Question on the proof of Tutte Theorem in the West's book I have questions on the Tutte's theorem, and its proof from the West's book:
"Introduction to Graph Theory". On page 137, the theoreom is quoted:
A graph $G$ has a 1-factor if and only if $o(G-S)\leq |S|$ for every $S\subseteq V(G)$
And for the proof:
Sufficency: When we add an edge joining two components of $G-S$, the number of odd components does not increase (odd and even together become one odd component, two components of the same parity become one even components). Hence, Tutte's condition is preserved by addition of edges: if $G'=G+e$ and $S\subseteq V(G)$ then $o(G'-S)\leq o(G-S)\leq |S|$. Also, if $G'=G+e$ has no 1-factor, then $G$ has no 1-factor.
Therefore, the theorem holds unless there exists a simple graph $G$\footnote{i.e. having no loop or multiedge} such that $G$ satisfies Tutte's condition, $G$ has no 1-factor, and adding any missing edege to $G$ yields a graph with a 1-factor\footnote{Soit l'ensemble des graphes ayant la propriété: 'ne pas avoir de 1-factor et qui satisfont la condition de Tutte'. On considère ensuite le graphe maximal verifiant cette propriété}. Let $G$ be such a graph. We obtain a contradiction by showing that $G$ actually does contain a 1-factor.
Let $U$ be the set of vertices in $G$ that have degree $|G|-1$.
Case 1: $G-U$ consists of disjoint complete graphs
[...]
Case 2: $G-U$ is not a disjoint union of cliques
[...]
So my questions are:
How can we be sure that the set $U$ can exists? 
How can we say that we can split $G-U$ into two cases:
One in wich $G-U$ is a disjoinyy complete graph, and one in which it is not a disjoint union of cliques. Cliques and complete graph are not synonyms. 
Thank you
 A: First, we can be sure that the set $U$ exists because we defined it:

Let $U$ be the set of vertices in $G$ that have degree $|G|-1$.

It might be the case that $U= \varnothing$: this will happen if there are no vertices in $G$ with degree $|G|-1$. That's different from $U$ not existing. If $U = \varnothing$, then $G-U = G$, and we proceed with the proof exactly as we do in all other cases.

Second, cliques and complete graphs are synonyms: both are graphs in which all possible edges are present. Conventionally, the word "clique" is more likely to be used for a subgraph of a larger graph, whereas "complete graph" is more likely to be used when we're considering such a graph by itself. But they're interchangeable, so the two cases cover all possibilities.
More precisely, suppose that $G-U$ consists of $k$ connected components $G_1, G_2, \dots, G_k$. (And, as far as edge cases go, keep in mind that $k$ could be $1$.) Then the two cases are:


*

*Each of $G_1, G_2, \dots, G_k$ is a complete graph.

*There is some $i$, $1 \le i \le k$, such that $G_i$ is not a complete graph.

