The proofs are exercises from Zalduendo’s Calculus off the Beaten Path.
Write this in symbolic logic:
$$s=\sup A$$ if and only if s is an upper bound of A and for each ε > 0 we may find a ε ∈ A such that s − ε < a ε ≤ s.
A compact way to put the sentence entirely in first-order logical symbols is
$$ s=\sup A \;\;\Longleftrightarrow\;\; \Bigl[\;\forall x\,(x\in A\;\to\;x\le s)\Bigr] \;\land\; \Bigl[\;\forall\varepsilon\bigl(\varepsilon>0 \;\to\; \exists x\,(x\in A \land s-\varepsilon < x\le s)\bigr)\Bigr]. $$
- “$\forall x\,(x\in A\to x\le s)$” says that s is an upper bound of A.
- “$\forall\varepsilon>0\,\exists x\in A\,(s-\varepsilon < x\le s)$” formalizes that for every $\varepsilon>0$ we can find an element of A within $\varepsilon$ of s from below.
Prove that the sentence is true.
To prove the sentence is true, we want to prove the biconditional:
$$ s = \sup A \quad \Longleftrightarrow \quad \text{"s is an upper bound of } A\text{ and } \forall \varepsilon > 0, \exists x \in A \text{ such that } s - \varepsilon < x \le s" $$
✅ Definitions
Let $A \subset \mathbb{R}$, and assume $A$ is nonempty and bounded above.
-
The supremum (least upper bound) $\sup A = s$ is defined as:
- $s$ is an upper bound: $\forall x \in A, x \le s$
- $s$ is the least such upper bound: $\forall \varepsilon > 0, \exists x \in A \text{ such that } x > s - \varepsilon$
The second condition ensures that no number strictly less than $s$ is an upper bound.
▶️ Direction 1:
Assume $s = \sup A$. Prove the logical sentence holds.
-
By definition of supremum:
- $s$ is an upper bound of $A$: $\forall x \in A, x \le s$
- $\forall \varepsilon > 0, \exists x \in A$ such that $x > s - \varepsilon$
But since $x \in A$ and $s$ is an upper bound, $x \le s$, so:
$$ s - \varepsilon < x \le s $$
Thus, both parts of the sentence are satisfied.
✅ So, the logical sentence is true if $s = \sup A$.
◀️ Direction 2:
Assume the right-hand side is true:
- $s$ is an upper bound: $\forall x \in A, x \le s$
- $\forall \varepsilon > 0, \exists x \in A \text{ such that } s - \varepsilon < x \le s$
We must show $s = \sup A$.
Step 1:
We already know $s$ is an upper bound.
Step 2:
We want to prove that $s$ is the least upper bound.
Suppose for contradiction that there is some $s' < s$ that is also an upper bound.
Let $\varepsilon = s - s' > 0$. By assumption, there exists $x \in A$ such that:
$$ s - \varepsilon < x \le s \quad \Rightarrow \quad s' < x \le s $$
This contradicts that $s'$ is an upper bound, since $x > s'$ but $x \in A$.
Thus, no number less than $s$ can be an upper bound ⇒ $s$ is the least upper bound.
✅ Therefore, $s = \sup A$.
✅ Conclusion
The statement
$$ s = \sup A \quad \Longleftrightarrow \quad \left[ \begin{array}{l} s \text{ is an upper bound of } A \\ \text{and } \forall \varepsilon > 0, \exists x \in A \text{ such that } s - \varepsilon < x \le s \end{array} \right] $$
is true. Proven in both directions.
In analogy to the definition of $$\sup A$$ define $$i=\inf A$$ (infimum of A) setting (a) i is a lower bound of A, and (b) if i < b, b is not a lower bound of A.
Definition (infimum or greatest lower bound). Let $A\subset\mathbb R$ be non–empty and bounded below. A real number $i$ is called the infimum of $A$ (denoted $i=\inf A$) iff the following two conditions hold
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Lower-bound property – $i$ is a lower bound of $A$:
$$ \forall x\in A,\; i\le x. $$
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Greatest-of-all-lower-bounds property – every number strictly larger than $i$ fails to be a lower bound:
$$ \forall b\;(i<b\;\;\Longrightarrow\;\;\exists x\in A\,(x<b)). $$
Symbolic summary
$$ i=\inf A \;\;\Longleftrightarrow\;\; \Bigl[\;\forall x\,(x\in A\to i\le x)\Bigr] \;\land\; \Bigl[\;\forall b\,(i<b\to\exists x\,(x\in A\land x<b))\Bigr]. $$
Equivalent “$\varepsilon$” formulation
Because the reals are linearly ordered, condition (2) can be rewritten with an $\varepsilon$-closeness clause, mirroring the supremum definition:
$$ i=\inf A \;\;\Longleftrightarrow\;\; \Bigl[\;\forall x\in A,\; i\le x\Bigr] \;\land\; \Bigl[\;\forall\varepsilon>0,\;\exists x\in A\,(i\le x<i+\varepsilon)\Bigr]. $$
Indeed, given $i< b$, choose $\varepsilon=b-i>0$; then the second clause supplies an element $x\in A$ with $x<b$, so $b$ cannot be a lower bound. Conversely, if the “$b$” clause holds, putting $b=i+\varepsilon$ gives the $\varepsilon$-closeness statement.
Hence the two-part description you asked for,
(a) $i$ is a lower bound of $A$, and (b) if $i<b$ then $b$ is not a lower bound of $A$,
is exactly the standard definition of the infimum.
If B is a (upper and lower) bounded set, and A ⊂ B, determine (and prove) the order of $$\displaystyle \begin{aligned} \sup A \quad \quad \quad \quad \inf A \quad \quad \quad \quad \sup B \quad \quad \quad \quad \inf B \end{aligned}$$
Let $A\subset B\subset\mathbb R$. Assume both sets are non-empty and bounded above and below, so that all four real numbers
$$ \inf B,\;\; \inf A,\;\; \sup A,\;\; \sup B $$
exist. (The chain we obtain is meaningless only if one of the sets is empty, because the infimum or supremum of an empty set is not defined in the usual real-analysis sense.)
Claim
$$ \boxed{\;\inf B\;\le\;\inf A\;\le\;\sup A\;\le\;\sup B\;} $$
Proof
-
$\inf B\le \inf A$.
- Because $\inf B$ is a lower bound of $B$, we have $\inf B\le x$ for every $x\in B$.
- Since $A\subset B$, the same inequality holds for every $x\in A$, so $\inf B$ is a lower bound of $A$.
- By definition, $\inf A$ is the greatest lower bound of $A$; hence no lower bound of $A$ can exceed $\inf A$. Therefore $\inf B\le\inf A$.
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$\inf A\le\sup A$.
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Pick any element $a\in A$ (possible because $A\neq\varnothing$).
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By the defining properties of infimum and supremum,
$$ \inf A\;\le\;a\;\le\;\sup A, $$
whence $\inf A\le\sup A$.
-
-
$\sup A\le\sup B$.
- Because $A\subset B$, every upper bound of $B$ is an upper bound of $A$. In particular $\sup B$ is an upper bound of $A$.
- By definition, $\sup A$ is the least upper bound of $A$; therefore $\sup A\le\sup B$.
Combining the three inequalities yields the announced order
$$ \inf B\;\le\;\inf A\;\le\;\sup A\;\le\;\sup B. $$
Remarks
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The two end-points $\inf B$ and $\sup B$ frame every element of $B$—and therefore every element of $A$—so it is natural that the extremal values for the smaller set $A$ are squeezed between those of $B$.
-
Equalities can occur:
- If $A=B$, then $\inf A=\inf B$ and $\sup A=\sup B$.
- If $A$ is, say, a single point $\{c\}\subset B$, then $\inf A=\sup A=c$, still inside the same chain.
If − A = {−x : x ∈ A}, determine (and prove) the relationships between $$\displaystyle \begin{aligned} \sup (-A) \quad \quad \quad \quad -\sup A \quad \quad \quad \quad \inf (-A) \quad \quad \quad \quad -\inf A \end{aligned}$$
Let $A \subset \mathbb{R}$ be a nonempty set that is bounded above and below, so that both $\sup A$ and $\inf A$ exist. Define:
$$
- A := \{ -x : x \in A \} $$
We want to determine the relationships between:
$$ \sup(-A),\quad -\sup A,\quad \inf(-A),\quad -\inf A $$
🔍 Claim
$$ \boxed{ \sup(-A) = -\inf A \quad\text{and}\quad \inf(-A) = -\sup A } $$
✅ Proof
Let’s prove each identity:
1. $\sup(-A) = -\inf A$
Step 1: Show $-\inf A$ is an upper bound of $-A$
Let $x \in A$. Then $\inf A \le x$, so $-x \le -\inf A$. Since $-x \in -A$, this shows that every element of $-A$ is $\le -\inf A$. So $-\inf A$ is an upper bound of $-A$.
Step 2: Show it's the least upper bound
Let $u < -\inf A$. Then $-u > \inf A$. Since $\inf A$ is the greatest lower bound of $A$, there exists some $x \in A$ with $x < -u$. Then $-x > u$, and $-x \in -A$, so $u$ is not an upper bound of $-A$.
Thus, $-\inf A$ is the least upper bound of $-A$, i.e.
$$ \sup(-A) = -\inf A $$
2. $\inf(-A) = -\sup A$
Step 1: Show $-\sup A$ is a lower bound of $-A$
Let $x \in A$. Then $x \le \sup A$, so $-x \ge -\sup A$. Thus, $-\sup A$ is a lower bound of $-A$.
Step 2: Show it's the greatest lower bound
Let $l > -\sup A$. Then $-l < \sup A$. Since $\sup A$ is the least upper bound of $A$, there exists some $x \in A$ with $x > -l$. Then $-x < l$, and $-x \in -A$, so $l$ is not a lower bound of $-A$.
Thus, $-\sup A$ is the greatest lower bound of $-A$, i.e.
$$ \inf(-A) = -\sup A $$
🧠 Summary
If $A \subset \mathbb{R}$ is bounded, then:
$$ \boxed{ \begin{aligned} \sup(-A) &= -\inf A \\ \inf(-A) &= -\sup A \end{aligned} } $$
These identities follow from how negation reverses the order of the real numbers.
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