# Prove Supremum and Infimum

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!

• If $A$ goes to infinity, how can it have an upper bound, let alone a least upper bound? But I agree that $\inf A = 1$ and $\sup B = 2$. Aug 11, 2019 at 3:53
• What do you mean when you wrote A:= from 1 to infinity, the intersection of (1-1/n, 1+1/n) ? Aug 11, 2019 at 3:56
• Sorry the translation messed up the notation, I meant from n=1 to infinity for all natural numbers n, not from A to infinity Aug 11, 2019 at 4:05
• Okay, great. Then I agree that $\inf A = \sup A = 1$. Aug 11, 2019 at 4:11
• Okay I'm glad that I am not totally incompetent, but I am unsure how to exactly prove it since its so intuitive. Aug 11, 2019 at 4:15

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.

Try to attack the definitions of supremum and infimum. I will do it for $$A$$, you should try the same for $$B$$.

First, observe that $$1 \in \left( 1 - \dfrac{1}{n}, 1 + \dfrac{1}{n} \right)$$ for every $$n \in \mathbb{N}$$. Therefore, $$1 \in A$$. Also, for any $$\delta > 0$$, we can have an $$n_0 \in \mathbb{N}$$ such that $$n_0 \delta > 1$$ or $$\dfrac{1}{n_0} < \delta$$. Hence, for any $$\delta > 0$$, there is some $$n_0$$ such that $$1 \pm \delta \notin \left( 1 - \dfrac{1}{n}, 1 + \dfrac{1}{n} \right)$$. Hence, $$A = \left\lbrace 1 \right\rbrace$$.

Clearly, $$1$$ is both an upper bound and a lower bound for $$A$$. To prove that it is supremum, we will need to prove that any number smaller than $$1$$ is not an upper bound. Therefore, consider $$\epsilon > 0$$. As discussed, $$1 - \epsilon \notin A$$ and $$1 > 1 - \epsilon$$. Therefore, $$1 - \epsilon$$ is not an upper bound for $$A$$ so that $$1$$ is its supremum.

Similarly for infimum, $$1 + \epsilon \notin A$$ and $$1 < 1 + \epsilon$$ so that any number strictly bigger than $$1$$ is not a lower bound. Hence, $$1$$ is also the infimum.

To prove for $$B$$, a hint is that $$2$$ is an upper bound since $$2^3 = 8$$ and $$x^3$$ is increasing.