The author proved the following exercise using a proof by contradiction, but I think it can be accomplished using a direct proof, so I was wondering if someone would verify that my argument is sound.


Suppose $A$ is a nonempty set of reals that is bounded below. Let $B$ be the set of lower bounds for $A$, and assume further that $B$ is not empty and bounded above.

Prove that $\sup B \in B$.

Before proving the statement, I present the following lemma.

Lemma Let $x,y \in \mathbb{R}$ such that $x \leq y + \epsilon$ for every $\epsilon > 0$. Then $x \leq y$.


Since $A$ is not empty and bounded below, the completeness axiom implies that there exists a real number $\psi$ such that $\psi = \inf A$. Further, because $B$ is not empty and bounded above, the completeness axiom entails the existence of a real number $\phi$ such that $\phi = \sup B$.

Now choose any $\epsilon > 0$. Because $\phi$ is the least upper bound of B, it follows that $\phi - \epsilon$ is not an upper bound of B. Thus there exists an $x \in B$ such that

$$\phi - \epsilon < x$$

Since $x \in B$, it follows by definition of $B$ that $x$ is a lower bound of A. Due to the fact that $\psi$ is the greatest lower bound of A, it follows that

$$x \leq \psi$$

By transitivity, we deduce that $\phi - \epsilon < \psi$, which implies that

$$\phi < \psi + \epsilon$$

From the lemma we conclude that

$$\phi \leq \psi$$ From the last inequality, we conclude that $\phi$ is a lower bound of A, i.e. $\phi = \sup B \in B$, as desired.

  • $\begingroup$ Note: I understand that $\phi$ and $\psi$ are typically used to denote either angle degrees or as functions due to convention, but I like using them because of how they look haha. $\endgroup$
    – J. Dunivin
    Oct 2 '15 at 21:18
  • $\begingroup$ "Let B be the set of lower bounds for A, and assume further that B is not empty and bounded above." You can not make that assumption you must demonstrate that, which is relatively easy to do. (Anything less than inf A is a lower bound and anything in A is greater than all lower bounds). Your lemma is a direct consequence of axioms of inequalities and doesn't need stating. Otherwise your proof seems sound and valid. (Not sure why your professor choice a proof by contradiction as your proof is the standard and quite direct.) $\endgroup$
    – fleablood
    Oct 2 '15 at 21:41
  • $\begingroup$ @fleablood The exercise that is in my post is the second part of a three part exercise, where part one is to prove that $B \neq \emptyset$ and $B$ is bounded above. So, I included the assumption without proof for convenience. $\endgroup$
    – J. Dunivin
    Oct 2 '15 at 21:56
  • $\begingroup$ Votaire. Never mind then :) $\endgroup$
    – fleablood
    Oct 2 '15 at 22:58
  • $\begingroup$ Instead of saying $\phi - \epsilon$ just say "for any y < $\phi$". We know R isn't bounded below so we know y exists. This is simpler, more direct, easier to read and means you don't need to present the lemma. $\endgroup$
    – fleablood
    Oct 2 '15 at 23:05

The proof is correct except a possible minor modification. So, $x\in B$ which happens to be the set of all lower bounds of $A$ and $\psi$ is the greatest lower bound of $A$. This implies $x\leq \psi$ (contrary to strict inequality in your proof.) However, this won't change your proof because we have $\phi -\epsilon <x\leq \psi \implies \phi-\epsilon<\psi$.

  • $\begingroup$ ah, good point. Thank you! $\endgroup$
    – J. Dunivin
    Oct 2 '15 at 21:34
  • $\begingroup$ Also, you can use a stronger lemma which states that x=y with the same hypothesis. This will also prove that sup (B) =inf (A)! $\endgroup$
    – Kashi
    Oct 2 '15 at 21:40

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