Gracious Living

Limit Point and Sequential Compactness
December 30, 2010, 02:51
Filed under: Math, Topology | Tags: , , ,

Back to topology.  The interesting thing about compactness, as I see it, is that its definition isn’t very intuitive.  We want to talk about what are basically “closed and bounded” sets without really using closedness, which doesn’t behave well with subspaces, or boundedness, which doesn’t behave well with anything.  At the time this idea came about in the early part of this century, the mathematicians ssewho invented it (Borel, Weierstrass, and Bolzano, mainly, I think) messed around with a couple other definitions but they turned out to rely on intuition from Euclidean space, and so not be nearly as useful.  In this post, we’ll study those other definitions.

I debated with myself for a long time about what to do first: this, the separation axioms (which measure how good the topology is at distinguishing disjoint sets), or the countability axioms (which measure, well, how countable the topology is).  In a sense, each one depends on the other two for examples and theorems, and in this case, though it’s easy to show that all the properties I’m listing are different, we’ll need to refer to the other properties to discover when they are the same.  So I expect this post to be brief; after we have all the other properties defined, it’ll be easier to talk about what implies what.

So first recall what we did last time.

• A topological space is compact if every open cover has a finite subcover.  A subset of a topological space is compact if it’s compact as a subspace (equivalently, if every cover of some superset of the set by open sets in the larger space has a finite subcover).

The Heine-Borel theorem showed us that sets in Euclidean space are compact iff they’re closed and bounded.

Now, one other way of thinking about boundedness is that sets can’t “diverge into infinity.”  While topology doesn’t give us the language to talk about infinity, it does let us talk about divergence, in terms of limits!  So the next definition we could formulate is

• A topological space is limit point compact if every infinite set has a limit point.

Because closed sets are sets that include all their limit points, this sort of encompasses “closedness” as well.  Oh, what if we said “every finite set has a limit point” instead?  Well, this reduces to “every one-point set has a limit point,” which means that the intersection of all the neighborhoods of each point includes at least one other point.  We call such points topologically indistinguishable, because every open set containing one also contains the other.  This is a Bad Thing for most topologies (among other things, it means that points aren’t closed), and there’s also a fairly standard way of getting rid of it that we’ll see later.  So really, we’re concerned with infinite sets.

Okay, so let’s take a compact space and show that it’s also limit point compact.  Limit point compactness is equivalent to saying that the only sets without limit points are the finite ones, so let’s prove this.  Let $A$ be a set without a limit point in a compact space $X$; then there are neighborhoods of each point of $A$, say $U_x$ for $x\in A$, that don’t intersect any other points of $A$.  And $A$ contains all its limit points (since there are no limit points, there are none that it doesn’t contain), so $A$ is closed, and $X-A$ is open.  Then $\{X-A\}\cup\{U_x:x\in A\}$ is an open cover of $X$, and every point of $A$ is covered by only one of these sets.  So there’s no way to shrink this to a strictly smaller subcover, except possibly if $U_x=X-A$ for some $x$; thus, the cover must be finite already.  But then $A$ is finite, as desired.

On the other hand, it’s almost trivial to build limit point compact spaces that aren’t compact.  Let $X$ be a non-compact set and look at $X\times \{0,1\}$, where $\{0,1\}$ has the indiscrete topology (that is, $\{0,1\}$ and $\emptyset$ are open, but neither of the singletons is).  The open sets here are just those of the form $U\times\{0,1\}$, where $U$ is open in $X$, so obviously this new space is still not compact.  But every infinite set, indeed, every point has a limit point: the points $(x,0)$ and $(x,1)$ are topologically indistinguishable and thus limit points of each other.

Likewise, if you look at the first uncountable ordinal $\omega_1$ in the order topology, it’s not compact because it has no largest element.  But all countable subsets of it have limit ordinals as upper bounds, which are then limit points for those sets.  And any infinite set has a countable subset, so it’s got a limit point.

A little thinking shows the problem.  We want to sort of cap off our infinite sets “at the ends,” but limit point compactness only ensures they have a limit point, which could be off to the side (as in the first case) or in the middle (as in the second).  If we restrict our attention to sequences, we get another possible definition.  We don’t want to require that all sequences converge, because it’s easy for sequences to diverge even in small sets just by going in “two different directions.”

• A space is sequentially compact if every sequence in it has a convergent subsequence.

In $\mathbb{R}$, this would include bounded divergent sequences like $2.7, 3.1, 2.71, 3.14, 2.718, 3.141\dotsc$ — converging to several places at once.  But it would not include sequences like $1, 2, 3, 4,\dotsc$ that diverge because they’re unbounded.

Now, as I’ve said several times (and as I’ll resolve probably next post), sequences just aren’t good enough to detect all topological properties, since they’re only countable.  Thus, for example, $\omega_1$ is sequentially compact because given any sequence that doesn’t hit the same point infinitely many times, we can pick out a strictly increasing subsequence (well, that or a strictly decreasing one, which is impossible by the axiom of regularity), and then the limit of this is just its limit as an ordinal, that is to say, its union.  This is in $\omega_1$ because all ordinals less than $\omega_1$ are countable, and so the union of a countable set of them is also countable.  But as we’ve said before, $\omega_1$ is not compact.

Moving in a perpendicular direction, we can get a compact space that’s not sequentially compact.  With a little bit more machinery, we’ll be able to show that any product of compact spaces is compact.  (It’s actually pretty easy to show for finite products — try first showing, as Munkres does, that when $Y$ is compact, any open set around $\{x\}\times Y$ in $X\times Y$ contains $U\times Y$ for some neighborhood $U$ of $x$ in $X$.)  In particular, $[0,1]^A$ is compact for any set $A$, including $A=[0,1]$.  Recall this set $I^I$ can be viewed as the set of functions (continuous or not) from $I$ to itself, and a neighborhood of a function $f$ is the set of functions $g$ for which $g(x)$ is close to $f(x)$ for finitely many $x$ (and uncontrolled elsewhere).  So $(f_i)$ converges to $f$ if $(f_i(x))$ converges to $f$ for each $x$.

The sequence we’ll construct is quite artful: let $f_i(x)$ be the $i$th digit in the binary expansion of $x$.  (So if $x=1/3=0.0101\dotsc$, then $f_i(x)=0$ if $i$ is odd and $1$ if it’s even.  End terminating “bimals” in $\dotsc 000\dotsc$ rather than $\dotsc 111\dotsc$ to prevent ambiguity).  Suppose this does have a convergent subsequence $f_{i_k}$; then this subsequence converges when evaluated at any point.  But we can define a number $y$ by saying the $i_k$th digit of $y$ is zero if $k$ is odd and one if $k$ is even.  Then $f_{i_k}$ never converges; it just flips back and forth between $0$ and $1$.  So $I^I$ is not sequentially compact.

Oh, the example of $X\times\{0,1\}$ is also not sequentially compact, though it is limit point compact.  On the other hand, any sequentially compact space is of course limit point compact, since any infinite set contains a sequence, and then we can pick out a convergent subsequence, which converges to a point that is then a limit point for the set.

I think that’s enough for now… there are quite a few properties derived from compactness by changing a “finite” to “countable” somewhere, but these might be more at home in their own post.  I’m falling in love with Steen and Seebach’s Counterexamples in Topology for these very point-set sort of posts… it has something like 200 different spaces, each demonstrating that one property doesn’t imply another, or something like that.  I might feature certain counterexamples out of it, like I did $X\times\{0,1\}$, $\omega_1$, and $I^I$ today.  Those are all pretty standard, but some of the ones Steen and Seebach have are more interesting.  I don’t think they show anything too deep about topology, but they’re fun little puzzles, pretty much.  Maybe what I’ll do is short posts on one or two properties, with accompanying counterexamples. 