I have the set $A=\left \{1+\frac{1}{n}:n\in \mathbb{N} \right \}$ and I am trying to show that $sup(A)\neq 3$.
Attempt:
I) Showing that 3 is an upper bound of A:
$1+\frac{1}{n}\\\\\Leftrightarrow n>\frac{1}{2}$
which proves that 3 is indeed an upper bound.
II) Showing that for a number $3-\varepsilon$ where $\varepsilon > 0$ there exists a $1+\frac{1}{n}>3-\varepsilon$:
I do othe algebra and arrive at the following:
if $\varepsilon \geq 2$: $n>\frac{-1}{\varepsilon-2}$
if $\varepsilon < 2$: $n<\frac{-1}{\varepsilon-2}$
Which would indicate that 3-2= 1 would be the supremum of A but clearly sup(A) = 2.
What am I doing wrong here? An explanation is appreciated.