Doubt with proof of if $G$ is simple and $\delta\geq\frac{n-1}2$, then $G$ is connected

Question

Proposition. If $G$ is simple and $\delta\geq\frac{n-1}2$, then $G$ is connected.

Proof. Assume the contrary. Then $G$ has at least two components, say $G_1$, $G_2$. Let $v$ be any vertex of $G_1$. As $\delta\geq\frac{n-1}2$, $d(v)\geq\frac{n-1}2$. All the vertices adjacent to $v$ in $G$ must belong to $G_1$. Hence, $G_1$ contains at least $d(v)+1\geq\frac{n-1}2+1=\frac{n+1}2$ vertices.

Similarly, $G_2$ contains at least $\frac{n+1}2$ vertices. Therefore $G$ has at least $\frac{n+1}2+\frac{n+1}2=n+1$ vertices, which is a contradiction. $\square$

In the method of contradiction can we use the hypothesis? Usually we prove suppose $G$ is not connected $\implies$ it has atleast two connected components $\implies$ some consequence $1$ $\implies$ some consequence $2$...$\implies$ ... some consequence $n$, which contradict the hypothesis. But here they use the hypothesis $\delta\ge (n-1)/2$ for the proof.

Can you explain why did they use like this?

Answer 1

Yes, you can use the hypothesis in a proof by contradiction. This is actually one of the hallmarks of a "genuine" proof by contradiction, rather than a proof by contraposition.

In general, to prove $P \to Q$ by contradiction, you assume both $P$ and the negation of $Q$ and try to obtain some contradiction. This is sometimes easier than working by contraposition, where you only assume the negation of $Q$ and try to prove the negation of $P$. It can be easier because you have more hypotheses to work with.

In general, if $P$ and the negation of $Q$ are inconsistent, then whenever $P$ holds we see $Q$ has to hold as well, which means that $P$ implies $Q$.

Answer 2

The proof looks like circulus vitiosus. But it seems that $n$ denotes the number of vertices, and the assumption leads to $n+1$, which is indeed a contradiction,

since a finite graph can only have one finite number of vertices which is a natural number, and there is no natural number with $n = n+1$.

So I don't see a problem, since the proof follows this valid proof schema:

 G, ~A |- f
 ----------
   G |- A

The assumptions G need not be empty, here its the condition of the proposition.