Why is this reasoning on ordered sets not correct : " Element a in S has no immediate predecessor. Therefore, a is the first element in S"?. In _Theory and Problems of Set Theory_ , Lipschutz considers as an example of a well ordered set the set of natural numbers {1,2,3,4...} ordered by the relation R as follows : (1) if a is odd and b is even, aRb (2) if both a and b are odd, then aRb if a < b (3) if both a and b are even, then aRb if a < b. Lipschutz notes this set in an unusual fashion : { 1,3,5,7..... ; 2,4,6,8...} . The number 2 is given as an example of "limit" element. How to explain that 2 has no immediate predecessor? How can it be the case that, although an element has at least a predecessor, no predecessor is an immediate one? Is it possible to give another example of set having a limit element ( that is , an element with no immediate predecessor, without being however the first element) ?

While $2$ certainly has many predecessors, the point is that none of them is an _immediate_ predecessor.

The proof proceeds by contradiction: Supposing that $n R 2$ is the immediate predecessor of $2$. Then $n R (n+2)$ and $(n+2) R 2$. Therefore $n$ is not the immediate predecessor and it does not exist.

The reason why this type of element is called a "limit element" is maybe explained better using fractions:

In the set $\\{ \frac{-1}{n} \\} \cup \\{ 0 \\} $, it is more obvious that $0$ is the "limit element". We see that $\frac{-1}{n}$ is _getting ever closer_ to the limit, _without ever getting there_.

It is in this sense that "limit elements" come into play. It takes a bit of time to get used to them (there are more complex examples, like the first uncountable ordinal $\omega_1$) but by working through some more examples you will surely get there!