family of closed sets with empty intersection has a finite subfamily with empty intersection.
And then we could take the open-closed dual of the statement:
any open cover of $S$ has a finite subcover.
The last one is the most common formulation of compactness.
What the Bolzano-Weierstrass theorem really gives us is an interpretation of compact sets as "kinda" like finite sets -- not in terms of the number of points, but in terms of the amount of "space". You have only a finite amount of "space" to move around in, you can't just
escape to infinity, where
infinity is a point not in the set. The Bolzano-Weirstrass theorem is really a "generalisation of the infinite pigenhole principle" (which is precisely the statement of having "limited space"), and in fact yields the latter for the discrete topology where the only convergent sequences are the constant ones.
So in contexts where we want to "generalise" facts about finite sets, like in Lie theory, it's natural that compact sets are of importance. To really drive the point home, consider the following reformulation of the open cover definition:
For a family of sets, a compact union of some of these sets has any property that every finite union of them does.
Or further, defining an
extensive property as a property of open sets such that $P(S)\land P(T)\implies P(S\cup T)$, a
locally true property as a property satisfied by some neighbourhood of every point in a space and a
globally true property as a property satisfied by the universal set:
Every locally true extensive property is globally true iff the space is compact.
This formulation of the compactness leads to several well-known results about compact sets, including:
A continuous real-valued function on a compact set is bounded.
More on the local/global correspondence at
Terry Tao, "Compactness and Compactification".
So here's a list of things we've seen are equivalent to compactness:
- $S$ is a compact set.
- Every locally true extensive property on $S$ is globally true.
- If $S$ is a union of sets, it has any property that all finite unions of them have.
- All open covers of $S$ have a finite subcover.
- All families of closed sets with the FIP in $S$ have a nonempty intersection.
- All ultrafilters on $S$ converge.
- All filters on $S$ have a convergent refinement.
- All convergent nets in $S$ have a convergent subnet.
Compactification
The obvious question is how we can make a space compact by adding in some points, like we can in the "subset without its entire boundary" example (just take the closure). In this sense, compactification is a "generalisation" of closure, where instead of just adding points that already exist in the space, we add completely new points: points we call infinity. Making the correspondence precise, we are asking for an embedding of the space such that the closure of the embedded set is the "compactification" (the minimum such embedding is the compactification).
Obviously, we have some freedom as to how to make this compactification -- as to how we can "glue the loose ends" of the space. For example, we could assign a single point $\infty$ to both sequences increasing without bound and those decreasing without bound (compactifying $\mathbb{R}$ into a circle), or we could define $+\infty$ and $-\infty$ differently, compactifying it into a interval.
There are still questions on how we ensure, in general, that e.g. $(n)$ and $(n^2)$ both "converge" to the same point.
Are they necessarily the same point?
Suppose we add a point, call it $\infty$ to represent the limit of the sequence $(n)$. What does it mean to say that $(n)$ converges to $\infty$? It means that $(n)$ is eventually in every neighbourhood of $\infty$ -- every net corresponds to a filter, and in this case we have the neighbourhood filter of $\infty$ -- the set of all sets containing a cofinite number of positive integers. Similarly, $(n^2)$ generates the filter of all sets containing a cofinite number of squares.
These are different points. Could we
make them the same point? For this, we'd need both $(a_n)$ and $(b_n)$ to be eventually in every neighbourhood of $\infty$ -- this corresponds to the filter of sets containing all but a finite number of points in each sequence. So if we wanted two infinities, $+\infty$ and $-\infty$, the neighbourhoods of $+\infty$ would be the filter of sets containing all but a finite number of points of any sequence diverging to $+\infty$ -- equivalently:
$$\begin{array}{l}N( + \infty ) = \{ S\mid \exists r,\forall x > r,x \in S\} \\N( - \infty ) = \{ S\mid \exists r,\forall x < r,x \in S\} \end{array}$$
But if we just wanted a single point at infinity,
$$N(\infty)=\{S\mid\exists r,\forall |x|>r, x\in S\}$$
How would this
one-point compactification work in general? We'd like to assign $\infty$ as the limit in $X\cup\{\infty\}$ of every net that does not have a limit point in $X$ (this will automatically also create a limit point at $\infty$ for nets that diverge but also have a limit point in $X$, like $n\sin(n)$) -- i.e. we want a filter that every divergent net is eventually in. Well, the point of the Bolzano-Weierstrass theorem is that a divergent net
eventually escapes every compact set, i.e. it is eventually in every cocompact set. So in general, we have a neighbourhood
basis for the point at infinity:
$$B(\infty)=\{S\cup\{\infty\}\mid S\subseteq\Phi_X\land \mathrm{compact}\ S'\}$$
Or in terms of open sets,
$$\Phi_{X\cup\{\infty\}}=\Phi_X\cup\{S\cup\{\infty\}\mid\mathrm{compact}\ S'\}$$
This is called the
one-point compactification or the
Alexandroff compactification of $X$.
But the one with two infinities was cool too. The idea was that the two infinities at the far ends of $\mathbb{R}$ are somehow "disconnected" -- this is in contrast to e.g. $\mathbb{R}^{n>1}$, where the "limits" of any divergent sequence
must be connected by a giant loop at infinity. If we want each infinity to be a connected component of its own, what we need is a filter base of connected sets converging to a point at infinity.
What do I mean? In general, all compactifications are obtained by considering some
partitions by non-compact sets of co-compact sets -- in the case of the one-point compactification, these subsets are the trivial ones. In the case of the "plus infinity, minus infinity" compactification, the partition is the set of
connected components.
 |
| Illustration of why we need a partition of the co-compact sets -- the above is not a partition, and there are sequences escaping every one of the four filters drawn. This is actually an illustration of why the end-compactification of the plane has only one infinity. |
OK, so how do we actually make this construction? The thing is that we have co-compact sets like $(-\infty,-2)\cup(-1,1)\cup(1,\infty)$ -- but $(-1, 1)$ is not really a connected component we care about. We care about the
connected components of a co-compact set that are a part of a chain that every co-compact set has connected components in. One way to do this is as follows:
- Construct an infinite nested basis $U_1\supseteq U_2\supseteq \dots$ of co-compact sets -- i.e. so that every co-compact set contains one of these sets.
- Each chain $N_1\supseteq N_2\supseteq\dots$ of connected components is a neighbourhood basis for a point at infinity, called an end.
This is the
end compactification.
There are just three questions:
- Is this independent of the co-compact set basis we use?
- Does such a basis always exist?
- Is the result always compact (this includes questions you may have about the existence of connected component chains, etc.)?
To prove the first, it suffices to show that for any two co-compact bases $U$ and $V$, if a set contains some $N_i$, a set in some connected component chain of $U$, it contains an $M_j$, a set in some connected component chain of $U$ (this creates a natural bijection between the ends arising from $U$ and those from $V$). Because connected components are a partition, it suffices to show that for any two co-compact bases $U$ and $V$, if a set contains some $U_i$, it must contain a $V_j$ -- which is clearly true. So:
TRUE.
For the second -- this is equivalent to asking the dual question: is there a chain of compact sets such that every compact set is contained in a set in the chain? A simpler, equivalent way to phrase the question is: is there a chain of compact sets whose interiors cover $X$? This is called
exhaustion by compact sets, or
hemicompactness -- and as you may have guessed,
not all spaces have it. So:
FALSE.
TBC: why 3 is false, general version, Gromov boundary, Stone-Cech, why Hausdorff compactifications are preferred, add labels
A stupid joke -- the seven Cs: closed, compact, connected, convergent, continuous, covers, co-. Covers and closed are actually repeated though aren't they?
