Find the $\sup$ and $\inf$ of $E=\{x\in\mathbb{R}\mid 1-\frac{1}{n} < x < 2-\frac{1}{n}, n \in \Bbb{N}\}$. Justify your answers
I claim that $\sup E = 2$ and $\inf E = 0$. Let $(1-1/n,2-1/n]$, then $\sup E = 2-1/n$ and $\inf E = 1-1/n $.
Case 1: $n\leq 0$, then $\sup E$ is $2$ and $\inf E$ is $1$.
Case 2: $n \geq 0$, then $\sup E$ is $1$ and $\inf E$ is $0$.
Case 3: $0 < x < 1$, then $\sup E$ is $2$ and $\inf E$ is $0$.
or do I even need to do cases?
On
Let $\sup E = a$. First we have that $2 - \frac{1}{n} < 2, \forall \ n\in\mathbb{N}$. Then $2$ is an upper bound. Take any $\epsilon > 0$. There exists $n_0$ such that $n_0\ \epsilon>1 \Rightarrow \epsilon > \frac{1}{n_0} \Rightarrow 2-\epsilon < 2-\frac{1}{n_0} = x \in A $, then there exists $x\in E$ such that $2 - \epsilon < x$. You may conclude that $a = 2$ by definition of supremum.
Similar goes to the infimum, using definition of infimum.
First, note that as $n \to \infty$, then $\frac{1}{n} \to 0$.
Next, note that $2-\frac{1}{n}$ is bounded above by $2$, and as $n \to \infty$, $2 - \frac{1}{n} \to 2-0=2$. Thus $2$ is the supremum of $\{x_n\}_{n=1}^\infty$.
Next, note that $1-\frac{1}{n}$ is bounded below by $0$ (take $n=1$) and monotone increasing, with $1 - \frac{1}{n} \to 1$ as $n \to \infty$. Thus $0$ is the infimum of $\{x_n\}_{n=1}^\infty$.