Proofs Involving Infima

Two Parts of an Infimum Proof

In the previous tutorial, proofs involving suprema were organized around two tasks: establish an upper bound, then show that set elements approach it. For infima, the order is reversed. We establish a lower bound and then show that elements of the set come arbitrarily close to it from above.

We will use the Approximation Characterization of the Infimum from the earlier tutorial “Infimum.” If $S$ is nonempty and bounded below, and $\beta=\inf S$, then every $s\in S$ satisfies $\beta\leq s$, and for every $\varepsilon>0$ there is an $s\in S$ such that $s<\beta+\varepsilon$. The first statement says that $\beta$ is a lower bound; the second rules out any larger lower bound.

Proof strategy. To prove that $\beta=\inf S$, show that $\beta$ is a lower bound for $S$, and then show that for every $\varepsilon>0$, some $s\in S$ satisfies $s<\beta+\varepsilon$. The Approximation Characterization of the Infimum then identifies $\beta$ as the infimum.

These two steps have different purposes. If we only show that $\beta$ is a lower bound, we have not shown it is the greatest lower bound: a smaller number might also be a lower bound. Approximation shows that the set has elements close enough to $\beta$ to rule out every number strictly above it as a lower bound.

Infima of Pairwise Sums

Suppose each element of a set $S+T$ is formed by adding an element of $S$ to an element of $T$. Lower bounds for $S$ and $T$ add to give a lower bound for $S+T$. To prove this is the infimum, approximate each infimum closely enough that the two errors together are smaller than a prescribed tolerance.

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

Proof. Write $\alpha=\inf S$ and $\beta=\inf T$. Since $S$ and $T$ 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 lower-bound property of $\alpha$ and $\beta$ gives $\alpha\leq s$ and $\beta\leq t$. Adding these inequalities gives $\alpha+\beta\leq s+t$. Thus $\alpha+\beta$ is a lower bound for $S+T$, which also shows that $S+T$ is bounded below.

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

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

Since $s+t\in S+T$, for every $\varepsilon>0$ there is an element of $S+T$ less than $\alpha+\beta+\varepsilon$. Together with the lower-bound argument, the Approximation Characterization of the Infimum gives $\inf(S+T)=\alpha+\beta$. $\square$

The choices of $s$ and $t$ can be made independently because the definition of $S+T$ allows any pair $s\in S$, $t\in T$. The split $\varepsilon/2+\varepsilon/2=\varepsilon$ ensures that the two approximation errors combine to no more than the desired tolerance. More generally, any two positive error allowances whose sum is $\varepsilon$ would work.

Worked Example: The Infimum of a Sum Set

Let $S=\{x\in\mathbb{R}:x>2\}$ and $T=[-3,1]$. Determine $\inf(S+T)$.

The number $2$ is a lower bound for $S$. For every $\varepsilon>0$, $2+\varepsilon/2\in S$, so the Approximation Characterization of the Infimum gives $\inf S=2$. The number $-3$ belongs to $T$ and is a lower bound for $T$, so $\inf T=-3$. The sets are both nonempty and bounded below. The Infimum of a Sum Set theorem therefore gives

\inf(S+T)=\inf S+\inf T=2+(-3)=-1.

This value can also be checked directly. If $x>2$ and $y\geq-3$, then $x+y>-1$, so $-1$ is a lower bound for the sum set. Given any $\varepsilon>0$, choose $x=2+\varepsilon/2$ and $y=-3$. Then $x\in S$, $y\in T$, and $x+y=-1+\varepsilon/2<-1+\varepsilon$. Thus elements of the sum set approach $-1$ from above.

Uniform Perturbations and Infima

A second useful technique compares each element of a set with a corresponding perturbed value. When the difference is uniformly bounded, the infimum of the perturbed set cannot move very far from the original infimum. The proof uses lower bounds in both directions: first bound every perturbed value from below, then use the infimum of the image to bound every original value from below.

Theorem (Stability of the Infimum Under Bounded Perturbations). Let $S\subseteq\mathbb{R}$ be nonempty and bounded below, 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 below, and $$ \left|\inf f(S)-\inf S\right|\leq\delta. $$

Proof. Let $\alpha=\inf S$. The absolute-value bound implies $f(s)-s\geq-\delta$, so $f(s)\geq s-\delta$. Since $\alpha\leq s$ for every $s\in S$, it follows that $f(s)\geq\alpha-\delta$. Thus $\alpha-\delta$ is a lower bound for $f(S)$. The image is nonempty because $S$ is nonempty and $f$ is defined on $S$; it is therefore nonempty and bounded below, and its infimum $\gamma=\inf f(S)$ exists.

Because $\alpha-\delta$ is a lower bound for $f(S)$, we have $\alpha-\delta\leq\gamma$. For the other direction, the absolute-value bound also gives $f(s)-s\leq\delta$, or $s\geq f(s)-\delta$. Since $\gamma$ is a lower bound for $f(S)$, $\gamma\leq f(s)$ for every $s\in S$. Hence

s\geq f(s)-\delta\geq\gamma-\delta \qquad\text{for every }s\in S.

Thus $\gamma-\delta$ is a lower bound for $S$, so $\gamma-\delta\leq\alpha$. Combining the two bounds gives

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

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

Both comparisons matter. The first shows the perturbed infimum is not too small; the second shows it is not too large. A one-sided comparison would establish only one side of the estimate. Notice also that $\delta=0$ is allowed: in that case $f(s)=s$ for every $s\in S$, and the conclusion says the infima are equal.

Worked Example: Bounding a Perturbed Infimum

Let $S=[1,3]$, and define $f:S\to\mathbb{R}$ by $f(x)=x+\frac{1}{1+x^2}$. Use the bounded-perturbation theorem to bound $\inf f(S)$.

For $x\in[1,3]$, $x\geq1$, so $x^2\geq1$ and $1+x^2\geq2$. Therefore

0<\frac{1}{1+x^2}\leq\frac{1}{2}, \qquad |f(x)-x|=\frac{1}{1+x^2}\leq\frac{1}{2}.

The set $S$ is nonempty and has infimum $1$. Apply the theorem with $\delta=1/2$. The image $f(S)$ is nonempty and bounded below, and

\left|\inf f(S)-1\right|\leq\frac{1}{2}, \qquad\text{so}\qquad \frac{1}{2}\leq\inf f(S)\leq\frac{3}{2}.

This is a guaranteed estimate, not a claim that either endpoint is the exact infimum. The theorem uses only the uniform bound on the perturbation, so it can be applied without first determining the minimum or infimum of the formula for $f$.

Pointwise Strict Inequalities and a Common Pitfall

Knowing that every element of a set is strictly greater than a number does not necessarily mean the infimum is strictly greater than that number. The positive gaps may shrink toward zero. To prove a strict lower bound for an infimum, it is enough to find a uniform margin: a number $c$ strictly greater than the proposed bound such that every set element is at least $c$.

Worked Example: Strictly Above Does Not Mean a Strict Infimum Bound

Let $C=\{-4+1/n:n\text{ is a positive integer}\}$. Every $c\in C$ satisfies $c>-4$, since $1/n>0$. Nevertheless, $\inf C=-4$.

First, $-4$ is a lower bound because $1/n>0$ implies $-4<-4+1/n$ for every positive integer $n$. Now let $\varepsilon>0$. By the Archimedean Property, choose a positive integer $n$ with $n>1/\varepsilon$. Both sides are positive, so this inequality implies $1/n<\varepsilon$. Consequently,

-4+\frac{1}{n}<-4+\varepsilon.

The element on the left belongs to $C$. Thus, for every $\varepsilon>0$, an element of $C$ is less than $-4+\varepsilon$. The Approximation Characterization of the Infimum now gives $\inf C=-4$. The set has no minimum because none of its elements equals $-4$.

A uniform margin changes the conclusion. If every $s\in S$ satisfies $s\geq a+\eta$ for some fixed $\eta>0$, then $a+\eta$ is a lower bound for $S$, and therefore $\inf S\geq a+\eta>a$. By contrast, separate inequalities $s>a$ provide no such fixed margin unless one is established.

Key takeaway. An infimum proof requires a lower bound and an approximation argument from above. For sums, split the allowed error between the two sets. For uniformly perturbed values, compare lower bounds in both directions. Pointwise strict inequalities alone do not guarantee a strict inequality for the infimum.

Check Your Understanding

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

In the Infimum of a Sum Set theorem, why does $\inf S+\inf T$ give a lower bound for every element of $S+T$?
If $\inf S=4$ and $\inf T=-7$, what is $\inf(S+T)$, assuming both sets are nonempty and bounded below?
Suppose $|f(s)-s|\leq 2$ for every $s\in S$ and $\inf S=5$. What interval must contain $\inf f(S)$?
Why does $s>-2$ for every $s\in S$ not by itself prove that $\inf S>-2$?
In the perturbation proof, why does showing $s\geq\gamma-\delta$ for every $s\in S$ give an inequality involving $\inf S$?

What You'll Learn