Prove Supremum and Infimum

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I am trying to find and prove the supremum and infimum for each of the sets
$$A = \bigcap_{n=1}^{\infty} \, \Big(1-\frac{1}{n}, \, 1+\frac{1}{n}\Big), \hspace{5mm} B = \{x \in \mathbb{R} \, : \, x^3 < 8\}$$
For $A$, I think the $\sup A =\inf A=1$, but I am not sure how to prove it.

For B, I think the supremum is $\sup B = 8^{1/3} = 2$, and it is not bounded below, but I am unsure how to prove them again. Could someone help me out on the proofs? Thanks in advance!

Best Answer

Our intuition shows us that $\sup A = \inf A = 1$, and one way to demonstrate this is to show that $A = \{1\}$. However, we can proceed in a different manner.

We will proceed in two steps:

  1. First, we will show that $1$ is an upper bound for $A$.
  2. Then we will assume that $\sup A < 1$ and derive a contradiction.

To show that $1$ is an upper bound, we must have for every $x \in A$, $x \leq 1$. To this end, let $x \in A$. By definition, we have $$\text{For EVERY } n \in \mathbb{N}, x \in \Big(1-\frac{1}{n}, 1+\frac{1}{n}\Big).$$ This is important, because we will show that, in fact, $x$ does not belong to every interval of that form as follows.

Since $x \in \Big(1-\frac{1}{n}, 1+\frac{1}{n}\Big)$ it follows that either $x \leq 1$ or $1 < x$. If $x \leq 1$, then we are done, because we want to show that $1$ is an upper bound for $A$.

Now suppose that $1 < x$. Then $0 < x-1$. By the Archimedean Principle, there exists a positive integer $M$ such that $$\frac{1}{M} < x-1,$$ which implies that $$1 + \frac{1}{M} < x.$$ But $1-\frac{1}{M} < 1+\frac{1}{M} < x$, which means that $$x \notin \Big(1-\frac{1}{M}, 1+\frac{1}{M}\Big),$$ which is a contradiction, because $x$ belongs to every interval of the form $\big(1-\frac{1}{n}, 1+\frac{1}{n}\big)$.

Consequently, we must have $x \leq 1$ for every $x \in A$, and thus $1$ is an upper bound for $A$.

Since $A$ is a nonempty set bounded above, the Completeness axiom asserts that $\sup A$ exists. For sake of contradiction, suppose that $\sup A < 1$.

What we need to do is prove that under the assumption $\sup A < 1$, we can show that $\sup A$ is not an upper bound for $A$. But if $\sup A$ is not an upper bound for $A$, then we should be able to find an $x \in A$ such that $\sup A < x$.

And if such an $x \in A$ exists, then we must have $x \in \big(1-\frac{1}{n}, 1+\frac{1}{n}\Big)$ for every positive integer $n$.

Can you think of a number $x$ that belongs to every interval $\big(1-\frac{1}{n}, 1+\frac{1}{n}\big)$ such that $\sup A < x$? If you can, continue reading by hovering your mouse over the yellowish area below.

The $x$ we seek is $1$. Since for every positive integer $n$ $$1-\frac{1}{n} < 1 < 1+\frac{1}{n}$$ it follows that $1 \in A$. Since $\sup A$ is an upper bound for $A$, we must have $1 < \sup A$. But this contradicts our assumption that $\sup A < 1$.

Therefore, $1 \leq \sup A$. Because $1$ is an upper bound for $A$, we conclude that $\sup A = 1$.

If you understand the proof above, it is possible to clean it up a bit.

To show that $\inf A = 1$, we will show that $1$ is a lower bound for $A$. To this end, let $x \in A$, so that $x \in \big(1-\frac{1}{n}, 1+\frac{1}{n}\big)$ for every positive integer $n$. Suppose that $x < 1$. Then $0 < 1-x$. By the Archimedean Principle, there exists a positive integer $m$ such that $$\frac{1}{m} < 1- x \implies x < 1 - \frac{1}{m}$$ But since $$1 - \frac{1}{m} < 1+\frac{1}{m},$$ it follows that $$x \notin \Big(1-\frac{1}{m}, 1+\frac{1}{m}\Big),$$ contradicting the assumption that $x$ belongs to every interval of the form $\Big(1-\frac{1}{n}, 1+\frac{1}{n}\Big)$. Therefore, $1 \leq x$ for all $x \in A$, so $1$ is a lower bound for $A$.

Since $A$ is bounded below and nonempty, $\inf A$ exists. Suppose $1 < \inf A$. Then since $1 \in A$ and $\inf A$ is a lower bound for $A$, we must have $\inf A \leq 1$, a contradiction. Therefore, $\inf A = 1$.

We have proven that $\sup A = \inf A = 1$.

Now onto the set $B$.

To show that $\sup B = 2$, we proceed as we did in the preceding proof: First, we demonstrate that $2$ is an upper bound for $B$, then we prove that $2$ is the smallest upper bound for $B$.

For every $x \in B$, we have $x^3 < 8$. Consequently, $x < 2$, so $2$ is an upper bound for $B$. Since $B$ is bounded above, $\sup B$ exists. Suppose that $\sup B < 2$. Then by the Archimedean Principle, there exists positive integer $k$ such that $$\frac{1}{k} < 2 - \sup B,$$ and thus $$\sup B + \frac{1}{k} < 2.$$ But this implies that $$\Big(\sup B + \frac{1}{k}\Big)^3 < 8,$$ so $\sup B + \frac{1}{k} \in B$.

Since $\sup B$ is an upper bound for $B$ and $\sup B + \frac{1}{k} \in B$, it follows that $$\sup B + \frac{1}{k} \leq \sup B,$$ which is a contradiction, because $1/k > 0$. We conclude that $\sup B = 2.$

To show that $B$ is not bounded below, we proceed by contradiction. Assume to the contrary that $B$ is bounded below by $L$. We state a couple observations

  1. Since $L \leq x$ for all $x \in B$, we have $L^3 \leq x^3$.
  2. Since $0 \in B$ and $L$ is a lower bound for $B$, we must have $L \leq 0$. Similarly, $L < 1$. Therefore $$L - L^2 = L(1-L) \leq 0.$$

We will now demonstrate that $L$ is not a lower bound for $B$ by showing that $L-1 \in B$. Obviously, $L-1 < L$, but we see from the two observations above that for every $x \in B$:

\begin{align} \big(L-1\big)^3 &= L^3 - 3L^2 + 3L - 1\\[5pt] &= L^3 + 3L\big(1-L\big) - 1\\[5pt] &\leq L^3 + 0 - 1\\[5pt] &< L^3 \\[5pt] &\leq x^3 \\[5pt] &< 8 \end{align} Therefore, $\big(L-1\big)^3 < 8$, so $L-1 \in B$. Because $L$ is a lower bound for $B$, we must have $L \leq L-1$, which is a contradiction.

Therefore, there are no lower bounds for $B$, and thus $B$ is unbounded below.

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