In a certain country ,$100$ roads lead out of each city and one can travel along those roads from one city to the other In a certain country ,$100$ roads lead out of each city and one can travel along those roads from one city to any other . One road is closed for repair work . Prove that one can still get from any city to another .
The solution given on the book is as follows:

If road $AB$  has closed then it is enough to prove that we can still get from $A$ to $B$. If this were not true, then in the connected component containing  $A$ all the  vertices other than $A$ would be even . The situation of having exactly one odd vertex in a connected component contradicts the statement "A graph must have an even number of odd vertices"

However, I am not getting the idea of the solution ...I dont get what does it mean by saying " If this were not true, then in the connected component containing  $A$ all the  vertices other than $A$ would be even " ...the path from A to B is deleted so...the degree of B is also odd...so, how does they draw this conclusion ...
 A: Let $G$ be the graph in question.
Suppose if we closed $AB$ we could not get from $A$ to $B$.
That means we could divide $G$ into two separate components: one containing $A$ and one containing $B$, just by deleting $AB$ (that is: closing the road between $A$ and $B$).
Anyway, suppose we could do this.
Let $S$ and $T$ be collections of vertices of the graph $G$, which are joined by a bridge $AB$, so that $A$ is a vertex of $S$ and $B$ is a vertex of $T$.
(In this context, a "bridge" is an edge of a graph that, if you remove it, separates the graph into two disconnected components.)
Thus when $AB$ is deleted, and we can no longer get from $A$ to $B$, that means $S$ and $T$ make two separate graphs.
Before we remove edge $AB$ (that is, close the road):

*

*All vertices in $S$ are even, including $A$ which is incident to the bridge $AB$.


*All vertices in $T$ are even, including $B$ which is incident to the bridge $AB$.
But when you delete $AB$:

*

*all the vertices in $S$ except $A$ are even, while $A$ is odd.


*all the vertices in $T$ except $B$ are even, while $B$ is odd.
Hence $S$ and $T$ are both graphs which do not have an even number of odd vertices.
Because that cannot happen, it cannot be the case that deleting $AB$ causes $S$ and $T$ to be disconnected.
Hence you must be able to get from $A$ to $B$ without needing to use the road $AB$.
