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Answer by piousbox. A previous answer, saved below after "-=-----5", is incorrect. The step that is incorrect is "surely since 1 ele of N and M is UB then (M-1) is ele of N". Why? Suppose that upper limit of N is 5, that is, 5 is in N, but 6 is not. I take M = 9. M is an upper bound of N by definition, and M-1 = 8 is still not in N, still an upper bound of N. The correct solution must involve the notion of *least* upper bound, which involves the notion of completeness (dedekind completeness):
1. Suppose N is bounded above
2. By Dedekind Completeness, there is a minimum upper bound of N, call it m
3. if n is in N then n+1 is in N, and n+1 <= m
4. n <= m-1
5. previous implies that m is not the least upper bound, reaching a contradiction.
-=-----5
sps there was then get a contradiction
what does it mean for N to be bounded above? (u write this defn urself using those cool math symbols)
surely 1 is ele of N therefore for N is not empty
therefore N is bounded Above
let M be an UB (write this meaning out w/ math symbols)
surely since 1 ele of N and M is UB then (M-1) is ele of N
there there exist n ele in N st n is strictly greater than that difference
so then
bring that -1 over to the other side of the inequality, hence n+1 is strcitly greater than M. HEY but wait one sec there buddy cause that's a contradiction of M UB on N
by defn nothing in N is strictly greater than M UB by the simple def of what it means to b UB ( on any set, but in this case the N)
there fore it is not the case that N is bounded above by M ( or by anthing for that matter)
therefore it is the case that N is not bounded above
u need to write this out using those really cool math symbols that you love all too much
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