# Recovering a topological space from its connected components

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 $\pi_0(X)$.

• $\pi_0(X)$ 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. Commented Nov 10, 2016 at 16:33
• It seems strange to me that $X$ is not always the union of its connected components since the connected components form abpartition of $X$ Commented Nov 10, 2016 at 16:36
• @ThomasAndrews, what do you consider to be the connected components of $\mathbb Q$? Commented Nov 10, 2016 at 16:37
• 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 Commented Nov 10, 2016 at 16:39
• @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). Commented Nov 10, 2016 at 16:47

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 $Y$.