Is this an equivalent definition for connectivity in graphs? Let $G=(V,E)$ be a simple graph. The usual definition of connectivity is:

$G$ is connected iff for each pair of vertices $u,v\in V$, there exists a path between $u$ and $v$.

I am fine with this definition, but I wanted to know if this definition is equivalent:

$G$ is connected iff there do not exist nonempty disjoint subsets $V_1,V_2$ of $V$ such that $V=V_1\cup V_2$ and there is no edge $uv\in E$ with $u\in V_1$ and $v\in V_2$.

My definition is more topologically motivated - we can think of $(V_1,V_2)$ as a separation of $G$. My intuition is that these definitions are equivalent, and I cannot see any reason why they would not be, but I do not believe I have seen the latter be used e.g. in a textbook or lecture. Is it the case that they are not equivalent, or just that the former is more intuitive?
 A: Suppose in addition that $V_1,V_2$ are non-empty and their union is the entire graph (as pointed out in the other answer, these assumptions are necessary).
To show that they are equivalent, first observe that if the "no cut" definition holds, then there can be no path from any vertex in $V_1$ to any vertex in $V_2$. Indeed, if there were such a path - call it $w_1,w_2,\ldots,w_n$ - then let $k$ be maximal such that $w_k$ is in $V_1$. It follows that $k<n$ and $w_{k+1}$ is in $V_2$, a contradiction.
Conversely, suppose the the first condition holds. Then, choose $v_1\in V_1$ and $v_2\in V_2$ arbitrarily and choose an arbitrary path from $v_1$ to $v_2$. As before, consider the maximal vertex in the path that is still in $V_1$. Then it is adjacent to a vertex of $V_2$, as desired.
A: You need to specify that both the subsets must be non-empty. Also, you need to specify that $V_1\cup V_2=V$. (These requirements both have analogous counterparts in the topological definition, so they shouldn't feel too foreign.)
Once you add these two requirements, then yes, they are equivalent.
