Question

How to prove ?
For all a,b (a >=b) <=> [a, +infinity) is subgroup of [b, +infinity)

Answer 1

since a is greater than or equal to b, the set of numbers from a to +infinity will be smaller than that of b to +infinity
such as if a=2 and b=1... [2, +infinity) is contained in [1, +infinity)

Answer 2

i'm not really sure how to do a formal proof though, sorry

Answer 3

\[a\geq b\quad,\quad [b,+\infty)=[b,a)\cup[a,+\infty)\]\[\Rightarrow\quad [a,+\infty)\subseteq[b,+\infty)\]