Why Does the Supremum of $(0, 1)$ Exist but the Maximum Doesn't?

Here is a sentence that sounds like a paradox on first read: the set (0, 1) has a supremum equal to 1, but it has no maximum. How can the "least upper bound" exist when "the largest element" does not? Aren't those just two names for the same thing?

They are not. They look similar because English trains you to read them as synonyms, but their mathematical definitions differ by one decisive clause — and the open interval (0, 1) is the one example that pins the difference down forever. This page walks through the two definitions, checks each against (0, 1), and shows why the completeness axiom had to be stated with \sup and not \max.

The two definitions, rewritten as membership questions

Both "maximum" and "supremum" answer the question "what is the biggest thing associated with this set?" But they disagree on where that biggest thing is allowed to live.

Maximum of S. An element m is the maximum of S if:

m \in S — m is itself a member of the set, and
x \le m for every x \in S — m is at least as large as every other member.

Both clauses are required. Drop clause 1 and you no longer have a maximum; drop clause 2 and you no longer have the largest element.

Supremum of S. A real number L is the supremum of S if:

x \le L for every x \in S — L is an upper bound, and
if M is any upper bound of S, then L \le M — L is the smallest such upper bound.

Notice what is missing. Clause 1 does not say L \in S. Clause 2 does not say L \in S either. The supremum is allowed to sit outside the set. It only has to be the tightest external ceiling that still sits at or above every element of the set.

That is the whole structural difference. Maximum demands membership; supremum does not. Everything else about the two concepts follows from this one asymmetry.

Is there a maximum of (0, 1)?

The set is (0, 1) = \{x \in \mathbb{R} : 0 < x < 1\} — every real number strictly between 0 and 1. The round brackets exclude both endpoints, so 0 \notin (0, 1) and crucially 1 \notin (0, 1).

Suppose, for contradiction, that m is the maximum of (0, 1). Then m \in (0, 1), so 0 < m < 1. Consider

m' = \frac{m + 1}{2}.

This is the midpoint of m and 1. Two things are true about it:

m' > m, because m < 1 implies \tfrac{m+1}{2} > \tfrac{m+m}{2} = m. Why: averaging m with something larger than m must produce a result larger than m.
m' < 1, because m < 1 implies \tfrac{m+1}{2} < \tfrac{1+1}{2} = 1.

So m' \in (0, 1) and m' > m. But we assumed m was the largest element of (0, 1), and here is a larger one inside the set. Contradiction.

Therefore no m can be the maximum of (0, 1). The set has no maximum — not because it runs off to infinity, but because whenever you try to put your finger on the largest element, the midpoint between your finger and 1 is already larger and still in the set. The set approaches 1 but never reaches it, and there is no "last stop" on the way.

Is the supremum of (0, 1) equal to 1?

Now the other concept. Call the candidate L = 1. Check both clauses of the supremum definition.

Clause 1 — is 1 an upper bound? For every x \in (0, 1), we have x < 1, so certainly x \le 1. Yes, 1 is an upper bound. (The fact that 1 is not in the set does not disqualify it from being an upper bound — an upper bound of S only needs to live in \mathbb{R}.)

Clause 2 — is 1 the smallest upper bound? Suppose M is another upper bound of (0, 1). We need to show 1 \le M. Argue by contradiction: suppose M < 1. Then M is a real number strictly below 1, and we can find an element of (0, 1) that exceeds it. Specifically, the number

\frac{M + 1}{2}

sits strictly between M and 1 (same midpoint trick as before). It is in (0, 1) because M \ge 0 implies \tfrac{M+1}{2} \ge \tfrac{1}{2} > 0, and M < 1 implies \tfrac{M+1}{2} < 1. And it exceeds M. So M is not an upper bound after all — contradiction. Therefore every upper bound M satisfies M \ge 1.

Both clauses hold. \sup (0, 1) = 1.

Notice the asymmetry. To show 1 is the supremum we never needed 1 \in (0, 1). In fact 1 \notin (0, 1) is exactly what disqualifies it from being the maximum — but it does nothing to disqualify it from being the supremum, because the supremum is allowed to live outside.

The diagram

The open interval $(0, 1)$ with hollow circles at both endpoints. The supremum $1$ is labelled with a red arrow above the line, outside the set. Interior dots sketch a sequence like $\tfrac{1}{2}, \tfrac{2}{3}, \tfrac{3}{4}, \tfrac{4}{5}, \ldots$ piling up against the hollow dot at $1$ — approaching the supremum without ever touching it. Because no element equals $1$ and no smaller element is $\ge$ all the others, the set has no maximum.

The philosophical reason this had to be true

Look at the proof of "no maximum" again. The key step was: given any candidate m < 1, the midpoint (m+1)/2 is closer to 1 than m is, and still in the set. That construction works for any m, no matter how close to 1. The set has no "last element before 1" because the real numbers have no "next real number" — between any two distinct reals there are infinitely many more. Why: averaging two distinct reals produces a third real strictly between them, and you can repeat this forever. This is the density of \mathbb{R}.

So on the open interval, the set gets arbitrarily close to 1 but can never contain a "last point." The ceiling 1 is approached but never occupied. The maximum does not exist because the set is dense and 1 is its excluded endpoint — the two facts conspire to leave no element at the top.

The supremum, by contrast, does not care about any of this. It is defined as a property of the external real line: "what is the smallest real number \ge every element?" Density of the set and excludedness of 1 both agree that 1 is the answer. No internal candidate is needed.

Why the completeness axiom uses \sup, not \max

The completeness axiom of \mathbb{R} says:

Every non-empty subset of \mathbb{R} that is bounded above has a least upper bound (a supremum) in \mathbb{R}.

What if the axiom said "every non-empty bounded subset has a maximum"? Then (0, 1) would be a counterexample — bounded by 1, but with no maximum. The axiom would be false for the real numbers as we know them, because open intervals are a perfectly normal kind of subset.

The genius of the supremum formulation is that it is strong enough to be useful — it guarantees a single precise ceiling, which is enough to prove theorems about limits and convergence — and weak enough to be true — it does not demand that the ceiling be reached. Every bounded set of reals has a supremum; only some of them have a maximum. By phrasing the axiom around \sup, mathematicians got a law that holds universally instead of one that holds only sometimes.

This is also why you see \sup all over analysis: in definitions of lim sup, in proofs of the extreme value theorem, in the construction of integrals. The supremum is the universal tool; the maximum is the lucky-case shortcut that happens to agree with it when the set is generous enough to contain its own ceiling.

A quick sanity check against [0, 1] and (0, 1]

To make the membership clause hit home, compare (0, 1) with two close cousins.

[0, 1] — both endpoints included. Supremum is 1, and 1 \in [0, 1], so \max [0, 1] = 1. Both exist and agree.
(0, 1] — right endpoint included, left excluded. Supremum is still 1. Now 1 \in (0, 1], so \max (0, 1] = 1 as well. Both exist and agree.
(0, 1) — both endpoints excluded. Supremum is still 1. But 1 \notin (0, 1), and as shown above no element of the set is the largest. Supremum exists, maximum does not.

Flipping the right bracket from round to square does not move the supremum — it just decides whether the supremum gets to double as the maximum. The supremum is a property of where the set lives on the number line; the maximum is a property of which elements the set owns.

One-line summary

(0, 1) is bounded above, so by completeness it has a supremum — which is 1. But 1 \notin (0, 1) and no element of (0, 1) is the largest, so the maximum does not exist. One concept needs membership; the other does not.

For more on this distinction, see Bounded vs. Has a Maximum — the Difference That Trips People Up. For an interactive feel of the supremum tracking the right endpoint regardless of membership, drag the set in Supremum on a Leash.