Proofs Involving Suprema

Two Parts of a Supremum Proof

The previous tutorial showed how absolute-value inequalities can control expressions without requiring a separate sign analysis. Here the main tool is different: to prove a statement about a supremum, we often need to establish both that a number is an upper bound and that elements of the set come arbitrarily close to it. The second part is what distinguishes a supremum from an ordinary upper bound.

We will use the Approximation Characterization of the Supremum from the earlier tutorial “Supremum.” In particular, if $S$ is nonempty and bounded above and $\alpha=\sup S$, then for every $\varepsilon>0$, there is an $s\in S$ such that $\alpha-\varepsilon<s$. Also, every element of $S$ is at most $\alpha$. These two facts form a useful proof pattern: first show that the proposed value bounds every element, then use approximation to show that no smaller value can be an upper bound.

Proof strategy. To prove that $\alpha=\sup S$, establish that $\alpha$ is an upper bound for $S$, and then show that for every $\varepsilon>0$, some element $s\in S$ satisfies $\alpha-\varepsilon<s$. The Approximation Characterization of the Supremum then identifies $\alpha$ as the supremum.

The two parts answer different questions. The upper-bound argument says that no set element lies above $\alpha$. The approximation argument says that the set does not stay a positive distance below $\alpha$. A proof that includes only the first part may identify an upper bound, but not necessarily the least one.

Suprema of Pairwise Sums

A common task is to bound every sum $s+t$, where $s$ comes from one set and $t$ comes from another. If the two sets have suprema, adding those suprema gives an upper bound. The approximation characterization shows that this bound is also the least possible one.

Theorem (Supremum of a Sum Set). Let $S,T\subseteq\mathbb{R}$ be nonempty and bounded above, and define $$ S+T=\{s+t:s\in S,\ t\in T\}. $$ Then $S+T$ is nonempty and bounded above, and $$ \sup(S+T)=\sup S+\sup T. $$

Proof. Write $\alpha=\sup S$ and $\beta=\sup T$. Since both sets are nonempty, choose $s_0\in S$ and $t_0\in T$; then $s_0+t_0\in S+T$, so $S+T$ is nonempty. For any $s\in S$ and $t\in T$, the upper-bound property of $\alpha$ and $\beta$ gives $s\leq\alpha$ and $t\leq\beta$. Adding these inequalities yields $s+t\leq\alpha+\beta$. Thus $\alpha+\beta$ is an upper bound for $S+T$.

It remains to prove that elements of $S+T$ can come arbitrarily close to $\alpha+\beta$ from below. Let $\varepsilon>0$. By the Approximation Characterization of the Supremum, there is an $s\in S$ with $\alpha-\varepsilon/2<s$, and there is a $t\in T$ with $\beta-\varepsilon/2<t$. Adding gives

\alpha+\beta-\varepsilon<s+t.

Since $s+t\in S+T$, the approximation characterization applies to $S+T$, whose upper bound $\alpha+\beta$ has just been established. It follows that $\sup(S+T)=\alpha+\beta$. $\square$

The proof relies on being able to choose the two elements independently. Choosing $s$ close to $\alpha$ does not constrain the choice of $t$, and vice versa. The split $\varepsilon/2+\varepsilon/2=\varepsilon$ ensures that the total shortfall is less than the desired tolerance.

Worked Example: The Supremum of a Sum Set

Let $S=\{x\in\mathbb{R}:x<3\}$ and $T=\{y\in\mathbb{R}:y\leq4\}$. Determine $\sup(S+T)$.

Both sets are nonempty and bounded above. The number $3$ is an upper bound for $S$, and for every $\varepsilon>0$, $3-\varepsilon/2\in S$; hence $\sup S=3$. The number $4$ is in $T$ and is an upper bound for $T$, so $\sup T=4$. The Supremum of a Sum Set theorem now gives

\sup(S+T)=\sup S+\sup T=3+4=7.

The value is consistent with the elements of the sum set: every $x+y$ is less than $3+4=7$. Moreover, for any $\varepsilon>0$, choose $x=3-\varepsilon/2\in S$ and $y=4\in T$. Then $x+y=7-\varepsilon/2>7-\varepsilon$. Thus sums approach $7$, even though $7$ itself is not in $S+T$.

Small Perturbations and Suprema

Another useful proof technique is to compare each element of one set with a corresponding element of another. If every value changes by at most a fixed amount, the supremum cannot change by more than that amount. This is a stability result: a uniform bound on all the individual errors gives a bound on the difference between the two suprema.

Theorem (Stability of the Supremum Under Bounded Perturbations). Let $S\subseteq\mathbb{R}$ be nonempty and bounded above, let $f:S\to\mathbb{R}$, and suppose there is a $\delta\geq0$ such that $$ |f(s)-s|\leq\delta\qquad\text{for every }s\in S. $$ Then $f(S)=\{f(s):s\in S\}$ is nonempty and bounded above, and $$ \left|\sup f(S)-\sup S\right|\leq\delta. $$

Proof. Let $\alpha=\sup S$. For every $s\in S$, the absolute-value bound implies $f(s)-s\leq\delta$, so $f(s)\leq s+\delta\leq\alpha+\delta$. Therefore $f(S)$ is bounded above. It is nonempty because $S$ is nonempty and $f$ is defined on $S$. Let $\gamma=\sup f(S)$.

The preceding inequalities show that $\alpha+\delta$ is an upper bound for $f(S)$, so $\gamma\leq\alpha+\delta$. For the other direction, the same absolute-value bound gives $s-f(s)\leq\delta$, and hence $s\leq f(s)+\delta\leq\gamma+\delta$ for every $s\in S$. Thus $\gamma+\delta$ is an upper bound for $S$. Since $\alpha$ is the least upper bound of $S$, $\alpha\leq\gamma+\delta$, or $\alpha-\delta\leq\gamma$. We have proved

\alpha-\delta\leq\gamma\leq\alpha+\delta.

This is equivalent to $|\gamma-\alpha|\leq\delta$, as required. $\square$

Notice the two upper-bound arguments in this proof. One compares every $f(s)$ with $\alpha+\delta$; the other compares every $s$ with $\gamma+\delta$. Together they give bounds in both directions. Establishing only the first would show that $\sup f(S)$ is not too large, but would not rule out its being much smaller than $\sup S$.

Worked Example: Controlling a Perturbed Supremum

Let $S=[-1,2]$, and define $f:S\to\mathbb{R}$ by $f(x)=x+\frac{1}{1+x^2}$. Use the bounded-perturbation theorem to bound $\sup f(S)$.

For every $x\in S$, $x^2\geq0$, so $1+x^2\geq1$. Consequently,

0<\frac{1}{1+x^2}\leq1 \qquad\text{and}\qquad |f(x)-x|=\left|\frac{1}{1+x^2}\right|\leq1.

The set $S$ is nonempty and has supremum $2$. Apply the theorem with $\delta=1$. It follows that $f(S)$ is bounded above and

\left|\sup f(S)-2\right|\leq1, \qquad\text{so}\qquad 1\leq\sup f(S)\leq3.

This estimate does not claim that either endpoint is the exact supremum. Its strength is that it obtains a guaranteed range for the supremum using only a uniform bound on the perturbation.

Worked Example: An Exact Supremum from Approximation

Determine the supremum of $A=\left\{1-\frac{1}{n}:n\text{ is a positive integer}\right\}$.

For every positive integer $n$, $1/n>0$, so $1-1/n<1$. Thus $1$ is an upper bound for $A$. Let $\varepsilon>0$. By the Archimedean Property, choose a positive integer $n$ with $n+1>1/\varepsilon$. Since both sides are positive, this implies $1/(n+1)<\varepsilon$, and therefore

1-\varepsilon<1-\frac{1}{n+1}\in A.

For clarity, the element displayed is obtained by using the positive integer $n+1$ in the definition of $A$. We have shown that for every positive $\varepsilon$, an element of $A$ is greater than $1-\varepsilon$. The Approximation Characterization of the Supremum now gives $\sup A=1$. The set has no maximum, since every one of its elements is strictly less than $1$.

Strict Bounds and a Common Pitfall

A frequent mistake is to infer that a supremum is strictly less than a number merely because every element of the set is strictly less than that number. The individual gaps may shrink toward zero. The preceding example illustrates this: every element $1-1/n$ is less than $1$, while the supremum is $1$.

Worked Example: Strictly Below Does Not Mean a Strict Supremum Bound

Let $B=\{5-1/n:n\text{ is a positive integer}\}$. Every element $b\in B$ satisfies $b<5$, because $1/n>0$. Nevertheless, $\sup B=5$.

Indeed, $5$ is an upper bound. Given any $\varepsilon>0$, choose a positive integer $n$ such that $n>1/\varepsilon$, using the Archimedean Property. Then $1/n<\varepsilon$, so

5-\varepsilon<5-\frac{1}{n}\in B.

The Approximation Characterization of the Supremum proves that $5=\sup B$. Thus the valid general conclusion from “every element is less than $5$” is that $5$ is an upper bound. To conclude that the supremum is strictly less than $5$, one needs a uniform gap: for example, a number $c<5$ that is itself an upper bound.

When proving a strict estimate for a supremum, look for a uniform margin rather than relying on pointwise strict inequalities. If there is a fixed $\eta>0$ such that every $s\in S$ satisfies $s\leq b-\eta$, then $b-\eta$ is an upper bound and $\sup S\leq b-\eta<b$. Without such a margin, elements may approach $b$ as closely as desired.

Key takeaway. A supremum proof has two jobs: establish an upper bound and show that elements approach the proposed bound. For sums, approximate each supremum closely enough that the combined error stays controlled. For perturbations, compare upper bounds in both directions. Pointwise strict inequalities alone do not guarantee a strict inequality for the supremum.

Check Your Understanding

Use the proof strategies in this tutorial to answer the following questions.

In the Supremum of a Sum Set theorem, why are nonemptiness and boundedness of both sets needed?
If $\sup S=6$ and $\sup T=-2$, what is $\sup(S+T)$, assuming $S$ and $T$ are nonempty and bounded above?
Suppose $|f(s)-s|\leq\frac{1}{4}$ for every $s\in S$, and $\sup S=3$. What interval must contain $\sup f(S)$?
Give a reason that knowing $s<8$ for every $s\in S$ does not by itself prove $\sup S<8$.
In the stability proof, why does showing $s\leq\gamma+\delta$ for every $s\in S$ give a lower bound for $\gamma=\sup f(S)$?

What You'll Learn