Proof verification: Show that $\max A=\frac{3}{2}$ and $\inf A = 0$ this question deals with $A:=\{2^{-n}+1/m: n,m\in \mathbb{N}\}$.
Claim 1: $\max A=3/2$
My proof:
Since $3/2=1/2+1\in A$ and $(\forall m,n \in \mathbb{N}): 1>\frac{1}{2^n}+\frac{1}{m}$, we can conclude that $\max A=3/2$.
Claim 2: $\inf A=0$
My proof:
$0\notin A$ and $0$ is a lower bound since $(\forall m,n \in \mathbb{N}):0<\frac{1}{2^n}+\frac 1{m}$. Furthermore $\forall \epsilon$ $\exists x\in A: x>\varepsilon$, just choose $x=\frac{1}{n}$, by Archimedes this is true for all $n\in \mathbb{N}$, thus we can conclude $\inf A = 0$
Is there anything I could improve or is something wrong? Thanks in advance!
 A: For the first claim, the $1>\frac{1}{2^n}+\frac{1}{m}$ should be $\frac{3}{2} \ge \frac{1}{2^n}+\frac{1}{m}$. Also, you should prove this by something as relatively simple as saying that both $\frac{1}{2^n}$ and $\frac{1}{m}$ are decreasing functions, so the maximum of their sum occurs at $n = m = 1$, giving $\frac{3}{2}$.
For your second claim, I'm not quite clear what you're saying as you want to show the sum is less than any fixed $\epsilon$, but you're using a fixed $x \gt \epsilon$. However, you're correct in the first part that $0$ is a lower bound. To show there's no lower bound greater than $0$, i.e., so that $\inf A=0$, you can use contradiction. Assume some $L \gt 0$ is a lower bound. However, for $m > \frac{2}{L}$ that $\frac{1}{m} \lt \frac{L}{2}$ and for $n \gt \log_2\left(\frac{1}{L}\right) + 1$ that $\frac{1}{2^n} \lt \frac{L}{2}$, you get the sum $\frac{1}{2^n} + \frac{1}{m} \lt L$. This shows that $L$ can't be a lower bound, proving that $0$ must be the greatest lower bound, i.e., $\inf A = 0$.
