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How does one find the supremum and infimum of an integer "sequenced" set such as

$$\bigcup_{i=1}^{\infty} [2+1/i, 5-1/i]$$

The problem that I'm having is that I use the following definition for e.g. $\inf$:

$x=\inf S$ if

(1) $\forall s \in S$, $x \leq s$ and
(2) $\forall \epsilon > 0$, $\exists s_{\epsilon} \in S$ s.t. $s_{\epsilon} \lt x+ \epsilon$.

Now since $\epsilon$ is a real number, then I'm having trouble seeing, how to pick $s_{\epsilon}$, when my set is sequenced in integers, rather than being continuous. Same thing for $\sup$.

In the continuous case I could pick $s_{\epsilon} = 2 + \frac{1}{\frac{2}{\epsilon}}$, but then I'd have trouble showing that $s_{\epsilon} \in \bigcup_{i=1}^{\infty} [2+1/i, 5-1/i]$, even if $s_{\epsilon} = 2 + \frac{\epsilon}{2} < 2 + \epsilon = x + \epsilon $.

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    $\begingroup$ Step 1: Use your intuition to guess what you think $x$ is. $\endgroup$
    – vadim123
    Nov 1, 2016 at 15:27
  • $\begingroup$ use $\lim\frac{1}{i}=0$ $\endgroup$ Nov 1, 2016 at 15:28
  • $\begingroup$ @AbdallahHammam I've had trouble connecting limits and unions together: math.stackexchange.com/questions/1994491/… $\endgroup$
    – mavavilj
    Nov 1, 2016 at 15:29
  • $\begingroup$ In (1) $<$ must be changed into $\leq$. In (2) you can leave out $x\leq s_{\epsilon}$ (which follows from (1) so is redundant) $\endgroup$
    – drhab
    Nov 1, 2016 at 15:34
  • $\begingroup$ s_e can be anything that is greater than x- e. The s_e is not important. The x is what's important. (In this case). $\endgroup$
    – fleablood
    Nov 1, 2016 at 15:39

2 Answers 2

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Hint :

let us prove that $\;inf S=2$.

we have

$$(\forall x\in S )\;\; 2\leq x$$

now let $\epsilon>0$.

we look for $s_\epsilon \in S$ of the form

$s_\epsilon=2+\frac{1}{n}$ such that

$2+\frac{1}{n}<2+\epsilon$

or

$n>\frac{1}{\epsilon}$.

so, we can take $n=\lfloor \frac{1}{\epsilon} \rfloor +1$.

you can prove that $\;sup S=5$

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  • $\begingroup$ Yeah, the part of using $2+\frac{1}{n}<2+\epsilon$ and then inferring $n=\lfloor \frac{2}{\epsilon} \rfloor +1$ (especially the use of the floor function) was something that I didn't figure out. $\endgroup$
    – mavavilj
    Nov 1, 2016 at 15:40
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If $2<x<5$, let $\epsilon=\min\{x-2,5-x\}>0$. There is an $n\in\Bbb Z^+$ such that $\frac1n<\epsilon$, so

$$2+\frac1n<2+\epsilon\le x\le 5-\epsilon<5-\frac1n\;,$$

and

$$x\in\left[2+\frac1n,5-\frac1n\right]\;.$$

This shows that

$$(2,5)\subseteq\bigcup_{n\in\Bbb Z^+}\left[2+\frac1n,5-\frac1n\right]\;,$$

and the opposite inclusion is clear.

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  • $\begingroup$ Yeah sure, but I think this is elementary. The problem was related to how to pick a "discrete" version of the continuous $s_{\epsilon}$. $\endgroup$
    – mavavilj
    Nov 1, 2016 at 18:01
  • $\begingroup$ @mavavilj: No, the problem is that you think that there is some essential difference between the two. There isn’t. If $x>2$, then there is an $n\in\Bbb Z^+$ such that $\frac1n<x-2$, and $2+\frac1n<x$. $\endgroup$ Nov 1, 2016 at 18:02
  • $\begingroup$ Now that I’ve a better idea of where the problem is, I’ve completely revised my answer. $\endgroup$ Nov 1, 2016 at 18:10

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