infimum and supremum finding

Question

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?

What is $n$? Is it a fixed positive integer, or is there some limiting/set operation going on? — J. David Taylor, Oct 25 '14 at 20:40
I gave an example of how to tex it. If you would modify the post to add in where the $n\in\mathbb{N}$ shows up, then I think I can help. — J. David Taylor, Oct 25 '14 at 20:46
Here's the tutorial: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — J. David Taylor, Oct 25 '14 at 20:47

score 0 · Answer 1 · answered Oct 25 '14 at 20:54

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 $\{x_n\}_{n=1}^\infty$.

Aaron Maroja · Answer 2 · 2014-10-25T21:10:28.633

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.

infimum and supremum finding

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