You see (3, 3) on a worksheet and freeze. The two endpoints are equal. Is this even legal notation? If it is, what numbers does it contain — zero, or one, or is this a typo the teacher missed?

Here is the short answer, and it is the one to lock in first:

(3, 3) is valid, legal, unambiguous notation. It denotes the empty set. No real number belongs to it.

The notation is fine. The set is just empty. Those two facts do not contradict — an empty set is still a set, and writing down a notation that names the empty set is no more strange than writing \{x \in \mathbb{R} : x^2 = -1\} or \{x : x \ne x\}. The name is well-formed; the thing it names happens to have nothing in it.

Unpack the definition

Every open interval (a, b) is defined as

(a, b) = \{x \in \mathbb{R} : a < x < b\}.

Plug in a = b = 3. You get

(3, 3) = \{x \in \mathbb{R} : 3 < x < 3\}.

Read the condition aloud: "x is strictly greater than 3 and x is strictly less than 3." Can any real number satisfy both at once? No. If x > 3 then x \ne 3 and x is bigger than 3, so x cannot also be smaller than 3. The condition is self-contradictory for every real x.

A set defined by a contradiction contains nothing. So

(3, 3) = \varnothing.

That is the entire derivation. The notation parses; the set is empty because the defining condition is unsatisfiable.

The four cases at a = b = 3, side by side

The interesting thing is how sensitive this is to the brackets. Swap one round bracket for a square bracket and the answer can change dramatically. All four cases with a = b = 3:

Notation Defining condition Elements Size
[3, 3] 3 \le x \le 3 \{3\} one element
(3, 3) 3 < x < 3 \varnothing empty
[3, 3) 3 \le x < 3 \varnothing empty
(3, 3] 3 < x \le 3 \varnothing empty

Only the fully closed interval [3, 3] contains anything — exactly the point 3 itself, because 3 \le 3 and 3 \le 3 both hold. The other three cases all require at least one strict inequality that the point 3 cannot satisfy against itself, so each is empty.

The four intervals with coincident endpoints at 3Four number lines stacked vertically, each showing only the point 3. Row one shows the closed interval 3 comma 3 in square brackets, drawn as a single filled circle at 3, and labelled on the right as containing one element, the number 3. Row two shows the open interval 3 comma 3 in round brackets, drawn as a single hollow circle at 3, labelled on the right as the empty set. Row three shows the half-open interval 3 closed to 3 open, also an empty set. Row four shows 3 open to 3 closed, also empty. All four rows have the interval notation on the left and the resulting set on the right. [3, 3] 3 = {3} (3, 3) 3 = ∅ [3, 3) 3 = ∅ (3, 3] 3 = ∅ Only [3, 3] contains the point 3. The other three all have at least one strict bracket at 3, so the point is excluded — from the only candidate — and the set is empty.
The four intervals with both endpoints at $3$. A filled circle represents "point included" (square bracket); a hollow circle represents "point excluded" (round bracket). Only when both brackets are square does any real number survive — the single point $3$. Any round bracket at $3$ kicks $3$ out of the set, and since $3$ is the only candidate, nothing remains.

Why [3, 3] is a point and (3, 3) is empty

The difference between [3, 3] and (3, 3) is a single-character change, but the sets they name are as different as they can possibly be: one has one element, the other has zero. Why is the behaviour so sharp?

Think of the brackets as two independent questions about the endpoints:

When a < b the two endpoints are different points, and each question has its own answer. The "body" of the interval — everything strictly between — does not care about either bracket.

When a = b the two endpoints are the same point, and the two questions collapse into one: "Is the single point 3 allowed?" For 3 to survive, the answer must be yes on both sides simultaneously. That only happens when both brackets are square. Any round bracket kicks the only candidate out, leaving nothing.

When a = b there is no "body" — nothing strictly between 3 and 3 — so the endpoint is the only thing the interval could contain. Exclude it, and \varnothing is what remains.

A set can be named without being inhabited

A source of confusion is the feeling that writing (3, 3) is somehow asking a question with no answer. But the empty set is a genuine, well-defined object: it has a name (\varnothing), it is a subset of every set, and it can be produced by perfectly legal set-builder expressions.

The notation (3, 3) is one such producer. It is the set-builder expression \{x : 3 < x < 3\} under a compact name. The fact that no element satisfies the condition does not invalidate the name any more than "the set of cricketers who bowled left-handed and right-handed in the same delivery" — a perfectly clear description that happens to pick out no one.

If you want more on this distinction, see Can an Interval Be Empty — and When?.

When does (a, b) with a = b actually show up?

It seems artificial, but coincident endpoints arise naturally in two places:

Solving inequalities that turn out to have no solution. If you solve |x - 3| < 0, the translation gives -0 < x - 3 < 0, which is 3 < x < 3, i.e., (3, 3) = \varnothing. And the answer is correct — |x - 3| is always \ge 0, so |x - 3| < 0 has no solutions.

Limiting arguments. If you shrink an interval (3 - \varepsilon, 3 + \varepsilon) by letting \varepsilon \to 0, the open intervals shrink towards (3, 3) = \varnothing. Meanwhile the closed intervals [3 - \varepsilon, 3 + \varepsilon] shrink towards [3, 3] = \{3\}. The fact that one limit is empty and the other is a single point is a hint at why closed intervals are the ones that have nice limiting behaviour (they are compact; open ones are not).

The self-test

Write the set described by each notation.

If all five felt automatic, you own this distinction. If any felt like a guess, re-read the rule at the top and rerun the set-builder translation.

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