If $X$ is a topological space, one can naturally view the set $\pi_0(X)$ of path-components of $X$ as a quotient space of $X$ by collapsing each path-component to a point by a quotient map $q:X\to \pi_0(X)$. Of course, $q$ is a homeomorphism if and only if $X$ is totally path-disconnected. I'd like to know of criteria that ensure $\pi_0(X)$ is totally path-disconnected.

Interestingly, there are lots of spaces $X$ for which $\pi_0(X)$ is not totally path-disconnected. For instance, if $X=[0,1]\times [0,1]$ has the lexicographical ordering and is given the resulting order topology, then $q:X\to \pi_0(X)\cong [0,1]$ is just the projection onto the first coordinate. In fact, a little known gem due to Douglas Harris is that for every space $Y$, one can construct a paracompact Hausdorff space $S(Y)$ such that $\pi_0(S(Y))\cong Y$.

D. Harris, Every space is a path-component space, Pacific J. Math. 91 no. 1 (1980) 95-104.

However, the ordered square and spaces $S(Y)$ are not metrizable...

One must also be wary of separation axioms since if $X$ is the closed topologists sine curve, $\pi_0(X)$ is the two-point Sierpinski space, which is not $T_1$ and therefore not totally path-disconnected.

Question: If $X$ is a metric space and $\pi_0(X)$ is $T_1$, must $\pi_0(X)$ be totally path-disconnected?

Note: I'm interested in other variations of the question too where $X$ is separable or perhaps Polish and/or $\pi_0(X)$ is Hausdorff.

  • $\begingroup$ That's an incredible result of Harris, it raises all sorts of questions. Does repeated application of π_0 stabilise at some point? Is it a monad? $\endgroup$ – David Roberts Jan 26 '18 at 5:13
  • 1
    $\begingroup$ Does $\pi_0$ stabilize at some finite stage? No. One can take a disjoint union of repeated applications $S^n(Y)$ of Harris's construction. However, it does stabilize at some transfinite stage by cardinal restrictions. Although $q:X\to\pi_0(X)$ is natural, I don't think there is a natural map $\pi_0(\pi_0(X))\to \pi_0(X)$ so I doubt it's a monad. For instance, when $X$ is the ordered square, there is no natural choice. $\endgroup$ – Jeremy Brazas Jan 26 '18 at 13:39

There exists the following metric counterpart of the Harris result:

Theorem (Banakh, Vovk, Wojcik): Each metric space $X$ can be (canonically) identified with the space $\pi_0(\circledast(X))$ of path-components of some complete metric space $\circledast(X)$.

The space $\circledast(X)$ is a closed subspace of the complete oriented graph with vertices in the set $X$.

The above theorem follows from Theorem 7.3 and Proposition 7.4 of this paper published in Fund. Math. 212 (2011), 145-173.

More generally, the Theorem holds for weakly first-countable spaces containing no countable connected subspaces. To formulate the general and more precise version of the Theorem, I need to recall some definition and results from this paper of Banakh, Vovk and Wojcik, which will be cited as [BVW]. So, I thank Jeremy Brazas for opportunity to advertise this my paper [BVW] :)

Unfortunately, writing the paper [BVW] we did not know about the paper of Harris. After looking at the Harris' paper, I discovered that his construction of the space $S(X)$ (with $\pi_0(S(X))\cong X$) is similar to our construction of $\circledast(X)$ with $\pi_0(\circledast(X))\cong X$. But our space $\circledast(X)$ always is complete metric. As a result, our construction works only for premetric (= weakly first-countable) spaces.

Let us recall that a topological space $X$ is weakly first-countable if to each point $x\in X$ one can assign a decreasing sequence $(B_n(x))_{n\in\mathbb N}$ of subsets containing $x$ such that a subset $U\subset X$ is open if and only if for each point $x\in U$ there exists $n\in\mathbb N$ such that $B_n(x)\subset U$.

By Proposition 3.7 in [BVW], a topological space $X$ is weakly first-countable if and only if the topology of $X$ is generated by a premetric $p$.

A premetric on a set $X$ is any function $p:X\times X\to[0,\infty)$ such that $p(x,x)=0$ for all $x\in X$. The topology generated by a premetric $p$ on $X$ consists of all sets $U\subset X$ such that for every $x\in X$ there exists $\varepsilon>0$ such that the $\varepsilon$-ball $B(x;\varepsilon):=\{y\in X:p(x,y)<\varepsilon\}$ is contained in $U$.

A premetric space is a pair $(X,p)$ consisting of a set $X$ and a premetric $p$.

To each premetric space $X$ we shall assign some special complete metric space $\circledast(X)$, called the cobweb of the premetric space $X$.

The cobweb space is a closed subset of the complete oriented graph $\Gamma X$ over the set $X$. The complete oriented graph $\Gamma X$ is the set $X\cup\{(x,y,t)\in X\times X\times(0,1):x\ne y\}$ endowed with the path metric $d$ (in which every oriented edge $[x,y]=\{x,y\}\cup\{(x,y,t):0<t<1\}$ has length 1.

For a premetric space $X$ endowed with a premetric $p$ the cobweb $\circledast(X)$ is defined as the closed subset $$\circledast(X):=X\cup\{(x,y,t)\in\Gamma(X):t\le 1-p(y,x)\}$$of the graph $\Gamma X$.

The compression map $\pi_X:\circledast(X)\to X$ assigns to each point $a\in \circledast(X)$ the point $a$ if $a\in X$ and the point $x$ if $a=(x,y,t)$ for some $x,y\in X$ and $t\in(0,1)$.

The following theorem follows from Theorem 7.3 and Proposition 7.4 of [BVW].

General Theorem (Banakh, Vovk, Wojcik) Let $X$ be a premetric space. Then:

  • The cobweb space $\circledast(X)$ is complete metric space of dimension $\le 1$.
  • The compression map $\pi_X:\circledast(X)\to X$ is a continuous quotient surjection with path-connected fibers $\pi_X^{-1}(x)$, $x\in X$.
  • The space $\circledast(X)$ is connected if and only if $X$ is connected.
  • If $X$ contains no countable connected subspaces, then the fibers of the compression map $\pi_X$ coincide with the path-components of $\circledast(X)$ and also with the separablewise components of $\circledast(X)$;

We recall that the separablewise component of a point $x$ in a topological space $X$ is the union of all separable connected subspaces of $X$ that contain $x$. It is clear that the path-component of $x$ is contained in the separablewise component of $X$.

Since the topology of a weakly first-countable space is generated by a premetric, the General Theorem implies the following corollary answering the question of Jeremy Brazas in (strong) negative.

Corollary 1. Each weakly first-countable space is homeomorphic to the space $\pi_0(M)$ of path-components of some complete metric space $M$.

Taking for $X$ any connected Hausdorff space, we obtaine the following surprising

Example. There exists a connected Polish space $P$ such that $P$ has countably many path-connected components and each path-connected component is closed in $P$.

It seems that for separable space $X$ the space $\pi_0(X)$ also can be homeomorphic to $[0,1]$. A possible counteexample can look as follows:

Let $$C=\big\{\sum_{i=1}^\infty \frac{x_i}{3^i}:(x_i)_{i\in\mathbb N}\in\{0,2\}^{\mathbb N}\big\}$$ be the standard Cantor set in the unit interval $I:=[0,1]$.

Let $\mathcal J$ be the set of connected components of the complement $I\setminus C$. It is clear that each set $J\in\mathcal J$ is an open interval of lentgth $\frac1{3^k}$ for some $k\ge 1$.

For $n\in\{0,1\}$ let $\mathcal C_n$ be the subset of $\mathcal C$ consisting of intervals of length $\frac1{3^{2i+n}}$ for some $i\ge 0$. Let $J_n=\bigcup\mathcal J_n$.

Now consider the following compact subset $$X:=(C\times[0,1])\cup(\{0\}\times J_0)\cup(\{1\}\times J_1).$$ I hope that the space $\pi_0(X)$ of path-components of $X$ is homeomorphic to $[0,1]$.

  • $\begingroup$ @JeremyBrazas You are welcome. Thank you for the nice question. $\endgroup$ – Taras Banakh Jan 27 '18 at 0:57
  • $\begingroup$ I do still wonder about separability - this is actually an important case to me and would be a nice question. $\endgroup$ – Jeremy Brazas Jan 27 '18 at 1:28
  • $\begingroup$ @JeremyBrazas I added a possible compact metrizable counterexample to my answer (at the very end). $\endgroup$ – Taras Banakh Jan 27 '18 at 1:55
  • $\begingroup$ Yes, your last example works by the Bing Shrinking Criterion. Thank you again! $\endgroup$ – Jeremy Brazas Jan 27 '18 at 2:39
  • $\begingroup$ Combining your last example with some of my previous work, I was able to answer some questions I had about path component spaces and fundamental group/shape group relationships. I am finishing the project now. Please let me know if you did not receive my previous email messages to you regarding this. $\endgroup$ – Jeremy Brazas Feb 22 '18 at 19:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.