Question

Don't think my proof is correct - Real Analysis

Answer 1

If there is an error, I couldn't find it.  Nice work writing a shorter proof!  This was always one of my favorite parts of math.

Answer 2

Another thing I noticed

>Since x is sup(E), nx (n an integer, positive) is greater or equal to x. So nx is an upper bound of E.

This requires x to be positive, else we could have nx<x, so nx is not necessarily a upper bound.

However, we don't require it to be a upper bound, so it doesn't matter, since we require only that m is an integer and m/n is an upper bound.

Answer 3

Looks OK (except that your value isn't an integer, but that's pretty easy to fix). Here's a similarly short proof (denoting nE = \{nx | x ∈ E\}):

The set \{x ∈ ℕ | x is an upper bound for nE \} is non-empty, so has a least element by the well-ordering theorem (if you prefer not to assume that, \{x ∈ 1, ..., K | x is an upper bound for nE\} is finite, so clearly has a smallest element). That smallest element (call it y) is still an upper bound for nE, so y/n is an upper bound for E. But y - 1 is not an upper bound for nE, so (y - 1)/n is not an upper bound for E.

Answer 4

The problem calls for an integer m or did I misunderstand?

If yes, in your proof, how do you guarantee m=nx is an integer?