Assumption that graph is connected to prove Ore's Theorem On Wikipedia, the Ore's Theorem says that if $G$ is a finite simple graph with $n \ge 3$ vertices and for all non-adjacent vertices $u$ and $v$, the sum of their degrees is greater than or equal to $n$, then $G$ is Hamiltonian.
On Proof Wiki, this theorem is stated almost in the same manner, but there is an additional hypothesis that $G$ must be connected.
I am wondering which of the two versions is more accurate.
It seems that this degree condition already implies that the graph is connected, therefore, we do not need to suppose that.
So, is this assumption really needed?
 A: If a simple graph of $n$ vertices is such that any non-adjacent pair of vertices $x$ and $y$ are such that $\text{deg}(x) + \text{deg}(y) \geq n$, then the graph is connected. So the connectedness condition is not need. 
To see this, suppose that the graph is not connected. Then there are vertices $x$ and $y$ in different components of sizes $\ell_x$ and $\ell_y$ ($\ell_x + \ell_y \leq n$ of course). Now $x$ and $y$ are not adjacent, and not that $\text{deg}(x) \leq \ell_x-1$ and $\text{deg}(y) \leq \ell_y - 1$. Hence,
$$
\text{deg}(x) + \text{deg}(y) \leq \ell_x + \ell_y -2 \leq n -2,
$$
which is a contradiction with Ore's condition. 
A: The assumption on degree sums implies connectivity.
To see it, let us suppose otherwise, i.e. $G$ is disconnected but every non-adjacent vertex pair has degree sum at least $n$.
Let $C_1, C_2$ be two connected components of $G$.
Let $x \in V(C_1)$ and $y \in V(C_2)$.  Clearly, $x$ and $y$ are non-adjacent.
The degree of $x$ is at most $|V(C_1)| - 1$ and that of $y$ is at most $|V(C_2)| - 1$.
Because $C_1$ and $C_2$ are vertex disjoint, $|V(C_1)| + |V(C_2)| \leq n$.
But then, $deg(x) + deg(y) \leq |V(C_1)| - 1 + |V(C_2)| - 1 < |V(C_1)| + |V(C_2)| \leq n$, contradicting our initial hypothesis.
