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)
- For any two finite points
- 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
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
is open
- 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:
and
are distinct path components. That is, there is no path from a point in
to a point in
.
- 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.
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)
?
This was a construction I tried early on in the problem, but I have yet to find a way to make get arbitrary infinite spaces as
for Hausdorff
. I don’t know of any reason it shouldn’t be possible; I just can’t find a way to do it.
In place of a topologist’s sin curve, there are a few other spaces you can use to connect points. You can connect points on your graph with long lines; it won’t embed in
anymore, but it may be easier to picture. You can also use a space like
. This example is due to my friend Sam. Of course, you’d have to embed the graph in
instead to make this work.
Some ad hoc clever arguments can get you particularly nice spaces as
of a Hausdorff space. For instance, we can get
as follows:
Playing with similar arguments, you might be able to build any CW complex, although I haven’t found a way to get
this way yet.
Another sort of ad-hoc argument I found was, if you consider the order topology on
(so like the really really long line), the path components correspond to partitions of
at ordinals of the form
, so the resulting topology is isomorphic to the order topology on
. I think a similar game can get you many order topologies, although I haven’t pursued this idea much. I should note this is likely a special case of your construction for “locally finite” topological spaces, if the assumptions you need are spelled out appropriately.
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.
This is a very nice argument. I’m glad to know there’s a completely different way to solve this problem.
Getting the indiscrete topology on
is not that hard, given the machinery you have from earlier. Consider the complete directed graph (so edges go both ways) with
as vertices, and replace the edges with long lines, with the endpoints identified with the vertices (or an analogous construction with sin curves can work). It’s not hard to see this works.
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!
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?
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!
There are some problems with taking wedge sums of <<bad>> spaces. I am not sure that the homotopy groups will behave as expected. However, I’m not sure about anything that is not CW either 🙂
I ran through the details and I’m fairly sure nothing Sam said is wrong, but I know it’s easy to wave your hands with abstract nonsense and say false things. Here are the details behind Sam’s claims.
We can let
be pretty stupid:

is continuous, we need to show that
is open for each
. We have 2 easy cases:
To check that
In fact, we didn’t use any properties of
here; it turns out
is strongly contractible by a very similar argument
A similar sort of trick shows you don’t have issues taking wedge sums with CW complexes. Let
be a space with base point
. Suppose we have
representing a homotopy class. Let
be defined by
if
and
if
.
is the composite of
with the wedge sum of the identity on
with the map contracting
to 0, so
is also continuous. I claim that
and
are homotopic by the following homotopy:

is continuous: let
be open.

is open, as
is continuous.
because if
then
, since
is the only neighborhood of
in
. Therefore, we don’t add any new points by saying
![Rendered by QuickLaTeX.com H^{-1}(U) = f^{-1}(U)\times [0,1) \cup f'^{-1}(U)\times [0,1]](https://thuses.com/wp-content/ql-cache/quicklatex.com-1dd96c9e64651921521949879f5ef681_l3.png)
is continuous. It follows that 
We can check
Clearly
This is a union of open sets, hence open, so
By the way, you have a typo in the link to Konig’s theorem
good catch; fixed now
I would like to point out that the topology on
is used when
is a spectral space, in which case
is a profinite set. See, for example, Bhatt-Scholze, The pro-étale topology for schemes, Section 2.1.