Let $A = (x_n)$ be a sequence that defined as $x_n = \frac{1}{3^{n+5}}$.
Show that $A$ is bounded and find it's supremum and the infimum.
Attempt:
First, I claim that $A$ is a decreasing sequence. I show this by induction as follows.
To show: $x_{n+1} < x_n$ for all $n \in \Bbb Z^+$. Indeed, this is true for $n=1$. Now, assume that for $n=k$, it's also true; that's, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Then,
$x_{k+2} = \frac{1}{3^{(k+2)+5}} = \frac{1}{3^{k+7}} < \frac{1}{3^{k+6}} = \frac{1}{3^{(k+1)+5}} = x_{k+1}$. Hence, $x_{k+1} < x_k$ for some $k \in \Bbb Z^+$. Therefore, $x_{n+1} < x_n$
for all $n \in \Bbb Z^+$. Thus, $A$ is a decreasing sequence. $\Box$
Now, back to the problem. To show: A is bounded. Since $A$ is a decreasing sequence and tending to $0$, then $A$ is bounded below by $0$. Next, it's clear that for all $s \in A$, we have
$s \le \frac{1}{3^6}$, which is happenned if and only if $n=1$. Hence, by definition, $A$ is bounded above by $\frac{1}{3^6}$. Therefore, $A$ is bounded.
Next, we find the supremum and the infimum of $A$. Indeed, $\inf A = 0$ and $\sup A = \frac{1}{3^6}$.
To show: $\inf A = 0$. Let $m$ be the another lower bound of $A$ and suppose that $0 \lt m$.
Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $r$,
such that $r \lt m$. A contradiction with the definition of a lower bound. Hence, we must have
$m \le 0$ and thus, $\inf A = 0$.
To show: $\sup A = \frac{1}{3^6}$. Let $M$ be the another upper bound of $A$ and suppose that $M \lt \frac{1}{3^6}$. Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $s$, such that $M \lt s$. A contradiction with the definition of a upper bound. Hence, we must have
$\frac{1}{3^6} \le M$ and thus, $\sup A = \frac{1}{3^6}$.
Am I true? Anyone please correct me. Thanks in advanced.
Best Answer
For induction part, base case is OK. But then inductive hypothesis case should also mention that $n=k\gt 1$. Then Induction part is OK.
To prove $\inf A=0$, you have mentioned that "Let $m$ be the another lower bound of A and suppose that $0<m$. Then, since $A$ is a decreasing and bounded sequence, there exist an element of $A$, say $r$, such that $r<m$..." Question is why does such $r$ exist? Example: The sequence $(y_n)=(2+1/n)$ is also decreasing and $0$ and $2$ are lower bounds but there is no $k\in \mathbb N$ such that $y_k\lt 2$. Similarly for supremum case.
In order to find $\sup A$ and $\inf A$, you may proceed as below also:
Let $m\gt 0$ be a lower bound for $A$, then by Archimedean property of real numbers $\exists N $ such that $1/N\lt m$. Now choose $n$ so large that $3^{n+5}\gt N$ and therefore, $1/3^{n+5}\lt m$, which is a contradiction. Therefore, $\inf A=0$.
Let $l\lt 1/3^6$ be an upper bound for $A$. Hence, $1/3^{n+5}\le l\lt 1/3^6$ for all $n$. Clearly, $n=1$ gives $l=1/3^6$, which is a contradiction.Therefore, $\sup A=1/3^6$ as $1/3^6$ is an upper bound for $A$.