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I am not sure how to approach these supremum and infimum sets. I also have to give a min or max

$$A = \Big\{(-1)^n+\frac{1}{n+1}\ :\ n\in \Bbb{N}\Big\}$$ $$B = \Big\{\frac{x}{x+1}\ :\ x\in [0,\infty)\Big\}$$ $$C = \bigcup_{n=1}^\infty\Big[\frac{1}{2n},\frac{1}{2n-1}\Big]$$

Any help is much appreciated!

Teddy38
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MMoe
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  • Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? – 5xum Nov 13 '17 at 14:22
  • Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Nov 13 '17 at 14:22
  • Third question with zero context, out of three. That you received answers to your two first questions (unfortunately) may have left you with a wrong impression of the site. – Did Nov 13 '17 at 14:25

1 Answers1

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For $A$, note that $\frac{1}{n+1}\xrightarrow{n\to\infty} 0$ and furthermore, it is monotonically decreasing. $(-1)^n$ just cycles between $-1$ and $1$.

For $n = 1$ we obtain $(-1)^1 + \frac12 = -\frac12$.

For $n=2$ we obtain $(-1)^2 + \frac13 = \frac43$.

Now, notice that for $n \ge 2$ we have:

$$(-1)^n + \frac1{n+1} \le 1 + \frac13 = \frac43$$

Since the value for $n = 1$ is smaller than $\frac43$ we conclude that $\frac43$ is the supremum of $A$. Since it is attained for $n = 2$, it is also the maximum of $A$.

To obtain the infimum, notice that $A$ is bounded from below by $-1$:

$$-1 < (-1)^n + \frac{1}{n+1}$$

Futhermore, elements of $A$ approach arbitrarily close to $-1$. To show this, take $\varepsilon > 0$. Since $\frac{1}{n+1}\xrightarrow{n\to\infty} 0$, there exists an odd integer $n_0 \in \mathbb{N}$ large enough so that $\frac1{1+n_0} < \varepsilon$.

We have:

$$(-1)^{n_0} + \frac1{n_0 + 1} = -1 + \varepsilon$$

We conclude that $\inf A = -1$. It is not a minimum because $\frac1{n+1} > 0$ so it cannot be attained for any $n \in \mathbb{N}$.

For $B$, we have $\frac{x}{x+1} \ge 0$ and for $x = 0$ obviously $\frac{0}{0+1} = 0$. Thus, $\min B = 0$.

$B$ is bounded from above by $1$:

$$\frac{x}{x+1} \le \frac{x+1}{x+1} = 1$$

Now recall that $\lim_{x\to\infty} \frac{x}{x+1} = 1$, so there exists $x > 0$ large enough so that $\frac{x}{x+1} > 1- \varepsilon$ for any $\varepsilon > 0$. Thus, $\sup B = 1$. It is not a maximum since:

$$\frac{x}{x+1} < \frac{x+1}{x+1} = 1$$

Can you manage $C$ on your own?

mechanodroid
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  • Hi! Thanks for the answer, it really helped me. One question though: Can you explain how the equation for n>=2 is true in A? It's only 4/3 for n=2 but not n>2 isn't it ? – MMoe Nov 13 '17 at 16:46
  • Nevermind I got it now ! – MMoe Nov 13 '17 at 17:00