The definition of supremum has two halves, and proofs have to check both. The first half, "$s$ is an upper bound," is usually easy. The second half, "nothing smaller is an upper bound," has a convenient form called the ε-test, and it is the first place in this course where the letter $\varepsilon$ does real work.
Key idea
Why it works
Here is a full proof in this style. Claim: $\sup\left\{1 - \frac{1}{n} : n \ge 1\right\} = 1$. Proof. Every element is $1 - \frac{1}{n} < 1$, so $1$ is an upper bound. Let $\varepsilon > 0$. By the Archimedean property choose a positive integer $n$ with $\frac{1}{n} < \varepsilon$. Then $1 - \frac{1}{n} > 1 - \varepsilon$, so $1 - \varepsilon$ is not an upper bound. Since $\varepsilon$ was arbitrary, $1$ is the least upper bound. $\blacksquare$
Worked examples
Worked example: Which element passes the test?
- We need $1 - \frac{2}{n} > 0.99$, that is, $\frac{2}{n} < 0.01$, or $n > 200$.
- The smallest whole number with $n > 200$ is $201$.
Answer: $n = 201$.
Worked example: An element of an interval
- We need $x \in A$ with $4.75 < x < 5$.
- Any such $x$ works, for example $4.9$.
Answer: $4.9$ (any number in $(4.75, 5)$ is right).
Worked example: The same test for an infimum
- Every element is $2 + \frac{3}{n} > 2$, so $2$ is a lower bound.
- For the ε-test turned around, we need an element below $2 + \varepsilon$: $\frac{3}{n} < \varepsilon$ holds once $n > \frac{3}{\varepsilon}$, which the Archimedean property supplies. So $2$ is the greatest lower bound.
- With $\varepsilon = 0.01$: $n > 300$, so the first such element is at $n = 301$.
Answer: $\inf A = 2$; the first element below $2.01$ is $2 + \frac{3}{301}$, at $n = 301$.
Common mistake