It's straightforward to prove that for a topological space, connected $\iff $ having one component.
It then follows by contra-positive that disconnected $\iff$ having more than one component.

In the case of a finite number of components of a space $\{C_i\}_{i = 1, n}$ a direct proof $X$ is disconnected is that $X$ = $C_1 \cup (\cup _{i = 2, n} C_i)$. Then with each component being closed, the finite union $\cup _{i = 2, n} C_i$ is a closed set and $X$ is the union of two non-empty disjoint closed sets $C_1$, and $ \cup _{i = 2, n} C_i$, which is a standard definition that $X$ is disconnected (and by the way also shows in the finite case that each component is both closed and open).

For the infinite case, the infinite union is not necessarily closed and this argument fails - is there a direct proof for the infinite case ?

Later clarification after many comments....

In other words, as per one of the comment below, given an infinite number of components can one construct a separation (disconnection) as one can in the finite case ?

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    $\begingroup$ I don't see how you can discuss the number of components being greater than 1 without already knowing it is disconnected. If you assume you have infinitely many components, it is already disconnected. $\endgroup$ – Randall Aug 16 at 14:48
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    $\begingroup$ A Space that is a union of infinitely many pairwise disjoint closed sets is not necessarily disconnected. Take for example $[0, 1]$, which is the union over all sets $\{x\}$, where $x \in [0, 1]$. $\endgroup$ – Dominik Aug 16 at 14:53
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    $\begingroup$ Your concept of a "direct proof" is unclear to me. For example, I would not call your proof of the finite case a "direct proof" because it uses a theorem about components, namely that each component of $X$ is a closed subset of $X$. To me it's more direct to simply apply the definition of component: if there exists a component $C \subset X$ such that $C \ne X$ then $X$ is not connected, because by definition of component the subset $C$ is a maximal connected subset. $\endgroup$ – Lee Mosher Aug 16 at 15:18
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    $\begingroup$ There are certainly interesting constructive issues: the existence of a component containing each point; the fact that the set of components is pairwise disjoint and hence partitions the space; and others. But once you admit the definition of a component, then the existence of a component which is a proper subset immediately implies $X$ is disconnected. $\endgroup$ – Lee Mosher Aug 16 at 15:26
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    $\begingroup$ If I've understood you directly, I think there's a perfectly reasonable question here, which is phrased a bit differently than you gave it: given that I know the infinitely many components of a space, can I explicitly construct a disconnection? $\endgroup$ – Kevin Carlson Aug 16 at 16:40

A "direct proof" is clear: if $X$ is connected it has one unique component (where a component is defined as a maximal connected subset: $C$ is a component of $X$ when $C$ is non-empty, connected and $C \subsetneq C' \subseteq X$ implies $C'$ is not connected, or equivalently: $C \subseteq C'\subseteq X$ and $C'$ connected implies $C'=C$). Because when $X$ is connected then $C=X$ fulfills the definition I gave trivially, and no proper subset can be a component as witnessed by $C'=X$ as well.

So if $X$ has more than one component (the number is irrelevant) we already know $X$ is not connected, as you say by contrapositive.

The other argument you gave for finitely many components is irrelevant and not needed anyway, so the infinite case is a "fake problem". The first paragraph is all you need. A contrapositive implication is no problem.

  • $\begingroup$ All true and thanks for the answer. I now have a better understanding of what I meant in my question, as in the comment from @KevinCarlson saying can you in fact construct a disconnection given an infinite number of components ? $\endgroup$ – Tom Collinge Aug 16 at 18:11

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