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Find the supremum and infimum of the following sets:

A={$\frac{m}{n+1}$| $m<2n$, $m\in N$ and $n \in N$}

B=({$1,\sqrt 3$ }$\cap Q$)$\cup${$(-1)^n+2|n \in N$}

There are some problems with the notation but I think it's relatively clear.

The first set if we fix $m=1$ we get the following sequence $\frac {1}{n+1}$ . The first couple of terms are : $\frac{1}{2}.\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7}.....$ which will eventually approach $0$.

For the supremum the following inequality holds :

$\frac{m}{n+1}< \frac{2n}{n+1}\le\frac{2n}{n} = 2$

Writing couple of terms : $\frac{1}{2},\frac{2}{3},\frac{3}{3},\frac{1}{4},\frac{2}{4},\frac{3}{4},\frac{4}{4},\frac{5}{4}...$ which approaches $2$.

$sup A = 2$ and $inf A=0$.

Now the second set

B is union of two sets: C= {$(1,\sqrt3)\cap Q$} $= 1$ and D= {$(-1)^n+2$} By definition the supremum and infimum of $B$ will be equal to:

inf $B=min${inf $C$,inf $D$}

sup $B=max${sup $C$,sup $D$}

Obviously sup $C$=inf $C = 1$. Now set $D$:

When $n$ is odd we have: $1,1,1,1,1.....$ When $n$ is even we have ; $3,3,3,3,3...$

So inf $D = 1$ and sup $D = 3$. Finally inf $B = 1$ and sup $B = 3$

Finally the official question : Are these proofs enough? All I did was write bunch of terms use couple of theorems and made guesses about which value the sequences approach. I'm really unsure about whether this is enough as an official proof on exam.

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  • $\begingroup$ I think you're using the $m<2n$ incorrectly there, maybe confusing it with $n < 2m$ For example the fraction $\frac{5}{3}$ is valid because $m=5, n=3$. So the supremum is 2, but the idea is correct, you can make it go as close as you want to 2 although it will never be 2. $\endgroup$
    – Hendrata
    Oct 10, 2017 at 21:00
  • $\begingroup$ Ah yes, good catch and makes sense why I was getting sup A = 2 using certain inequalities. $\endgroup$
    – DreaDk
    Oct 10, 2017 at 21:06

1 Answer 1

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Those are proofs enough, yes, although you made minor error: $\sup C$ is actually $\sqrt{3}$, because there are rational numbers close to $\sqrt{3}$ and you can make it as close as possible. It does not change the final answer though.

If you're worried about the "Validity" of your proof, just think of supremum as the lowest upper bound. So you have to prove: 1. it's an upper bound 2. that any other number lower than that does not qualify as an upper bound.

So in your example for $A$, you can say 2 is an upper bound... and for any number less that 2, say $2 - x$ (no matter how small $x$ is), you can find $m,n$ such that $m/(n+1) > 2-x$ (or in other words, arbitrarily close to 2). The trick is by using $n$ really large, and $m = 2n-1$ and see what happens.

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  • $\begingroup$ Thanks but in set $C$ we have intersection with the rational numbers and $\sqrt3$ doesn't belong in $Q$, therefore we are left with only one element ( $1$). Unless I misinterpret what was written since as I already mentioned it is not very clear. $\endgroup$
    – DreaDk
    Oct 11, 2017 at 10:06
  • $\begingroup$ If you write $(1, \sqrt{3}) \cap Q$ it means the intersection of the interval $1 < x <\sqrt{3}$ with $Q$. If you write $\{ 1, \sqrt{3} \} \cap Q$ then you're right, the intersection of a set with 2 elements (1 and $\sqrt{3}$) with $Q$. $\endgroup$
    – Hendrata
    Oct 11, 2017 at 14:44

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