## What Topological Spaces are π₀?

This was a fun question I thought about once.  My answer is at the end, in case you’d like to try solving the problem yourself.  The question is likely more interesting than my solution.

A well known theorem says that every group occurs as for some topological space .  It’s not a hard construction; is a CW-complex with a 1-cell for each generator and a 2-cell for each relation.  You can generalize this to higher if you want (although you only get abelian groups).  Sadly, little attention is paid to lower .

is in many ways the easiest homotopy “group” to understand.  Following the general definition, is the set of homotopy classes of maps from into is 2 disconnected points .  If we send to a fixed base point (which one we choose won’t matter), then there is a homotopy between iff there is a path from to ; thus is the set of path components of (this is the definition I usually use).  has no group structure, and is usually just understood as a set.  Obviously every set occurs as the set of path components of a topological space; namely a discreet space, so this appears trivial.

However, has a bit more structure than just a set.  Path componenthood is an equivalence relation on a topological space, so the set of path components is a quotient of a topological space, and such sets come with a quotient topology (this is the finest topology so that the quotient map is continuous).  So, really, is a topological space.  For nice spaces (e.g. CW-complexes), is always a discreet space.  However, some not-too-pathological examples, can have some interesting topology.

Examples: Topologist’s sin curve: Let be the closure of the graph of , so .  We want to compute .

• has 3 path components, corresponding to , , and .  I will call them , respectively.
• From the definition of the quotient topology, a set is open iff the union of it’s members is open.  So, for instance:
• is open, since is an open subset of
• is not open, since any open set containing a point in also contains points in
• The topology is completely described by saying and are the only open points

Long line: Let be the long ray with a point at infinity, that is endowed with the order topology (using lexicographic order).

• has 2 path components:
• For any two finite points and , the interval is isomorphic to for some countable ordinal .  An easy transfinite induction shows is isomorphic to , and that isomorphism defines a path from to .
• On the other hand, no path connects the point at infinity to any finite point; if such a path existed, it would send to a dense subset of .  In particular, such a set is cofinal (has a member bigger than any fixed element), but has uncountable cofinality (cofinality is the smallest cardinality of a cofinal set)
• The set of finite points is open in , so this corresponds to an open point in is not open in , so it’s not open in .  The resulting space is sirpinski space.

If avoiding spoilers, read no further

Theorem: Every topological space occurs as , for some other topological space .

As an informative special case, let’s build so is the trivial topology on ; that is, the only open sets are and .

First, fix a well ordering of , and let be the corresponding ordinal (or any to have big enough cofinality will do).  We will give the topology where only sets of the form are open.  This is equivalent to taking intervals to be basic open sets (i.e. upwards closed sets are open), as every open set must have a minimal element.  The following lemma might give some intuition for this topology.

Lemma: With this topology, is path connected: proof: We need to understand when a function is continuous.  By definition, is continuous if is open for every .  This means is open in .  Notice is uniquely determined by the sequence , as is the minimum so , so a path in this space is easy to picture as a length< shrinking sequence of open sets. To find a path from to , with , we can use the sequence of open sets where

that is to say, and otherwise.

Each path component of will be homeomorphic to a copy of with this topology; we will call the components and , and .  Open sets in are defined by

1. is open
2. If and are non-empty open subsets of and , respectively, then is open in .

It not hard to see this defines a topology, and and are homeomorphic to in the subspace topology.  This means and are both path connected, so has at most 2 path components.  It remains to show:

1. and are distinct path components.  That is, there is no path from a point in to a point in .
2. Neither nor are open in .

For 1, we will make use of the largeness of .  Suppose is a path with and .  Let and .  Clearly ; we will get a contradiction by showing and are both open.

Lemma: is bounded in . proof: Since , .  However, has cofinality bigger than by König’s theorem, so any size continuum set is bounded in

Let be an upper bound for .  Let open in , and since , .  As is continuous, this shows is open.  A symmetric argument shows is open, thus and are both clopen, and [0,1] is disconnected.

2 is more straighforward; any open set in containing also contains points from , thus any open set in containing also contains .  This completes our special case.

For the general case, fix a topological space , and we will build a space such that .  Let be a disjoint copy of for each .  The space will be .  We define a  topology on by:

• For every which is open and family of non-empty open sets, the set is open in

Notice our first example was a special case of this; the only possibilities for were and .  As before, this does define a topology, and each is a path connected subspace homeomorphic to .

To see each is a distinct path component, suppose is a path with and for some .  Let .  We’ll show is clopen in , but clearly and , so this will be a contradiction.

As before, we know for any , is bounded in .  Let .  To see is closed, let .  Since is open in , is an open set in , so is open.  To see is open, let is also open, and , so is open.  Thus must be clopen, so such a path can’t exist.

Now we know these path components are distinct, so , so as a set is .  To see the map is a homeomorphism,  we need to show is an open set in iff is an open set in ; this follows immediately from the definition.

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1 year ago

Nice post. In the first two examples you give, the spaces are Hausdorff, while your general construction only gives non-Hausdorff spaces. I think one can get any finite space occurring as of a closed subspace of using the following construction. First, consider as a directed graph with vertices given by the points of and edges given as follows: for any points , attach an edge from to if lies in the closure of . Note that this graph determines the space. Since is a finite graph, it embeds into . Replace every vertex with a small closed interval, and replace every edge from to with a compact topologist’s sine curve, attaching the boundary point to the source of the edge and the boundary line to a small subinterval of the target of the edge. The resulting closed subspace of then satisfies . (This is essentially a generalization of your first example, in which is the Sierpinski space.) By gluing arbitrary path-connected CW complexes to the small intervals, one can find such with arbitrary homotopy groups for the path components, answering Grisha’s question when is finite.

So anyway, it seems that every finite space is of a Hausdorff (even normal) space. The above construction probably also applies to countable posets with an appropriate local finiteness condition. Do you know whether in fact every space is for some Hausdorff (or normal) ?

1 year ago

Beautiful construction for . To avoid the extra step of identifying points, one could also let .

Rather than make a new construction for , I claim that for arbitrary spaces , we have via the natural map from left to right. The fact that the map is a bijection is straightforward. To show that it is continuous, it suffices to show that each map is continuous, and using the universal property of quotient maps, this follows from the fact that is continuous. It remains to show that the map is open; for this, it suffices to note that the projection maps from a product are open, and that under the projection a path component of is sent onto a path component of .

So in particular, once one has the construction for , one obtains Hausdorff with for any cardinal . One also obtains Hausdorff with any subspace of by taking the preimage under , so for example all normal spaces occur as for Hausdorff (Urysohn). In particular, one gets all CW complexes. Note that this construction will produce regular but not necessarily normal , since infinite products of normal spaces need not be normal, and also subspaces of normal spaces need not be normal. (I forgot how delicate normality is as a condition; probably it is too much to ask that be normal, but it would be interesting to know whether one could actually *prove* that normal cannot occur for some particular space.)

In fact, I believe one can get all spaces using this idea, but there is a small gap at the end of my argument. First, note that maps from to the Sierpinski space correspond to open subsets of , so the map gives rise to an injective map , where is the largest quotient of (i.e., identify points which cannot be distinguished topologically). Since this latter map is always continuous and open onto its image, it identifies as a subspace of . If denotes with the indiscrete topology, then the map is a homeomorphism onto its image (where is the identity). Thus if and are Hausdorff spaces satisfying and then one can obtain a subspace of satisfying . So, using the long line or the topologist’s sine curve for , we have reduced, amusingly, to showing that any indiscrete space may be obtained as of a Hausdorff space. (Even without this, the above proves it for any space.)

I think I have a proof of this last statement, but there are some details I haven’t checked and this comment is already very long. If you see a nice construction, I would be interested to know it.

1 year ago

Lovely. I had a similar idea to get indiscrete spaces but implemented it in a rather complicated way; your construction is very clean. Thanks for the cool problem!

1 year ago

A nice result! I wonder is there any relation between homotopy groups of different components if is not discrete? For example, do they form a sheaf over ? Or can we construct an arbitrary with arbitrary fibers?

1 year ago

I’m guessing there is not much communication between higher homotopy groups based in different path components, even if isn’t discrete. In particular, I think we can construct with arbitrary with such that the fibers of the path-connectedness projection have arbitrary (based, path-connected) weak homotopy type. As a lemma, we note that Brian’s space is weak contractible (I’m actually not sure about this; non-CW complexes are hard for me!). Then wedge-sum each fiber with a (based, path-connected) space of your choice!

Last edited 1 year ago by moomitten
1 year ago