By definition a topologic space $X$ is connected $\iff$ there are no $U,V\subseteq X$, such that both are open, disjoint and non empty.
On the other hand, by wikipedia: "The components of any topological space X form a partition of X: they are disjoint, non-empty, and their union is the whole space"
So we can say in general that a topologic space $X$ is connected if the cover can be build by connected components? It seems a bit like recursive defination
Best Answer
A topological space is connected iff we cannot write it as $X=U \cup V$ where $U \cap V = \emptyset$, both $U,V \neq \emptyset$ and $U,V$ are both open.
This definition also applies to subspaces of any space $X$, a subspace $A$ is connected iff it is a connected space (in its own right) in the subspace topology from $X$.
A space $X$ that is not connected can still have many connected subspaces $A$. E.g. $\Bbb R\setminus \{0\}$ is not connected ($U= (-\infty,0), V = (0,+\infty)$ disconnect it) but $(0,1)$ is a connected subspace of it and $(1,2)$ too. The maximal connected subspaces of $X$ ($A$ is connected and a strict superset of $A$ is not) is a component of $X$. So there is nothing circular about that.
For $x \in X$ consider $\mathcal{C}_x = \{C \subseteq X\mid C \text{ connected and } x \in C\}$ which is always nonempty as $\{x\}$ is connected in any space and contains $x$. Its union $C_x$ is connected by a standard theorem (because all members intersect in $x$) and if $C$ is connected and $C_x \subseteq C$ then by definition $C \in \mathcal{C}$ and so $C \subseteq C_x$ and so $C=C_x$ so $C_x$ is a maximal connected subspace of $X$ and so a component of $X$.
We then observe that for $x \neq y$ in $X$ the sets $C_x$ and $C_y$ (thus constructed) obey $$C_x = C_y \text{ or } C_x \cap C_y = \emptyset$$
which explains the remark in the text that components form a disjoint partition of $X$.
The components of $\Bbb R \setminus \{0\}$ are precisely the $U$ and $V$ that disconnected it (they are connected and maximally so). A space can have one component (when it is already connected) or uncountably many ones too (in the irrationals the components are the singletons).