Show that $\bigcup_{n=1}^{\infty} (0, 2+\frac{1}{n}] = (0, 3]$ Show that: $$ \bigcup_{n=1}^{\infty}  (0, 2+ \frac{1}{n}] = (0, 3]$$
The above statement seems quite obvious but i'm unsure how one would prove it.
So far i've done this:
$A_n := (0, 2+ \frac{1}{n}]$  for $n \in N$
$A_{n+1} \subset A_n $ $ ,\forall$ $ n \in N$
So $\bigcup_{n=1}^{\infty} A_n = A_1 = (0, 3] $
I'm not sure if this is the way i'm meant to answer the question.
 A: You asked how we might solve a more general case without relying on the increasingness of the left-hand side.
So: $x \in \bigcup_{n = 1}^{\infty} \left(0, 2+\frac{1}{n}\right]$ if and only if there is $n$ such that $0 < x \le 2 + \frac{1}{n}$.
On the other hand, $x \in (0, 3]$ if and only if $0 < x \le 3$.
$0 < x \le 2 + \frac{1}{n} \implies 0 < x \le 3$: because all $n$ have $0 < 2 + \frac{1}{n} \le 3$, so the left-hand interval is contained in the right-hand interval.
$0 < x \le 3 \implies 0 < x \le 2+\frac{1}{n}$ for some $n$: this is trivial by setting $n = 1$.

The general technique is to identify precisely when something is in the left-hand side, identify precisely when something is in the right-hand side, and then show that the two conditions are equivalent. Your answer takes a handy shortcut by noting that the left-hand side has a sufficiently nice structure that it can be immediately simplified to $(0, 2+1]$.
A: Yes your proof is perfect. Considering that each $A_{n+1} \subset A_{n}$ for all natural numbers, the largest 'parent' set that we can observe from this union is when we can choose the smallest possible natural number: $n=1$. Thus $A_{1}=\bigcup_{n=1}^{\infty} A_n = (0,3]$.
