Consider the following:
A = The set of all numbers (if it’s a non-contradictory number, then it is included in this set)
B = The set of all numbers except the number 19
It seems that both A and B encompass an endless number of numbers. However, one cannot deny that A is greater than B in terms of quantity or cardinality. I will attempt to show: 1) B is semi-infinite in quantity, whilst A is infinite in quantity, and 2) Semi-infinites come in various sizes, but there is only one infinity (so there aren’t infinities of various sizes).
If you tell me “there is no end to the number of numbers that B encompasses”, and I ask you “does B encompass the number 19?”, you will say “no”. To which I will say “clearly, there is an end to the number of numbers that B encompasses. Had you said ‘excluding 19, there is no end to the number of numbers that B encompasses’ I might have believed you”. I say might because I’m not sure if B encompasses infinity. Either we say A is infinity (in which case B does not encompass infinity), or we say A is not infinity (in which case both A and B encompass infinity, but only A encompasses/is an infinity of numbers).
Whilst there absolutely/truly is no end to the number of numbers that A encompasses, there is an end to the number of numbers that B encompasses in an absolute sense. Having said that, the number of numbers that B encompasses is not finite in quantity (hence the term semi-infinite). Furthermore, B is one possible maximally large semi-infinite set of numbers (because it encompasses all numbers but one, and there are an endless number of semi-infinite sets that do this. A semi-infinite set that encompasses all numbers but two is smaller than the aforementioned semi-infinite set).
Hopefully, the above proves that whilst there are many semi-infinite sets of varying sizes, there is only one infinite set.