Let S be a non-empty subset of N. Suppose S has an upperbound (not necessarily in S) b e N. Show that S is finite.
How would I show this?
clearly $\displaystyle S \subseteq B$, where $\displaystyle B = \{ n \in \mathbb{N} \mid n \le b \}$.
note that $\displaystyle B$ is finite, since it has b (or, depending on your definition of natural numbers, b + 1) elements. since $\displaystyle S \subseteq B$, $\displaystyle S$ must be finite as well
S has an upper bound. that is, all its elements are less than or equal to b. so what i did was to construct the set of ALL elements of natural numbers that are less than or equal to b. showing S is a subset is not that difficult, for assume x is some element in S, then that means it is a natural number less than or equal to b, but then, that means x is in B, since B is the set of ALL such elements. thus, S is a subset of B.