Real analysis — rigorous epsilon-delta, metric spaces: Worked examples

4.10.22 · D3 · Maths › Advanced Topics (Elite Level) › Real analysis — rigorous epsilon-delta, metric spaces

Scenario matrix

Har $ε$ – $δ$ / metric problem in cells mein se ek hoti hai. Hum cells ko tags Cell A … Cell I se label karte hain (yeh tags sirf is table ki rows ke naam hain — page par kisi aur cheez ke liye use nahi hote). Hum neeche har ek ko kam se kam ek baar cover karenge.

Cell	Case class	Ise kya alag banata hai	Example
Cell A	Linear $f$ (constant slope)	multiplier $M$ already constant hai — koi caging nahi	Ex 1
Cell B	Quadratic / polynomial (variable slope)	multiplier bound karne ke liye $∥ x - a ∥ < 1$ cap karna padta hai	Ex 2
Cell C	Root / reciprocal-type (algebra zaroori)	$∥ x - a ∥$ expose karne ke liye rationalise ya invert karo	Ex 3
Cell D	Degenerate limit point ( $a = 0$ )	dono sides par signs; punctured $0 < ∥ x ∥$	Ex 4
Cell E	Non-uniform continuity	dikhaao ki koi bhi single $δ$ kaam nahi karta (negation)	Ex 5
Cell F	Sequence limit in $R$ ( $δ$ ki jagah $N$ )	discrete index, "large $n$ ke liye"	Ex 6
Cell G	Non-Euclidean metric (taxicab / discrete)	distance axioms, unusual balls	Ex 7
Cell H	Real-world word problem	tolerance → $ε$ , setting → $δ$ translate karo	Ex 8
Cell I	Exam twist (limit exist nahi karta)	non-existence contradiction se prove karo	Ex 9

Intuition Lagbhag har cell ke peeche ek hi recipe

Zyaadatar cells (A, B, C, D, H) exactly usi five-move loop ko follow karti hain. Ise abhi ek baar padho; neeche har example bas is loop ka specialized version hai. Flow mein plain step names use hote hain (cell tags nahi), isliye kuch bhi reuse nahi ho raha:

graph LR
  S1["pick epsilon"] --> S2["factor out distance"]
  S2 --> S3["bound the multiplier M"]
  S3 --> S4["set delta from epsilon over M"]
  S4 --> S5["verify implication holds"]

Ex 1 mein "bound M" step skip hota hai (M already constant hai); Ex 2 full loop hai; Ex 3 mein "factor" step rationalising se hota hai; Ex 8 same loop hai jo ek word problem ki tarah dressed hai. Cells E, F, G, I deliberately loop ko tod dete hain aur unhe apne methods se handle kiya jaata hai.

Cell A — linear, constant slope

Worked example

Ex 1. Prove karo ki $x \to 5 lim (4 x + 2) = 22$ . Forecast: slope $4$ hai. Guess karo: kya output error input error se $4 \times$ tezi se badhti hai, toh $δ = ε /4$ ?

Target gap likho. $∣ f (x) - L ∣ = ∣ (4 x + 2) - 22∣ = ∣4 x - 20∣$ . Yeh step kyun? Poora game $∣ f (x) - L ∣$ ki size ke baare mein hai; pehle hum ise simplify karte hain.
$∣ x - 5∣$ factor out karo. $∣4 x - 20∣ = 4 ∣ x - 5∣$ . Kyun? $δ$ sirf $∣ x - 5∣$ ko control karta hai, isliye hume sab kuch uske terms mein express karna hai. Bacha hua multiplier constant $M = 4$ hai — koi caging zaroori nahi (Cell A ki defining feature).
$ε$ se neeche force karo. Chahiye $4∣ x - 5∣ < ε$ , yaani $∣ x - 5∣ < ε /4$ . Kyun? Desired inequality ko multiplier $4$ se divide karo.
$δ = ε /4$ choose karo. $min$ kyun nahi? Koi preliminary restriction nahi thi ( $M$ global tha), isliye intersect karne ke liye kuch nahi hai.

Verify: agar $0 < ∣ x - 5∣ < ε /4$ toh $∣ f (x) - 22∣ = 4∣ x - 5∣ < 4 \cdot \frac{ε}{4} = ε .$ ✅ $ε = 0.1$ lo $\Rightarrow δ = 0.025$ ; $x = 5.02$ par, $∣ f - 22∣ = 4 (0.02) = 0.08 < 0.1$ . ✔

Cell B — quadratic, variable slope (caging)

Worked example

Ex 2. Prove karo ki $x \to 2 lim x^{2} = 4$ . Forecast: $x = 2$ ke paas $x^{2}$ ka slope lagbhag $2 x = 4$ hai, isliye $δ$ roughly $ε /4$ guess karo — lekin slope change hoti hai, isliye hume ise cap karna hoga.

Target gap aur factor. $∣ x^{2} - 4∣ = ∣ x - 2∣ ∣ x + 2∣$ . Kyun? $x^{2} - 4 = (x - 2) (x + 2)$ se controllable $∣ x - 2∣$ alag ho jaata hai. Bacha hua $∣ x + 2∣$ constant nahi hai — yahi Cell B hai.
Preliminary restriction $∣ x - 2∣ < 1$ se $∣ x + 2∣$ cage karo. Tab $1 < x < 3 \Rightarrow 3 < x + 2 < 5 \Rightarrow ∣ x + 2∣ < 5$ . Pehle cap kyun? $∣ x + 2∣$ poore $R$ par unbounded grow karta hai; hum ise sirf $2$ ke paas rehkar bound kar sakte hain.
Combine karo. Us cap ke saath, $∣ x^{2} - 4∣ < 5 ∣ x - 2∣$ . $ε$ se beat karne ke liye: chahiye $∣ x - 2∣ < ε /5$ . Kyun? Ab multiplier fixed $M = 5$ hai; Cell A ki tarah divide karo.
$δ = min (1, ε /5)$ choose karo. $min$ kyun? Dono requirements — $∣ x - 2∣ < 1$ (cage valid rakhne ke liye) aur $∣ x - 2∣ < ε /5$ — ek saath hold karni chahiye.

Neeche figure. Lavender curve $f (x) = x^{2}$ hai. Butter-yellow horizontal band target tube $∣ f - 4∣ < ε$ hai; $x = 1$ aur $x = 3$ par do coral dashed lines cap $∣ x - 2∣ < 1$ mark karti hain; mint vertical band actual $δ$ -band hai jiska half-width $min (1, ε /5)$ hai. Dhyaan do ki mint band dono ke andar hai — coral cap ke bhi aur us region ke bhi jahan curve yellow tube mein rehti hai — yahi visual "inside both" exactly reason hai ki answer $min$ kyun hai.

Verify: $ε = 1$ ke liye, $δ = min (1, 0.2) = 0.2$ . $x = 2.2$ par: $∣ x^{2} - 4∣ = ∣4.84 - 4∣ = 0.84 < 1$ . ✔ $ε = 25$ ke liye, $δ = min (1, 5) = 1$ — cap $1$ le leta hai, exactly isliye $min$ zaroori tha.

Cell C — root type (distance expose karne ke liye rationalise karo)

Worked example

Ex 3. Prove karo ki $x \to 9 lim x = 3$ . Forecast: $x$ $9$ ke paas gentle hai (slope $\frac{1}{2 9} = \frac{1}{6}$ ), toh ek generous $δ$ guess karo, roughly $6 ε$ .

Pehle domain guard. $x$ ko $x \geq 0$ chahiye. Isliye hum restriction $δ \leq 9$ ko choice mein shuru se hi build karenge, taaki $0 < ∣ x - 9∣ < δ$ se $x > 0$ force ho. Pehle kyun? Agar hum $δ$ ko $9$ se zyaada hone dein, toh $9$ ke aas paas punctured ball $x \leq 0$ tak pahunch sakta hai jahan $x$ undefined hai — proof un points ke baare mein baat karega jo exist hi nahi karte. Abhi guard karna baad ki har line ko legal rakhta hai.
Target gap. $∣ x - 3∣$ . Abhi koi obvious $∣ x - 9∣$ nahi hai. Kyun tension? $δ$ $∣ x - 9∣$ control karta hai, $∣ x - 3∣$ nahi. Hume $∣ x - 9∣$ manufacture karna hoga.
Rationalise karo. $\frac{x + 3}{x + 3}$ se multiply karo: $∣ x - 3∣ = \frac{∣ ( x - 3 ) ( x + 3 ) ∣}{x + 3} = \frac{∣ x - 9∣}{x + 3} .$ Yeh trick kyun? $(x - 3) (x + 3) = x - 9$ — difference of squares messy root ko plain $∣ x - 9∣$ mein badal deta hai jo hum control kar sakte hain.
Denominator ko neeche se bound karo. Step 1 se hamein $x \geq 0$ milta hai, isliye $x \geq 0$ aur $x + 3 \geq 3$ , hence $∣ x - 3∣ = \frac{∣ x - 9∣}{x + 3} \leq \frac{∣ x - 9∣}{3} .$ Denominator ko lower-bound kyun karo? Bada denominator fraction ko chhota banata hai; sabse chhota denominator ( $3$ ) use karne se sabse safe (largest) upper bound milta hai.
$ε$ se neeche force karo aur $δ$ choose karo. Chahiye $\frac{∣ x - 9∣}{3} < ε$ , yaani $∣ x - 9∣ < 3 ε$ . Domain guard $δ \leq 9$ ke saath combine karke, airtight choice hai $δ = min (9, 3 ε) .$ $min$ kyun? Hume $∣ x - 9∣ < 3 ε$ aur $∣ x - 9∣ \leq 9$ (taaki $x > 0$ rahe) ek saath chahiye — exactly yahi situation $min$ resolve karta hai.

Verify: $ε = 0.3 \Rightarrow δ = min (9, 0.9) = 0.9$ . $x = 9.9$ par: $∣ 9.9 - 3∣ \approx ∣3.1464 - 3∣ = 0.1464 < 0.3$ . ✔ Hamara guaranteed bound tha $\frac{∣ x - 9∣}{3} = \frac{0.9}{3} = 0.3$ , aur sach mein true error $0.1464$ usse neeche hai. ✔

Cell D — degenerate limit point $a = 0$ (dono signs)

Worked example

Ex 4. Prove karo ki $x \to 0 lim x^{2} sin \frac{1}{x} = 0$ (squeeze, $ε$ – $δ$ se). Forecast: $sin (1/ x)$ $x \to 0$ par dono sides se wildly oscillate karta hai, phir bhi $x^{2}$ use crush kar deta hai. Guess karo $δ = ε$ .

Target gap. $∣ x^{2} sin \frac{1}{x} - 0∣ = x^{2} ∣ sin \frac{1}{x} ∣$ . Kyun? $L = 0$ hai, isliye gap sirf function ki magnitude hai.
Wild part ko bound karo. $∣ sin θ ∣ \leq 1$ saare real $θ$ ke liye, isliye $x^{2} ∣ sin \frac{1}{x} ∣ \leq x^{2} = ∣ x ∣^{2} .$ Yeh key move kyun hai? Hum $sin (1/ x)$ ko directly kabhi control nahi kar sakte, lekin karne ki zaroorat bhi nahi — hume sirf ek upper bound chahiye, aur $1$ ek universal bound hai. Yeh $x$ ke dono signs ko ek saath handle karta hai kyunki $x^{2} \geq 0$ regardless.
$ε$ se neeche force karo. Chahiye $∣ x ∣^{2} < ε$ , yaani $∣ x ∣ < ε$ . Square root kyun? Hum $∣ x ∣^{2} < ε$ ko $∣ x ∣$ ke liye solve karte hain; dono sides positive hain isliye root monotone hai.
$δ = ε$ choose karo, aur dhyan do $0 < ∣ x - 0∣ < δ$ (punctured — hum $x = 0$ kabhi use nahi karte, jahan $sin (1/ x)$ waise bhi undefined hai; exactly isliye definition point ko puncture karti hai).

Verify: $ε = 0.04 \Rightarrow δ = 0.2$ . Kisi bhi $x$ ke liye worst case jab $∣ x ∣ < 0.2$ : $∣ x^{2} sin \frac{1}{x} ∣ \leq x^{2} < 0.04$ . Check karo $x = - 0.15$ : $x^{2} = 0.0225 < 0.04$ regardless of sine value. ✔ (Negative $x$ identically kaam karta hai — degenerate two-sided case fully covered hai.)

Cell E — non-uniform continuity (failure case)

Worked example

Ex 5. Dikhaao ki $f (x) = \frac{1}{x}$ $(0, 1)$ par uniformly continuous nahi hai. Forecast: $x \to 0^{+}$ par graph vertically steep ho jaata hai — guess karo ki koi single $δ$ saare points serve nahi kar sakta, kyunki $0$ ke paas tiny input moves huge output jumps cause karti hain.

Poori definition likho, phir negate karo. Uniform continuity yeh hai: $\forall ε > 0 \exists δ > 0 \forall x, y : ∣ x - y ∣ < δ ⟹ ∣ f (x) - f (y) ∣ < ε .$ Iska logical negation (jo hume prove karna hai) yeh hai: $\exists ε > 0 \forall δ > 0 \exists x, y : ∣ x - y ∣ < δ aur ∣ f (x) - f (y) ∣ \geq ε .$ Dono kyun likho? Hume exact negation chase karni hai: hum ek $ε$ choose kar sakte hain, lekin phir har $δ$ ko defeat karna hoga bad points $x, y$ produce karke.
$ε = 1$ fix karo. Ab, koi bhi arbitrary $δ > 0$ given ho, $0$ ke paas ek saath close do points pick karo: $x = \frac{1}{n}, y = \frac{1}{n + 1}, jab n bada ho.$ Yeh kyun? Dono $(0, 1)$ mein hain, aur unki distance $∣ x - y ∣ = \frac{1}{n ( n + 1 )} \to 0$ — isliye large $n$ ke liye $∣ x - y ∣ < δ$ possible hai.
Output gap compute karo. $∣ f (x) - f (y) ∣ = ∣ n - (n + 1) ∣ = 1$ . Punchline kyun? Consecutive $1/ n$ ke reciprocals exactly $1$ se differ karte hain — chahe $∣ x - y ∣$ kitna bhi tiny ho, output gap $\geq ε = 1$ rehta hai.
Conclude karo. Har $δ$ ke liye humne $x, y$ find kiye jahan $∣ x - y ∣ < δ$ lekin $∣ f (x) - f (y) ∣ = 1 \geq 1 = ε$ . Toh koi bhi $δ$ kaam nahi karta ⇒ uniformly continuous nahi. ∎

Neeche figure. Lavender curve $(0, 1)$ par $f (x) = 1/ x$ hai. Do coral dots $x = \frac{1}{3}$ aur $y = \frac{1}{4}$ par hain; mint dotted verticals axis par drop karte hain yeh dikhane ke liye ki inputs kitne paas hain (short horizontal green arrow, "tiny $∣ x - y ∣$ "), jabki right par coral vertical arrow output gap measure karta hai, jo poori ek unit oocha hai. Do inputs ek baal jitne paas, outputs poori ek unit door — yahi failure hai.

Verify: $n = 100$ : $x = 0.01$ , $y = \frac{1}{101} \approx 0.009901$ , $∣ x - y ∣ \approx 9.9 \times 1 0^{- 5}$ (tiny), lekin $∣ f (x) - f (y) ∣ = ∣100 - 101∣ = 1$ . ✔ (Contrast: closed bounded set par yeh failure nahi ho sakti.)

Cell F — sequence limit ( $δ$ ki jagah $N$ )

Worked example

Ex 6. Prove karo ki $n \to \infty lim \frac{2 n + 1}{n} = 2$ . Forecast: term $2 + \frac{1}{n}$ hai, aur $\frac{1}{n} \to 0$ . Guess karo: tail $ε$ ke andar rahne ke liye, $n$ ko $1/ ε$ se aage jaana hoga.

Target gap. $\frac{2 n + 1}{n} - 2 = \frac{2 n + 1 - 2 n}{n} = \frac{1}{n}$ . Kyun? Sequence limits $d (x_{n}, L)$ use karte hain; $R$ par yeh absolute value $∣ x_{n} - 2∣$ hai.
$ε$ se neeche force karo. Chahiye $\frac{1}{n} < ε ⟺ n > \frac{1}{ε}$ . Flip kyun? $\frac{1}{n} < ε$ jab dono sides positive hain toh invert hoke $n > \frac{1}{ε}$ ban jaata hai.
$N = ⌈ 1/ ε ⌉$ choose karo (round up karo taaki $N$ ek integer ho jahan $N > 1/ ε$ , hence $\frac{1}{N} < ε$ ). $δ$ ki jagah $N$ kyun? Index $n$ discrete hai; "limit ke close" ka matlab hai "kaafi aage," isliye challenge–response variable ek threshold $N$ hai.
Conclude karo: saare $n \geq N$ ke liye, $\frac{2 n + 1}{n} - 2 = \frac{1}{n} \leq \frac{1}{N} < ε$ . ∎

Verify: $ε = 0.01 \Rightarrow N = 100$ . $n = 100$ par: $\frac{201}{100} - 2 = 0.01 \leq 0.01$ ; $n = 101$ par: $\frac{1}{101} \approx 0.0099 < 0.01$ . ✔

Cell G — ek non-Euclidean metric (taxicab vs discrete)

Worked example

Ex 7. $R^{2}$ mein taxicab metric $d_{1} (x, y) = ∣ x_{1} - y_{1} ∣ + ∣ x_{2} - y_{2} ∣$ ke saath, open ball $B (0, 1)$ describe karo; aur discrete metric ke under $B (0, 1)$ describe karo. Forecast: taxicab "circles" usually diamonds ki tarah draw hote hain — $4 5^{\circ}$ rotated square guess karo. Discrete metric ke liye, kuch strange guess karo (sirf centre?).

Taxicab ball — defining inequality padho. $B (0, 1) = {(x_{1}, x_{2}) : ∣ x_{1} ∣ + ∣ x_{2} ∣ < 1}$ . Diamond kyun? Boundary $∣ x_{1} ∣ + ∣ x_{2} ∣ = 1$ char straight segments hai; jaise first quadrant mein yeh line $x_{1} + x_{2} = 1$ hai. Char aise segments $4 5^{\circ}$ turn kiya hua unit square enclose karte hain — ek diamond. (Hum $d_{1}$ ke liye triangle inequality parent note se already known maante hain; yahan hume sirf shape chahiye.)
Discrete ball — defining inequality padho. $d (x, y) = 1$ agar $x \neq = y$ , warna $0$ : $B (0, 1) = {x : d (x, 0) < 1} = {x : d (x, 0) = 0} = {0} .$ Sirf centre kyun? $1$ se strictly less sirf $0$ distance hai, jo sirf $x = 0$ par achieve hoti hai. ( $r > 1$ radius hoti toh poori space milti.)

Neeche figure (do panels). Left: mint diamond taxicab $B (0, 1)$ hai; interior coral dot $(0.5, 0.4)$ ka $d_{1} = 0.9 < 1$ hai isliye andar hai, jabki $(0.7, 0.7)$ ka $d_{1} = 1.4$ hai isliye bahar. Right: discrete $B (0, 1)$ origin par sirf ek lavender dot hai; baaki har point (hollow coral circles) exactly $1$ distance par hai, isliye excluded. Same radius, wildly different balls — metric geometry decide karta hai.

Verify: point $(0.5, 0.4)$ : $d_{1} = 0.9 < 1$ ⇒ diamond ke andar ✔; point $(0.7, 0.7)$ : $d_{1} = 1.4 \neq < 1$ ⇒ bahar ✔. Discrete: $d ((0.5, 0.4), 0) = 1 \neq < 1$ ⇒ discrete $B (0, 1)$ mein nahi, sirf $0$ hai. ✔

Cell H — real-world word problem

Worked example

Ex 8. Ek machine rods ko target length par cut karti hai ek dial $x$ (cm) set karke. Finished length $L (x) = 3 x$ (cm) hai. Ek customer ki tolerance target $30$ cm ke aas paas $\pm 0.06$ cm hai. Dial ko $x = 10$ par kitna precisely ( $δ$ ) set karna hoga? Forecast: "gain" $3 \times$ hai, isliye dial tolerance output tolerance se $3 \times$ tighter honi chahiye: $0.02$ cm guess karo.

Translate karo. Output target $L = 30$ dial $x = 10$ par. "Tolerance ke andar" ka matlab hai $∣ L (x) - 30∣ < ε$ jahan $ε = 0.06$ cm. Dial precision hamara $δ$ hai: $∣ x - 10∣ < δ$ . $ε$ – $δ$ kyun? "Dial precisely enough set karke length sahi rakhna" literally challenge–response game hi hai.
Factor karo. $∣3 x - 30∣ = 3∣ x - 10∣$ (Cell A structure — constant gain $M = 3$ ).
Solve karo. $3∣ x - 10∣ < 0.06 \Rightarrow ∣ x - 10∣ < 0.02$ . Toh $δ = 0.02$ cm. $3$ se divide kyun? Machine dial error ko teen guna amplify karti hai, isliye hume input error teen guna shrink karna hoga.

Verify (units): dial $0.02$ cm off ⇒ length $3 \times 0.02 = 0.06$ cm $= ε$ off, exactly boundary par. $0.02$ cm ke andar dial length ko strictly $0.06$ cm ke andar rakhta hai. ✔ Units: cm·(dimensionless gain)=cm ✔.

Cell I — exam twist: prove karo ki limit exist NAHI karta

Worked example

Ex 9. Dikhaao ki $x \to 0 lim \frac{∣ x ∣}{x}$ exist nahi karta. Forecast: $∣ x ∣/ x$ $x > 0$ ke liye $+ 1$ aur $x < 0$ ke liye $- 1$ hai. Guess karo: dono sides disagree karte hain, isliye koi single $L$ game nahi jeet sakta.

Assume karo ki limit $L$ exist karta hai aur $ε = \frac{1}{2}$ se challenge karo. Definition ke anushaar koi $δ > 0$ hona chahiye taaki $0 < ∣ x ∣ < δ \Rightarrow ∣ f (x) - L ∣ < \frac{1}{2}$ . Contradiction kyun? $\forall ε \exists δ$ ko disprove karne ke liye, hum ek aisa $ε$ dikhate hain jise koi $δ$ satisfy nahi kar sakta; ek $L$ assume karne se hum ek impossibility derive kar sakte hain.
Koi bhi $δ > 0$ dono sides par points allow karta hai. $x_{+} = δ /2 > 0$ aur $x_{-} = - δ /2 < 0$ pick karo; dono $0 < ∣ x ∣ < δ$ satisfy karte hain, isliye dono ko $\frac{1}{2}$ -bound maanna hoga. Dono signs kyun? $0$ ke aas paas punctured ball hamesha left aur right dono points contain karta hai — yahi $a = 0$ ki degenerate two-sided nature hai.
Har side par evaluate karo. Kyunki $f (x_{+}) = + 1$ aur $f (x_{-}) = - 1$ , $\frac{1}{2}$ -bound se force hota hai: $∣1 - L ∣ < \frac{1}{2} aur ∣ - 1 - L ∣ < \frac{1}{2} .$ Dono kyun hold karne chahiye? Dono $x_{+}$ aur $x_{-}$ punctured $δ$ -ball mein hain, isliye definition dono par apply hoti hai.
Contradiction derive karo (triangle inequality). Yeh kehte hain ki $L$ ek saath $+ 1$ se $\frac{1}{2}$ ke andar aur $- 1$ se $\frac{1}{2}$ ke andar hai — lekin $+ 1$ aur $- 1$ poore $2$ apart hain: $2 = ∣1 - (- 1) ∣ = ∣ (1 - L) + (L - (- 1)) ∣ \leq ∣1 - L ∣ + ∣ L + 1∣ < \frac{1}{2} + \frac{1}{2} = 1,$ yaani $2 < 1$ — impossible. Triangle inequality kyun? Yeh ek hi tool hai jo " $L$ , $+ 1$ ke paas" aur " $L$ , $- 1$ ke paas" ko ek single false statement mein fuse karta hai, exactly wahi distance axiom jo theory ko power deta hai.
Conclude karo. "Limit $L$ exist karta hai" assumption ne $2 < 1$ lead kiya. Toh aisa koi $L$ exist nahi karta ⇒ limit exist nahi karta. ∎

Verify: $x = + 0.001$ par: value $= + 1$ ; $x = - 0.001$ par: value $= - 1$ ; dono one-sided values $2$ se differ karte hain, kisi bhi $ε < 2$ (khas taur par $ε = \frac{1}{2}$ ) se unbeatable. ✔

Recall Self-test: cell ko method se match karo

Kaun si technique pehle $∣ x - a ∣ < 1$ cap karti hai ::: Variable-slope polynomial case (Cell B), non-constant multiplier bound karne ke liye Limit exist nahi karta yeh prove kaise karte ho ::: Ek $ε$ aur do approaches find karo jo separated values dein; triangle inequality se contradiction nikalo (Cell I) Discrete-metric ball $B (0, 1)$ hai ::: sirf ${0}$ — sirf $0$ distance $< 1$ hai Sequence limit kaun sa threshold use karta hai ::: ek index $N = ⌈ 1/ ε ⌉$ $δ$ ki jagah Ex 3 mein $δ = min (9, 3 ε)$ kyun chahiye na ki sirf $3 ε$ ::: cap $9$ punctured ball ko domain $x \geq 0$ ke andar rakhta hai jahan $x$ defined hai