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Let $X$ be a topological space and denote by $\pi_0(X)$ its space of connected components i.e. the set of connected components of $X$ with the quotient topology induced by the topology of $X$. Note that $\pi_0(X)$ is discrete if and only if every connected component of $X$ is open in $X$ (this is always the case if $X$ has only a finite number of connected components but not true in general).

My question is : can $X$ be reconstructed from the set of its connected components (view as topological spaces with the induced topology) and $\pi_0(X)$ as a topological space.

Here are two example where it is possible : $\pi_0(X)$ is discrete because in this case $X$ is just the disjoint union of its connected components, $X$ is totally disconected because in this case $X$ is homemorphic to $\pi_0(X)$.

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    $\begingroup$ $\pi_0(X)$ is usually the path-connected components of $X$. It is different from the general topology notion of "connected" component. A "connected component" is, in fact, always open, but that is not true for path-connected components. $\endgroup$
    – Thomas Andrews
    Commented Nov 10, 2016 at 16:33
  • $\begingroup$ It seems strange to me that $X$ is not always the union of its connected components since the connected components form abpartition of $X$ $\endgroup$
    – marwalix
    Commented Nov 10, 2016 at 16:36
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    $\begingroup$ @ThomasAndrews, what do you consider to be the connected components of $\mathbb Q$? $\endgroup$
    – Mees de Vries
    Commented Nov 10, 2016 at 16:37
  • $\begingroup$ Just because there are not always a partition of spaces into connected components doesn't mean that it doesn't make sense to talk about the connected components of a space. @MeesdeVries $\endgroup$
    – Thomas Andrews
    Commented Nov 10, 2016 at 16:39
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    $\begingroup$ @MeesdeVries The intersection of all clopen sets containing a point is usually called the quasicomponent of that point. The connected component of a point is the maximal connected subset containing that point. Connected components are always closed, but generally not open (they are open for example in locally connected spaces). $\endgroup$
    – Daniel Fischer
    Commented Nov 10, 2016 at 16:47

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The answer to this question is no. Consider two subspaces of $\mathbb R^2$, $X, Y$ with $$ X = \left(\bigcup_{n \in \mathbb N} [0,1] \times \{\tfrac1n\}\right) \cup \{(0,0)\} $$ and $$ Y = \left(\bigcup_{n \in \mathbb N} [0,\tfrac1n] \times \{\tfrac1n\}\right) \cup \{(0,0)\}. $$ In both cases, the pathconnected components are infinitely many intervals and a single point. Also, in both cases, the topology on the space of connected components is that of $\{1,\tfrac12,\tfrac13,\ldots,0\}$. But these spaces are not homeomorphic, as there is an open set around the origin in $X$ which does not contain an entire other pathcomponent, but the same thing is not true for $Y$.

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  • $\begingroup$ Thanks a lot, that's great ! $\endgroup$
    – cannonball
    Commented Nov 10, 2016 at 16:58
  • $\begingroup$ In fact, this is a counterexample to a stronger question: even if you include in the data the bijection between the connected components (as topological spaces) and the points in the component space, you still can't recover the original space. $\endgroup$
    – tomasz
    Commented Nov 11, 2016 at 12:42

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