I encountered that a topological space $X$ is connected if no separation exists. Here a separation is a pair of disjoint non-empty open sets whose union is $X$. Such a separation can only exist if $X$ contains two distinct elements, so $\emptyset$ is supposed to be connected (right?). But what about its components? Does it have $\emptyset$ as unique component or are there in this case no components at all? They should form a partition of $\emptyset$, and I was taught that elements of a partition are non-empty.

  • $\begingroup$ No separation exists in the case of the empty set because there is no pair of disjoint non empty open sets whose union is the empty set. Hence the empty set is connected $\endgroup$ – Shahab Sep 24 '13 at 10:39
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    $\begingroup$ According to the relevant Wikipedia page there's no universal convention. It's just a matter of semantics anyway. Regarding the components, I think all authors would agree the empty set has zero components - the only partition of the empty set is itself empty. $\endgroup$ – Anthony Carapetis Sep 24 '13 at 10:41
  • $\begingroup$ @AnthonyCarapetis. Especially the existence of an empty partition (as only partition of the empty set) is enlightening. Thanks. $\endgroup$ – drhab Sep 24 '13 at 11:04
  • $\begingroup$ Is $1$ prime? Take the typical definition, but without adding the explicit exception of $1$, and $1$ satisfies it. But there are reasons we exclude $1$ from being prime. Similarly, there are reasons why $\varnothing$ could be worth excluding from being connected. $\endgroup$ – Jonas Meyer Feb 6 '17 at 23:15

The set of components is $\emptyset$, i.e., there are no components. The statement "every component is non-empty" is then trivally true.


Firstly, let us suppose that the space $X$ is non-empty. The condition of being connected then becomes equivalent to any of the following: (a) if $X = Y \coprod Z$, then either $Y$ or $Z$ is canonically isomorphic to $X$, (b) if $X \xrightarrow{f} (Y \coprod Z)$ then $f$ factors through one of the coproduct injections, (c) the hom-functor $\mathbf{Top} \xrightarrow{\mathrm{hom}(X, -)} \mathbf{Set}$ preserve coproducts.

The statements are nice and should then be taken as a definition. However, when the empty space $\emptyset$ is taken into account one needs to add to the statements (a) the additional clause $X \neq \emptyset$ to make the equivalence work.

Thus, the empty space $\emptyset$ should not be connected!


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