Intervals and the Real Line
Intervals and the Real Line
The humble interval — a connected piece of the real line — is the building block of all of analysis. Every limit, derivative, and integral is ultimately a statement about what happens on some interval. In this post we define all six types rigorously, draw them on the number line, connect them to absolute value inequalities, and prove the celebrated Nested Interval Property that depends on the completeness of $\mathbb{R}$.
🇮🇳 हिंदी में पढ़ेंDefinition — All Interval Types
Let $a, b \in \mathbb{R}$ with $a < b$. The four bounded intervals with endpoints $a$ and $b$ are:
Open: $(a,b) = \{x \in \mathbb{R} : a < x < b\}$ Closed: $[a,b] = \{x \in \mathbb{R} : a \leq x \leq b\}$
Half-open: $[a,b) = \{x \in \mathbb{R} : a \leq x < b\}$ $(a,b] = \{x \in \mathbb{R} : a < x \leq b\}$
The unbounded intervals: $(a,\infty)$, $[a,\infty)$, $(-\infty,b)$, $(-\infty,b]$, and $\mathbb{R} = (-\infty,\infty)$.
Degenerate cases: $[a,a] = \{a\}$ (singleton, length 0); $[a,b] = \emptyset$ if $a > b$.
| Notation | Name | Endpoints included | Bounded? | Length |
|---|---|---|---|---|
| $(a,b)$ | Open | Neither | Yes | $b-a$ |
| $[a,b]$ | Closed | Both | Yes | $b-a$ |
| $[a,b)$ | Half-open (left-closed) | Left only | Yes | $b-a$ |
| $(a,b]$ | Half-open (right-closed) | Right only | Yes | $b-a$ |
| $(a,\infty)$ | Unbounded open | Neither | No | $\infty$ |
| $[a,\infty)$ | Unbounded closed | Left only | No | $\infty$ |
| $(-\infty,b)$ | Unbounded open | Neither | No | $\infty$ |
| $(-\infty,b]$ | Unbounded closed | Right only | No | $\infty$ |
| $\{a\}=[a,a]$ | Degenerate | Both (same) | Yes | $0$ |
← Sets and Basic Notation — intervals are subsets of $\mathbb{R}$; set-builder notation used throughout.
← Order Relations and Ordered Sets — intervals are defined using the order $<$ on $\mathbb{R}$.
← Absolute Value and the Real Line — $\lvert x-c \rvert < r$ translates directly into the interval $(c-r, c+r)$.
Visualisation and the Nested Interval Property
An interval is the simplest possible "connected" subset of the real line — cut the line at two points and take what's between. The only question is whether to include the cut-points (endpoints) or not, giving the four bounded types. On a number line, open endpoints are drawn as hollow circles and closed endpoints as filled circles.
Absolute value to interval: For $c \in \mathbb{R}$ and $r > 0$:
$$\lvert x - c \rvert < r \;\Longleftrightarrow\; x \in (c-r,\; c+r)$$
$$\lvert x - c \rvert \leq r \;\Longleftrightarrow\; x \in [c-r,\; c+r]$$
This translation is the key step in every $\varepsilon$-$\delta$ argument in analysis.
Nested Interval Property (Bartle & Sherbert, Ch. 2, §2.5, Thm. 2.5.2):
If $[a_1,b_1] \supseteq [a_2,b_2] \supseteq [a_3,b_3] \supseteq \cdots$ are closed bounded intervals with $b_n - a_n \to 0$, then there exists a unique $\xi \in \mathbb{R}$ with $\displaystyle\bigcap_{n=1}^{\infty}[a_n,b_n] = \{\xi\}$.
Key: The property requires closed intervals. Counterexample for open: $\displaystyle\bigcap_{n=1}^{\infty}\!\left(0,\frac{1}{n}\right) = \emptyset$.
This property holds in $\mathbb{R}$ but fails in $\mathbb{Q}$ — it is equivalent to the Completeness Axiom.
Origin Story. Bernard Bolzano (1817) was the first to use a nested bisection argument with closed intervals to prove the Intermediate Value Theorem rigorously — one of the earliest genuinely rigorous proofs in analysis. He showed that if $f(a) < 0 < f(b)$, bisecting $[a,b]$ repeatedly produces nested closed intervals converging to a root. Georg Cantor and Richard Dedekind formalised interval theory as a set-theoretic concept in the 1870s while constructing $\mathbb{R}$ from $\mathbb{Q}$.
The Problem It Solved. Before Bolzano, the Intermediate Value Theorem was considered "geometrically obvious" but had no rigorous proof. The nested interval method gave analysis its first tool for proving existence of points purely algebraically, without appeal to geometric intuition.
Interesting Fact. The Nested Interval Property is equivalent to the Completeness Axiom of $\mathbb{R}$. In $\mathbb{Q}$, consider $[1.4, \sqrt{2}+\tfrac{1}{n}]$ — these nest and shrink to $\sqrt{2}$, which is not in $\mathbb{Q}$. So the intersection is empty in $\mathbb{Q}$, showing that $\mathbb{Q}$ is fundamentally incomplete. The same bisection algorithm that finds roots in $\mathbb{R}$ would "fail" in $\mathbb{Q}$.
Why It Matters Today. The bisection method is a standard root-finding algorithm in numerical analysis and computer science; its correctness relies directly on the Nested Interval Property of $\mathbb{R}$, and it is a standard CSIR NET / IIT JAM result.
अंतराल (Interval) वास्तविक रेखा का एक सम्बद्ध भाग है। खुला (Open) $(a,b)$: दोनों अन्त-बिंदु शामिल नहीं; बंद (Closed) $[a,b]$: दोनों शामिल; अर्ध-खुला (Half-open): एक शामिल, एक नहीं। असीमित (Unbounded): $[a,\infty)$, $(-\infty,b)$ आदि। परम मान से अंतराल: $\lvert x-c \rvert < r \Leftrightarrow x \in (c-r, c+r)$। नेस्टेड अंतराल गुण (Nested Interval Property): यदि $[a_n,b_n]$ बंद अंतरालों का घटता अनुक्रम हो और $b_n-a_n \to 0$, तो उनका प्रतिच्छेदन एकल बिंदु है — यह $\mathbb{R}$ की पूर्णता का परिणाम है।
Solved Examples
(a) $\{x : 2 < x \leq 5\} = (2,5]$ — half-open (right-closed), bounded, length $3$.
(b) $\{x : x \geq -3\} = [-3,\infty)$ — unbounded closed (on left), unbounded.
(c) $[-1,-1] = \{-1\}$ — degenerate (singleton), bounded, length $0$.
(d) $\lvert x \rvert \leq 4 \Leftrightarrow -4 \leq x \leq 4$, so $[-4,4]$ — closed, bounded, length $8$. $\blacksquare$
$\lvert 2x - 3 \rvert < 5$
$\Leftrightarrow -5 < 2x - 3 < 5$
$\Leftrightarrow -2 < 2x < 8$
$\Leftrightarrow -1 < x < 4$
The set is the open interval $(-1, 4)$, with centre $c = \frac{-1+4}{2} = \frac{3}{2}$ and radius $r = \frac{5}{2}$, confirming $\lvert x - \frac{3}{2} \rvert < \frac{5}{2}$. $\blacksquare$
$(1,5) \cap [3,7]$: We need $x > 1$, $x < 5$, $x \geq 3$, and $x \leq 7$. So $3 \leq x < 5$, giving $[3,5)$. The left endpoint $3$ is included (from $[3,7]$); right endpoint $5$ is excluded (from $(1,5)$).
$[2,4] \cup [3,6]$: The intervals overlap on $[3,4]$. The union spans from $2$ (included, from $[2,4]$) to $6$ (included, from $[3,6]$), giving $[2,6]$.
$(0,2) \cap [2,5)$: $(0,2)$ requires $x < 2$; $[2,5)$ requires $x \geq 2$. No $x$ satisfies both, so the intersection is $\emptyset$. $\blacksquare$
Part (a) — Nested: $a_n = \frac{1}{n}$ is decreasing ($a_{n+1} = \frac{1}{n+1} < \frac{1}{n} = a_n$) and $b_n = 1-\frac{1}{n}$ is increasing ($b_{n+1} = 1-\frac{1}{n+1} > 1-\frac{1}{n} = b_n$). Therefore $[a_{n+1}, b_{n+1}] \supseteq [a_n, b_n]$, i.e. the sequence is expanding. So in the standard NIP direction ($I_{n+1} \subseteq I_n$) the sequence is NOT nested — it expands.
Part (b) — Intersection: Since the sequence is expanding, $\bigcap_{n=2}^{\infty} I_n = I_2 = \left[\frac{1}{2}, \frac{1}{2}\right] = \left\{\frac{1}{2}\right\}$.
Part (c) — NIP: The standard Nested Interval Property requires $I_{n+1} \subseteq I_n$. Here the containment is reversed, so NIP does not apply. However, the intersection is still a single point $\{\frac{1}{2}\}$ because the expanding intervals share $\frac{1}{2}$ (the midpoint of all $I_n$) as their common element from $I_2$ onwards.
Note: The union $\bigcup_{n=2}^{\infty} I_n = (0,1)$ — as $n\to\infty$, the intervals expand to fill $(0,1)$ but never reach the endpoints $0$ or $1$. $\blacksquare$
Quick Revision Cards
📊 A — Key Notation & Formulae
- $(a,b)$: open, $a < x < b$
- $[a,b]$: closed, $a \leq x \leq b$
- $[a,b)$, $(a,b]$: half-open
- $(a,\infty)$, $(-\infty,b]$: unbounded
- $\lvert x-c \rvert < r \Leftrightarrow x \in (c-r,c+r)$
- $\lvert x-c \rvert \leq r \Leftrightarrow x \in [c-r,c+r]$
- NIP: $\bigcap [a_n,b_n] = \{\xi\}$ if closed, nested, lengths $\to 0$
⚙️ B — Edge Cases & Conditions
- $[a,b]=\emptyset$ if $a>b$; $[a,a]=\{a\}$
- NIP needs closed intervals; $\bigcap(0,1/n)=\emptyset$
- NIP fails in $\mathbb{Q}$ (completeness required)
- $\infty$ is NOT a real number: $\infty \notin [0,\infty)$
- $(a,b)\cap(c,d)$ is open or empty
- $[a,b]\cap[c,d]$ is closed or empty
🎯 C — Exam Tips
- 🔵 CSIR NET: Convert $\lvert f(x) \rvert < \varepsilon$ to interval form as first step in any $\varepsilon$-$\delta$ proof
- 🟢 GATE: NIP fails for open intervals — classic MCQ
- 🟠 IIT JAM: For intersection/union: draw number line with filled/hollow dots
- 🔴 B.Sc. Raj.: Name each attribute (open/closed, bounded/unbounded) explicitly
Common Mistakes
❌ Errors to Avoid
Wrong: "$[5,2]$ contains real numbers between $2$ and $5$." | Correct: $[a,b] = \emptyset$ whenever $a > b$. The notation $[5,2]$ is an empty set. Interval notation requires $a \leq b$ for a meaningful non-empty interval.
Wrong: "$\{(0, 1/n)\}$ is nested with lengths $\to 0$, so $\bigcap (0,1/n) = \{0\}$." | Correct: $\bigcap_{n=1}^{\infty}(0,1/n) = \emptyset$. The Nested Interval Property requires closed intervals. Open intervals can have empty intersection even when nested and shrinking.
Wrong: "$\infty \in [0,\infty)$." | Correct: $[0,\infty) = \{x \in \mathbb{R} : x \geq 0\}$. The symbol $\infty$ indicates unboundedness; it is not a real number and is not a member of the set.
Wrong: "$(2,5]$ excludes both $2$ and $5$." | Correct: $(2,5]$ excludes $2$ (open, round bracket) and includes $5$ (closed, square bracket). Always: square bracket $[$ or $]$ means included; round bracket $($ or $)$ means excluded.
Real-World Applications
Analysis — $\varepsilon$-$\delta$ Proofs
"$f(x)$ is within $\varepsilon$ of $L$" translates to $f(x) \in (L-\varepsilon, L+\varepsilon)$. Every limit and continuity proof is an interval containment statement at its core.
Bisection Method
The bisection root-finding algorithm generates nested closed intervals $[a_n, b_n]$ halving at each step. By NIP, they converge to the unique root — the algorithm's correctness is the Nested Interval Property.
Probability Theory
For a continuous random variable $X$, $P(a \leq X \leq b) = \int_a^b f(x)\,dx$. Events are unions of intervals; the entire sample space is $\mathbb{R} = (-\infty,\infty)$.
Interval Arithmetic (CS)
IEEE 754 interval arithmetic represents each computed value as $[a,b]$, guaranteeing the true result lies within. Nested interval refinement underpins certified numerical computation.
Summary Table & Key Result
| Concept | Statement | Condition | Reference |
|---|---|---|---|
| Open interval | $(a,b) = \{x: a| $a < b$ | Rudin §1.20 | |
| Closed interval | $[a,b] = \{x: a\leq x\leq b\}$ | $a \leq b$ | Rudin §1.20 |
| Abs. val. form | $\lvert x-c\rvert < r \Leftrightarrow x\in(c-r,c+r)$ | $r>0$ | Remark |
| Length | $b - a$ | Bounded interval | Standard |
| NIP (closed) | $\bigcap[a_n,b_n] = \{\xi\}$ (unique) | Nested, $b_n-a_n\to0$ | BS §2.5 |
| NIP (open) fails | $\bigcap(0,1/n) = \emptyset$ | Counterexample | Remark |
Nested Interval Property:
$$[a_1,b_1]\supseteq[a_2,b_2]\supseteq\cdots \text{ (closed, bounded)},\quad b_n-a_n\to0 \;\Longrightarrow\; \exists!\,\xi:\;\bigcap_{n=1}^{\infty}[a_n,b_n]=\{\xi\}$$
Holds in $\mathbb{R}$ (completeness) | Fails in $\mathbb{Q}$ | Fails for open intervals
Cross-References & Related Posts
← Prerequisites: Sets and Basic Notation — intervals as subsets | Order Relations and Ordered Sets — intervals defined via order on $\mathbb{R}$ | Absolute Value and the Real Line — absolute value inequalities are interval statements.
→ Next Topic: Bounded Sets, Supremum and Infimum — intervals are bounded sets; the supremum of $(a,b)$ is $b$ (not attained); the LUB property connects directly to the Nested Interval Property.
📖 Further Reading: Rudin, Ch. 1, §1.20; Apostol, Ch. 1, §1.5–1.6; Bartle & Sherbert, Ch. 2, §2.5.
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