Exercises — Real analysis — rigorous epsilon-delta, metric spaces

Idea. "For all $\epsilon$, there exists $\delta$, such that (implication)" wali statement ko negate karne ke liye, har quantifier ko flip karo aur andar wali cheez negate karo.

Positive statement yeh hai: $\forall \epsilon >0\exists \delta >0:(0<|x-a|<\delta ⟹|f(x)-L|<\epsilon ).$

Negate karne par: $\forall$ ban jaata hai $\exists$, $\exists$ ban jaata hai $\forall$, aur implication "$P⟹Q$" tab fail hoti hai jab "$P$ aur not-$Q$" hold kare. Toh: $\exists \epsilon >0\forall \delta >0\exists x:(0<|x-a|<\delta \text{ and }|f(x)-L|\ge \epsilon ).$

Plain words mein: ek aisi buri challenge $\epsilon$ exist karti hai jiska jawab kabhi nahi diya ja sakta — chahe kitna bhi chhota $\delta$ chuno, punctured $\delta$-neighbourhood ke andar koi na koi point $x$ milega jo $\epsilon$-tube ke bahar phenka jaata hai.

Definition se $B(a,r)=\{x:d(x,a)<r\}$. Yahan $a=3$, $r=\frac{1}{2}$: $B(3,\frac{1}{2})=\{x:|x-3|<\frac{1}{2}\}=(3-\frac{1}{2},3+\frac{1}{2})=(\frac{5}{2},\frac{7}{2}).$ Number line pe open ball bas ek open interval hota hai jo point pe centred ho, radius $=r$. Figure dekho — endpoints hollow hain (included nahi).

Goal: $|(5x+1)-11|<\epsilon$ banana hai. Factor karo. $|(5x+1)-11|=|5x-10|=5|x-2|$. Multiplier constant slope $M=5$ hai — caging ki zaroorat nahi kyunki yeh $x$ pe depend nahi karta. Solve karo. $5|x-2|<\epsilon ⟺|x-2|<\epsilon /5$. Choose karo $\boxed{\delta =\epsilon /5}$. Tab $0<|x-2|<\delta \Rightarrow 5|x-2|<5\cdot \frac{\epsilon }{5}=\epsilon .$ ✅

Goal: $|x^2-1|<\epsilon$. Factor karo. $|x^2-1|=|x-1||x+1|$. "Stuff" $|x+1|$ globally unbounded hai, isliye hume $x=1$ ke paas use cage karna hoga. Cap karo. Restrict karo $|x-1|<1\Rightarrow 0<x<2\Rightarrow |x+1|<3$. Combine karo. $|x^2-1|<3|x-1|$. $\epsilon$ se beat karne ke liye $|x-1|<\epsilon /3$. Choose karo $\boxed{\delta =\min (1,\epsilon /3)}$. $\min$ dono — cap aur tolerance — ko simultaneously force karta hai. ✅

Goal: $|\sqrt{x}-3|<\epsilon$. $|x-9|$ expose karo. $\sqrt{x}-3$ ko directly factor nahi kar sakte, toh rationalize karo: $\frac{\sqrt{x}+3}{\sqrt{x}+3}$ se multiply karo: $|\sqrt{x}-3|=\frac{|(\sqrt{x}-3)(\sqrt{x}+3)|}{\sqrt{x}+3}=\frac{|x-9|}{\sqrt{x}+3}.$ Rationalize kyun karte hain? Yeh ugly $\sqrt{x}-3$ ko $x-9$ mein convert karta hai, jise $\delta$ directly control karta hai. Denominator ko neeche se bound karo. $x>0$ ke liye, $\sqrt{x}\ge 0$ toh $\sqrt{x}+3\ge 3$, isliye $\frac{1}{\sqrt{x}+3}\le \frac{1}{3}$. Therefore: $|\sqrt{x}-3|\le \frac{|x-9|}{3}.$ Choose karo $\boxed{\delta =3\epsilon }$ (hume $x>0$ bhi chahiye; kyunki hum $9$ ke paas kaam kar rahe hain, jaise $\delta \le 9$ se $x>0$ rahega, lekin chhote $\epsilon$ ke liye $3\epsilon$ kaafi hai). Tab $|x-9|<3\epsilon \Rightarrow |\sqrt{x}-3|<\epsilon$. ✅

Negation recall karo (Exercise 1.1 ki logic ko uniform continuity pe apply karke): $\exists \epsilon >0\forall \delta >0\exists x,y:|x-y|<\delta \text{ and }|f(x)-f(y)|\ge \epsilon .$ Bura $\epsilon$ choose karo. $\epsilon =1$ lo. Arbitrary $\delta >0$ ke liye, pair banao. $n$ ko bada lo aur set karo: $x=\frac{1}{n},y=\frac{1}{n+1}.$ Tab $|x-y|=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}\to 0$, toh $n$ itna bada lo ki $|x-y|<\delta$. Lekin outputs door rehte hain: $|f(x)-f(y)|=|n-(n+1)|=1\ge \epsilon .$ Toh same $\epsilon =1$ ka jawab kisi bhi single $\delta$ se nahi diya ja sakta. Geometrically: $x\to 0^{+}$ jaane par graph arbitrarily steep ho jaata hai, toh equally-close inputs ever-farther outputs dete hain. Figure dekho — fixed separation wale do inputs ko origin ke paas aur door compare kiya gaya hai. ✅

Parent ke metrics ki table use karke:

• Euclidean: $d_2=\sqrt{3^2+4^2}=\sqrt{25}=5$.
• Taxicab: $d_1=|3|+|4|=7$.
• Chebyshev: $d_{\infty }=\max (|3|,|4|)=4$.

Check karo: $d_{\infty }=4\le d_2=5\le d_1=7$. ✅ General fact (${\mathbb{R}}^n$ mein sach) yeh hai ki max-coordinate distance sabse chhoti hoti hai, sum-of-coordinates sabse badi, Euclidean beech mein.

Goal: $\left|\frac{1}{x}-\frac{1}{3}\right|<\epsilon$. $|x-3|$ factor karo. $\left|\frac{1}{x}-\frac{1}{3}\right|=\left|\frac{3-x}{3x}\right|=\frac{|x-3|}{3|x|}.$ Multiplier $\frac{1}{3|x|}$ hai, jo $x\to 0$ ke paas blow up karta hai — toh $|x|$ ko $0$ se door rakhne ke liye cap karna zaroori hai. Cap karo. Restrict karo $|x-3|<1\Rightarrow 2<x<4\Rightarrow |x|>2$, toh $\frac{1}{3|x|}<\frac{1}{6}$. Combine karo. $\left|\frac{1}{x}-\frac{1}{3}\right|<\frac{|x-3|}{6}$. $\epsilon$ se beat karne ke liye: $|x-3|<6\epsilon$. Choose karo $\boxed{\delta =\min (1,6\epsilon )}$. Tab $0<|x-3|<\delta \Rightarrow \left|\frac{1}{x}-\frac{1}{3}\right|<\frac{6\epsilon }{6}=\epsilon .$ ✅

Definition recall: $x_n\to L$ iff $\forall \epsilon >0\exists N:n\ge N\Rightarrow d(x_n,L)<\epsilon$. Discrete metric mein $d(x,y)=0$ agar $x=y$ aur $1$ otherwise.

($\Leftarrow$) Eventually constant $\Rightarrow$ converges. Agar $n\ge N_0$ ke liye $x_n=L$ hai, tab $d(x_n,L)=0<\epsilon$ har $\epsilon >0$ ke liye jab $n\ge N_0$. Toh $x_n\to L$. ✅

($\Rightarrow$) Converges $\Rightarrow$ eventually constant. Specific challenge $\epsilon =\frac{1}{2}$ lo. Convergence se ek $N$ milta hai jahan $n\ge N$ ke liye $d(x_n,L)<\frac{1}{2}$. Lekin $d$ sirf values $0$ ya $1$ leta hai, aur $1<\frac{1}{2}$ false hai, toh zaroori hai $d(x_n,L)=0$, yaani $x_n=L$, $n\ge N$ ke liye.

Moral: discrete metric mein "close" ka matlab "equal" hota hai. Convergence collapse ho ke eventual equality ban jaata hai — sirf metric axioms se yeh force hota hai, koi real-line structure ki zaroorat nahi.

Vector $u$ ke liye $\Vert u\Vert =\sqrt{u\cdot u}=d_2(u,0)$ likho, taaki $d_2(p,q)=\Vert p-q\Vert$.

Step 1 — norm statement tak reduce karo. $u=x-y$ aur $v=y-z$ set karo. Tab $x-z=u+v$, aur goal: $d_2(x,z)\le d_2(x,y)+d_2(y,z)$ exactly ban jaata hai: $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert .(\star )$ Yeh substitution kyun? Yeh teen points ko do vectors mein convert karta hai aur triangle inequality ko single norm inequality $(\star )$ tak reduce karta hai.

Step 2 — squared left side expand karo. $\Vert w{\Vert }^2=w\cdot w$ aur dot product ki distributivity use karke: $\Vert u+v{\Vert }^2=(u+v)\cdot (u+v)=\Vert u{\Vert }^2+2(u\cdot v)+\Vert v{\Vert }^2.$

Step 3 — Cauchy–Schwarz apply karo. Cross term ko $u\cdot v\le \Vert u\Vert \Vert v\Vert$ se replace karo: $\Vert u+v{\Vert }^2\le \Vert u{\Vert }^2+2\Vert u\Vert \Vert v\Vert +\Vert v{\Vert }^2=(\Vert u\Vert +\Vert v\Vert {)}^2.$ Poore proof mein sirf yahi ek inequality hai — baaki sab algebra hai.

Step 4 — square roots lo. Dono sides $\ge 0$ hain aur $\sqrt{\cdot }$ increasing hai, toh: $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert ,$ jo $(\star )$ hai. Substitute back karne se milta hai $d_2(x,z)\le d_2(x,y)+d_2(y,z)$. ✅

Ingredients kahan hain: dot product ki symmetry ne clean cross term diya; Cauchy–Schwarz ne use bound kiya; norms ki non-negativity ne square roots lene ko justify kiya. Figure dekho — geometrically $u+v$ triangle ki teesri side hai, aur yeh kabhi do-leg path se zyada nahi ho sakti.

$\delta$ fill karo) ${\lim }_{x\to 2}(5x+1)=11$ ::: $\delta =\epsilon /5$ ${\lim }_{x\to 1}{x}^2=1$ ::: $\delta =\min (1,\epsilon /3)$ ${\lim }_{x\to 9}\sqrt{x}=3$ ::: $\delta =3\epsilon$ ${\lim }_{x\to 3}1/x=1/3$ ::: $\delta =\min (1,6\epsilon )$ $d_2,d_1,d_{\infty }$ from $(0,0)$ to $(3,4)$ ::: $5,7,4$