You stare at a problem. It says: find the supremum and, if it exists, the maximum of S = (0, 5). Your pen is already moving. You write \sup S = 5. You are about to write \max S = 5 and move on to the next question.

Stop.

There is exactly one reflex that belongs in your head at this moment, and it is worth more than any theorem you will memorise for this chapter. Before you equate the supremum with the maximum, before you write "max" at all, ask the one-line question:

Is 5 actually an element of S?

If the answer is yes, you are safe — maximum and supremum agree, write both. If the answer is no, you have just dodged a silent mark-loss: the supremum exists, but the maximum does not, and claiming otherwise is not a minor slip. It is a category error. It says you have not internalised what a maximum is.

Why the reflex exists

The definitions of supremum and maximum look almost identical, and that is precisely the trap. A supremum is the least upper bound of a set — the smallest real number that sits at or above every element of the set. It is a statement about \mathbb{R}: the supremum is allowed to live anywhere on the number line, whether or not the set itself contains it. A maximum is different: it is the largest element of the set. It must be inside S to count. That one word — element — is the whole distinction.

Every supremum is an upper bound. Not every supremum is a maximum. The only way to know which you have is to check membership explicitly. There is no shortcut, no theorem that removes the check, no shape of bracket that lets you skip it. You look at the candidate — the number you computed as \sup S — and you ask whether that number appears in the set. If yes, it is also the maximum. If no, the maximum does not exist at all.

For the parent concept — why "bounded" and "has a maximum" are genuinely different — see Bounded vs. Has a Maximum — the Difference That Trips People Up. This page is about the reflex; that page is about the underlying concept.

The contrast, crystallised

Put two intervals next to each other and the reflex becomes visible.

Two intervals: closed (0, 5] has a maximum; open (0, 5) does not Two horizontal number-line snippets stacked. The top line shows the half-open interval (0, 5] with a hollow dot at 0 and a filled dot at 5; a label reads max = sup = 5. The bottom line shows the open interval (0, 5) with hollow dots at both 0 and 5; a label reads sup = 5 but no max. (0, 5] — sup = max = 5 (5 is in the set) 0 5 5 ∈ S ✓ → max exists (0, 5) — sup = 5 but no max (5 is NOT in the set) 0 5 5 ∉ S ✗ → no max
Same supremum, different verdicts. The filled dot on the top line says "the boundary is an element" — so the maximum exists and equals the supremum. The hollow dot on the bottom line says "the boundary is not an element" — so the supremum still exists, but there is no maximum at all.

The two number-line pictures differ by exactly one dot: filled vs. hollow at the right endpoint. That single graphical difference is the entire algebraic difference between "max = 5" and "max does not exist." Once you can see the dot, you can apply the reflex.

A trickier case where the trap bites harder

Intervals make the trap obvious because the bracket notation telegraphs membership. The trap is sharper when the set is written implicitly — as a rule or a formula — because you have to work out the membership yourself.

Consider S = \{\,1 - \tfrac{1}{n} \;:\; n \in \mathbb{N}\,\} = \{0,\ \tfrac{1}{2},\ \tfrac{2}{3},\ \tfrac{3}{4},\ \tfrac{4}{5},\ \ldots\}.

Compute the supremum. As n \to \infty, the values climb toward 1 without ever arriving. For any candidate upper bound strictly less than 1 — say 0.9999 — you can find an n large enough that 1 - 1/n exceeds it (take n = 10^6 and you get 0.999999, which is larger than 0.9999). So no number below 1 is an upper bound. And 1 itself is an upper bound, because 1 - 1/n < 1 for every natural n. Hence \sup S = 1.

Now apply the reflex. Is 1 an element of S? To be in S it must equal 1 - 1/n for some n \in \mathbb{N}. That would force 1/n = 0, which no natural n satisfies. So 1 \notin S. The supremum exists, and the maximum does not. Writing "\max S = 1" here is wrong — the sentence type-checks, but the mathematical claim behind it is false.

Notice how much more temptation there is to make the error in this form. There is no bracket to warn you; you have to derive the fact that 1 is out of reach from the structure of the formula. The reflex is doing real work here.

How to execute the reflex in under ten seconds

On every problem that asks for "maximum" or "sup = max?", interleave these three steps between the computation and the answer:

  1. Compute your candidate ceiling L — the supremum.
  2. Ask: does L satisfy the defining condition of S? (Plug it in. Check the bracket. Check the formula.)
  3. If yes, write \sup S = \max S = L. If no, write \sup S = L and \max S does not exist.

Step 2 is the reflex. It is a membership check, and it takes a single line of scratch work. Skipping it is how careful students lose marks on problems they actually understood.

The diagnostic phrase

In your own handwriting, in the margin of rough work, write the phrase "is L in S?" every single time you are about to declare a maximum. It is the same phrase every time. It looks trivial. After three problems it will feel automatic, and automatic is exactly what you want — because once the reflex is automatic, you stop losing marks on this class of question entirely.

Related: Bounded vs. Has a Maximum — the Difference That Trips People Up · Real Numbers — Properties · Working With Suprema? Always Ask — Least Upper Bound or Just an Upper Bound? · Bounded Above and Non-Empty? Sup Exists — Completeness in One Line