Higher Mathematics 2nd Paper · Classes 11-12 · Chapter 1 · 1.2.3-1.2.5

Intervals, Bounds and Completeness.

All the real numbers between two numbers fit inside one pair of brackets. Whether the brackets are square or round tells you whether the end numbers are in. And for a set you can look for a roof and a floor: the supremum and the infimum.

1. Four kinds of interval

If a and b are real numbers with a < b, the set of all real numbers between them is an interval. There are four kinds: [a, b], (a, b), [a, b) and (a, b].

[a, b] is closed, {x : a ≤ x ≤ b}; (a, b) is open, {x : a < x < b}; [a, b) is closed-open, {x : a ≤ x < b}; (a, b] is open-closed, {x : a < x ≤ b}. A square bracket keeps the endpoint, a round one leaves it out. On the number line a closed end is a filled dot and an open end an empty circle. Pieces can be joined too: [5, 7) ∪ (7, 10] means 5 ≤ x ≤ 10 with x ≠ 7.

Move a and b below, and open or close each end. The brackets, the set and the integers inside all change together.

For integer x: x ∈ [−3, 5] = {−3, −2, −1, 0, 1, 2, 3, 4, 5}; x ∈ (−3, 5) = {−2, −1, 0, 1, 2, 3, 4}; x ∈ [−3, 5) = {−3, ..., 4}; and x ∈ (−3, 5] = {−2, ..., 5}. Now name the interval from its inequality.

2. Roofs and floors

Let S ⊂ ℝ be non-empty. If some real number m has x ≤ m for every x ∈ S, then m is an upper bound of S and S is upper bounded. Every real number larger than an upper bound is an upper bound too, so there is not one roof but endlessly many.

The smallest of the upper bounds is the supremum, the least upper bound (lub). The supremum of S = (−1, 1) is 1.

The other way round, if m ≤ x for every x ∈ S, then m is a lower bound and S is lower bounded. Every real number smaller than a lower bound is a lower bound as well.

The largest of the lower bounds is the infimum, the greatest lower bound (glb). The infimum of S = (−1, 1) is −1.

Pick a set and drag m. Watch when m is a roof, when it is a floor, and where it has to stop to be exactly the sup or the inf.

The book's table lines several sets up. ℕ = {1, 2, 3, ...} is lower bounded: infimum 1, no supremum. {..., −3, −2, −1} is upper bounded: supremum −1, no infimum. ℤ, Q, ℝ and Q′ are not bounded and have no bounds at all.

A = {x ∈ ℝ : 2 < x ≤ 5} has sup 5 and inf 2, yet 2 ∉ A. B = [2, 3) has sup 3 and inf 2; S = {x : 0 < x < 1} has sup 1 and inf 0. So the sup or the inf need not belong to the set.

3. Working out sup and inf

For an infinite sequence, write out the first few terms. {1 − 1/n : n ∈ ℕ} = {0, 1/2, 2/3, ...} has sup 1 and inf 0. {−(n+1)/n : n ∈ ℕ} = {−2, −3/2, −4/3, ...} has sup −1 and inf −2. {(3n+2)/(2n+1) : n ∈ ℕ} = {5/3, 8/5, 11/7, ...} has sup 5/3 and inf 3/2.

Turn a quadratic inequality into an interval first. D = {x ∈ ℝ : 3x² − 10x + 3 < 0} = (1/3, 3), so inf 1/3 and sup 3. Example 13: (x − 1)(x − 2) ≤ 0 holds when x − 1 ≥ 0 and x − 2 ≤ 0; the other case, x − 1 ≤ 0 with x − 2 ≥ 0, is impossible. So A = {x : 1 ≤ x ≤ 2} = [1, 2]. Its upper bounds are {x ≥ 2}, so sup 2; its lower bounds are {x ≤ 1}, so inf 1.

In a finite set of integers or naturals, the sup is the largest member and the inf the smallest. {x ∈ ℤ : 8 ≤ x³ ≤ 125} = {2, 3, 4, 5}: sup 5, inf 2, and its lower bounds are {x ∈ ℝ : x ≤ 2}. S = {x ∈ ℕ : 4 ≤ x² ≤ 81} = {2, 3, ..., 9}: sup 9, inf 2. {x ∈ ℤ : 8 ≤ x² ≤ 27} = {±3, ±4, ±5}: sup 5, inf −5. The book's table (row 11) gives sup 3 and inf 2 for this set; the correct answer is sup 5 and inf −5, since x² = 9, 16, 25 gives x = ±3, ±4, ±5.

Every member of {x ∈ ℝ : x < a} is less than a, so a is an upper bound. Take any c < a: the number (c + a)/2 is in the set and bigger than c, so c is not an upper bound. Hence sup = a, even though a is not a member. In the same way inf{x ∈ ℝ : x > b} = b.

In a compound inequality, work on every part at once. From −2 ≤ 3 − x ≤ 8 subtract 3 to get −5 ≤ −x ≤ 5, that is −5 ≤ x ≤ 5. So S = [−5, 5], Sup S = 5 and Inf S = −5. Now find some yourself: write the set first, then the sup and the inf.

4. The completeness property

The completeness property: every non-empty upper bounded subset of ℝ has a unique supremum in ℝ, and every non-empty lower bounded subset has a unique infimum in ℝ. It holds because the real line has no gaps. In the rationals it fails.

An example: S = {x ∈ Q : x > 0 and x² > 15}. 0 is a lower bound, so S is lower bounded, but it has no greatest lower bound in Q. In ℝ, Inf S = √15, which is not rational. The book writes S = {x ∈ Q : x² > 15}; the condition x > 0 is needed, because without it the set takes in every negative x < −√15, such as −4 and −5, and then it is not lower bounded at all.

5. Check yourself

Three questions in the exam's style. The first is about the supremum.

The next one asks you to test several statements.

The last puts two sets side by side.

The whole note on eight cards.

Source: Higher Mathematics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Real Numbers and Inequalities”, section 1.2.3-1.2.5.