# Proving a set is bounded and its supremum and infimum

Question: Let $$X = \{\frac{n-1}{n+1}:n\in\Bbb{N}^{>0}\}$$

1. Is $$X$$ bounded above?
• If yes, what is Sup$$X$$?
2. Is $$X$$ bounded below?
• If yes, what is Inf$$X$$?

My attempt:

I know that $$X$$ is bounded, such that Sup$$X=1$$ and Inf$$X=0$$, but my uni lecturer wants us to practise proving our answers.

So I need to show that:

1. $$\exists B\in\Bbb{R}: |x_n|\le B,\forall n\in\Bbb{N}^{>0}$$ (Bounded)

Or show both:

1. $$\exists B\in\Bbb{R}: x_n\le B,\forall n\in\Bbb{N}^{>0}$$ (Bounded above)
2. $$\exists B\in\Bbb{R}: B \le x_n,\forall n\in\Bbb{N}^{>0}$$ (Bounded below)

I don't know how to approach this though, and am not sure how to prove Sup$$X=1$$ and Inf$$X=0$$.

Any help would be greatly appreciated.

I’ll give you some pointers.

The easiest part is to show that $$\inf X=0$$: first show that in fact $$0\in X$$, and then show that $$x\ge 0$$ for each $$x\in X$$.

It’s also not hard to show that $$x<1$$ for each $$x\in X$$, so $$X$$ is bounded above by $$1$$. Then note that $$1-\frac{n-1}{n+1}=\frac{2}{n+1}$$ for each $$n\in\Bbb N^+$$, and use this to show that if $$y<1$$, there is an $$n\in\Bbb N^+$$ such that $$\frac{n-1}{n+1}>y$$; this will show that $$\sup X=1$$ (why?).

Well, a few things:

You know that $$0\le n-1 < n+1$$ so $$0\le \frac {n-1}{n+1} < 1$$.

So that shows there $$\exists B = 1$$ so that $$x_n \le B=1$$ for all $$n$$.

And there $$\exists B = 0$$ so that $$0 = B \le x_n$$ for $$n$$.

So that shows it is bounded above and below.

....

Also $$x_1 = \frac {1-1}{1+1} = \frac 0 2 = 0 \ge 0$$. So the set $$X$$ does achieve $$0$$ as a value.

If a set achieves a lower bound as a value that lower bound must be the $$\inf X$$. Why? Because for any $$y > 0$$ then $$x_1 < y$$ so $$y$$ can't be a lower bound and as $$0$$ is a lower bound and nothing larger can be, then $$0$$ is $$\inf X$$.

......

So that was $$\frac 34$$ of the problem with no actual work.

The final thing though is that $$X$$ never does acheive an $$x_n = 1$$ so to prove $$1$$ is $$\sup A$$ will require some work.

Prove that if $$y < 1$$ then there is some $$x_n$$ so that $$y < x_n \le 1$$.

That would mean that any $$y < 1$$ can't be an upper bound, so the upper bound $$1$$ must be the least upper bound.

So I'll leave to you. Show that for any $$y < 1$$ there will be an $$x_n \in X$$ so that $$y < x_n \le 1$$.