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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.

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2 Answers 2

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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?).

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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.

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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 $\inf X$.

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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 $y < x_n \le 1$.

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