Determine the upper and lower bound, maximum, minimum, supremum and infimum of a set.

Question

$A= (\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \ldots)$

The answers I've got are:

Upper Bound: $8$
Lower Bound: $-8$
Supreme: $1$
Infimum: --
Maximum: --
Minimum: $1\over2$

I just want to know if the answers are correct. I am a little confused with the meanings of infimum, supremum, maximum and minimum.

Answer 1

If you notice the elements of the set A, each element of the set can be represented as $x_i = \frac{i-1}{i} = 1-\frac{1}{i} \quad\quad ;2\le i \le \infty$

Now $\lim_{i\to \infty}$ $x_i = \frac{i-1}{i} = 1-\frac{1}{\infty}= 1+0 =1 $

$\therefore$ the supremum = maximum =1, while the minimum =infimum =$\frac{1}{2}$

Answer 2

Let's work through you answers one by one:

Upper Bound: Any number $ub$ such that for all $a \in A$, we have $a \leq ub$. So $8$ is an upper bound.

Lower Bound: Any number $lb$ such that for all $a \in A$, we have $a \geq lb$. So $-8$ is an lower bound.

Supremum: The least upper bound, the number $lub$, such that for all upper bounds $ub$ for $A$, we have $lub \leq ub$. Indeed, in this case this is $1$.

Infimum: The greatest lower bound $glb$. Here $\frac{1}{2}$ is the smallest number such that for any lower bound $lb $ of $A$, we have $lb \leq \frac{1}{2}$, so $\frac{1}{2}$ is the infimum.

If the $lub = a$ for some $a\in A$, we call $a$ the maximum. Here, $lub \not \in A$, so it has no maximum.

Similarly, if $glb = a$ for some $a\in A$, we call $a$ the minimum. Here, $glb \in A$, with $glb = \frac{1}{2} $, so $\frac{1}{2}$ is the minimum.

So they're all correct, except for the $\inf$, which is $\frac{1}{2}$

Answer 3

Any number that is greater than or equal to all the elements in set $A$ is an upper bound. Since each element in set $A$ has the form $$\frac{n}{n + 1}$$ for some positive integer $n$, any number larger than or equal to $1$ will serve as an upper bound. In particular, $8$ does, so your answer is correct.

Any number that is less than or equal to all the elements in set $A$ is a lower bound. Since each element in set $A$ is at least $1/2$, any number less than or equal to $1/2$ will serve as a lower bound. In particular, $-8$ does, so your answer is correct.

The supremum, if it exists, is the least upper bound of set $A$. In this case, $1$ is an upper bound. No smaller upper bound exists since given $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for each $n \in \mathbb{N}$ such that $n \geq N$, $$\frac{n}{n + 1} = 1 - \frac{1}{n + 1} \geq 1 - \frac{1}{N + 1} > 1 - \frac{1}{N} \geq 1 - \epsilon$$ since given $\frac{1}{\epsilon} > 0$, we can always find an integer $N \geq \frac{1}{\epsilon}$. Hence, the supremum of $A$ is $1$, as you found.

The infimum, if it exists, is the greatest lower bound of set $A$. For each positive integer $n$, $$\frac{n}{n + 1} = 1 - \frac{1}{n + 1} \geq 1 - \frac{1}{1 + 1} = \frac{1}{2}$$ Hence, $1/2$ is the infimum of set $A$.

The maximum of set $A$, if it exists, is the largest element of set $A$. Since $$\frac{n + 1}{n + 2} - \frac{n}{n + 1} = \frac{(n + 1)^2 - n(n + 2)}{(n + 1)(n + 2)} = \frac{n^2 + 2n + 1 - n^2 + 2n}{(n + 1)(n + 2)} = \frac{1}{(n + 1)(n + 2)} > 0$$ for each positive integer $n$, there is no largest element of $A$. Hence, there is no maximum. If it did exist, it would have to be $1$, but $$\frac{n}{n + 1} < 1$$ for each positive integer $n$.

The minimum of set $A$, if it exists, is the smallest element of $A$. The smallest element of $A$ is $1/2$, so the minimum is $1/2$, as you found.