# Let A = {z ∈ R : there exists n, m ∈ N such that z = 1/n − 1/m }. Prove, using definition(s) and/or result(s) from the lectures, that sup(A) = 1

This is what I have. Please let me know if I am on the wrong track?

Step 1: Let 𝑎 ∈ 𝐴. Then a = to 1/n - 1/m for some n.

a = 1/n - 1/m < 1/n - 1/m + 1 = 1

so 1 is the upper bound of A.

Step 2:

suppose x < 1, then there is some n such that:

x < 1/n - 1/m < 1

x + 1/m < 1/n < 1+ 1/m

m / xm < n

Since x < 1, there is some n in the natural numbers such than x is not the upper bound of A, hence sup(A) = 1

Is this right?

• Yes your reasoning is correct. If 1 is an upper bound while every $x < 1$ is not an upper bound, then it is the supremum. – Daniel Apsley Mar 26 at 4:53

You write $$a=\frac1n-\frac1m<\frac1n-\frac1m+1=1.$$ Of course the latter equality is false. Fortunately it is also unnecessary. To show that $$1$$ is an upper bound, note that $$a=\frac1n-\frac1m<\frac1n\leq1.$$ This indeed shows that $$\operatorname{sup}A\leq1$$. Your proof that equality holds is a bit unclear; indeed for every $$x<1$$ you want to find $$m,n\in\Bbb{N}$$ such that $$x<\frac1n-\frac1m.$$ This will show that $$x<\operatorname{sup}A$$, and hence that $$\operatorname{sup}A\geq1$$. You proceed with $$x + \frac1m < \frac1n < 1+ \frac1m,$$ which is correct, but then you state that $$\frac{m}{xm} which does not follow; consider $$x=\tfrac12$$ and $$n=1$$ and $$m=3$$. And what if $$x=0$$?
Instead, note that for all $$m,n\in\Bbb{N}$$ you have $$\frac1n-\frac1m\leq\frac11-\frac1m=1-\frac1m.$$ With this in mind, it suffices to find some $$m\in\Bbb{N}$$ such that $$x<1-\frac1m.$$