If $X$ has more than one components then $X$ is disconnected.

Question

I am reading chapter 4 of “Introduction to Topological Manifolds” by Lee. The book says that a space $X$ is said to be disconnected if there are two open, disjoint, non-empty subsets $U, V$ of $X$ such that $X=U\cup V$. It also defines component to be a maximal connected set. And proves that a component is always closed. And remarks that a component need not be open. For example, in the set of rational numbers with subspace topology, components are singleton which are closed but not open.

This brings me to the following question. The intuition suggests that one way of defining a disconnected space should be to say that $X$ is disconnected if it has more than one component.

Now, my question is pretty simple. Is this correct? Why am I asking this? Well, suppose I have a space $X$ which have more than one component but finitely many. Then I can say that all components has to be open as well. And I can construct $U,V$ as in the original definition, hence the space is disconnected. But, suppose $X$ has infinitely many components (think of rational numbers). Now each component is closed. I do not see a way to construct $U, V$ as in the first definition. Even in a simple example like $Q$, I can give the same topology as the subspace topology using a metric without any reference to the real line. In that case how would we prove that $\mathbb{Q}$ is disconnected?

Answer 1

You're right that if a space has finitely many components then the components are both open and closed: closed as they always are closed and open, because each complement of a component is a finite union of the other components (so also closed).

$\Bbb Q$ has infinitely many components (all singletons, so it is totally disconnected). It's trivial that then if $X$ has more than one component it is disconnected: if it were connected, it would have been the single component (nothing is bigger than $X$ in this space, and a component is by definition a maximal connected subset).

This highlights the problem with your "definition" of connectedness: it's circular: you define a space to be disconnected if it has more than one "component" but to define component you first need to know what "connected" is (for it is a maximal connected subset) and you're back to square one.

The usual definition avoids this: it's just defined in terms of open sets or closed sets:

$X$ is disconnected iff we can write $X$ as the disjoint union of two non-empty open sets if we an write $X$ as the disjoint union of two non-empty closed sets or iff we can write it as the disjoint union of two non-empty separated sets. (these are all clearly equivalent). $X$ is connected iff it's not "disconnected" in the previous sense.

So we define it by its negation, which is IMHO quite intuitive: being disconnected is "easier to visualise" and often easier to prove as well. You can formulate the $X$ is connected in a more positive way:

$X$ is connected when $X=U \cup V$ with $U,V$ non-empty and open implies $U \cap V \neq \emptyset$.

Components are more of a secondary notion: if a space is disconnected it allows us to talk about "how disconnected" it is. It's not the primary way to think about connectedness, I think.