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In the above proof the following equivalence is used at the last step:

A space $M$ is connected if and only if the only open and closed subsets of $M$ are $\emptyset$ and $M$ (ie. there are no proper open and closed subsets).

However I do not understand the claim that $X$ is the completment of the open union and feel like it is non-trivial. Can someone please elaborate?


1 Answer 1


The argument shows that that every path-connected component of your space is open. Since the space is the disjoint union of its path connected components, this implies that the complement of each path-connected component is a union of path-connected components —as each of these is open, the union is also open.

  • $\begingroup$ Is the statement that the space is the disjoint union of its path-connected components deemed trivial? I see this intuitively but this is essentially the question I'm trying to address. $\endgroup$
    – suarko
    Jan 2, 2015 at 17:28
  • $\begingroup$ Every point is in some path-connected component: namely, its own path connected component. $\endgroup$ Jan 2, 2015 at 17:29
  • $\begingroup$ On the other hand, two path connected components which share a point are equal: this is a trivial exercise. $\endgroup$ Jan 2, 2015 at 17:30
  • $\begingroup$ This clears it up, thank you. $\endgroup$
    – suarko
    Jan 2, 2015 at 17:31
  • 1
    $\begingroup$ (A sensible way to put this is to define a relation $R$ between points on your space, so that $xRy$ iff there is a path joining $x$ to $y$. Then you can easily show that $R$ is an equivalence relation and that the equivalent classes of $R$ are precisely the path-connected components of the space) $\endgroup$ Jan 2, 2015 at 17:31

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