Question bank — Real analysis — rigorous epsilon-delta, metric spaces

False. Ek fixed $\delta$ jo har $\epsilon$ ke liye kaam kare, woh $f$ ko $a$ ke paas constant hone par majboor kar deta; limit ke liye $\delta$ ko $\epsilon$ ke saath shrink karna padta hai, isliye $\delta$ ko $\epsilon$ ke baad choose karna hota hai.

False. Limit sirf punctured neighbourhood $0<|x-a|$ ko dekhti hai; $f(a)$ kuch bhi ho sakta hai (ya undefined bhi). $f(a)=L$ equality ek extra condition hai jise continuity kehte hain.

True. Agar $|x-a|<{\delta }_0$ se $|f(x)-L|<\epsilon$ guarantee hoti hai, toh koi bhi $\delta \le {\delta }_0$ bhi wahi guarantee karta hai, kyunki fewer inputs qualify hote hain. $\delta$ unique nahi hota — tumhe sirf ek kaam karne wala chahiye.

False. $f(x)=1/x$ on $(0,1)$ continuous hai lekin $0$ ke paas steepness itni badh jaati hai ki koi single $\delta (\epsilon )$ ek saath sabhi points par kaam nahi karta — yahi failure exactly non-uniform continuity hai.

True. Kisi bhi point $x$ ke liye, ball $B(x,\frac{1}{2})=\{x\}$ hoti hai (distance $1$ ke andar aur kuch nahi), aur $\{x\}\subseteq U$ tab hota hai jab $x\in U$, toh har set apne har point ke around ek ball contain karta hai.

False. Ek metric ka codomain $[0,\infty )$ hota hai; non-negativity axioms se follow hoti hai (triangle inequality mein $x=z$ set karo aur symmetry use karo). Distance ka koi sign nahi hota — koi "sides" nahi hote.

True. $\mathbb{R}$ mein, $d(0,2)=2=d(0,1)+d(1,2)$ jab $y$, $x$ aur $z$ ke "beech" hota hai. Equality allowed hai ($\le$), iska matlab sirf itna hai ki koi detour nahi liya gaya.

False. Ek $\epsilon$ ke liye prove karna kuch bhi prove nahi karta; ek function jo $L$ ke $0.1$ ke andar wiggle karta hai woh pass kar jaata hai lekin uski koi limit nahi hoti. Tumhe har $\epsilon >0$ ke liye jeetna hoga.

True. Numerically alag hain lekin ek ka har ball doosre type ka ek ball contain karta hai, isliye "har point ke paas andar ek ball hai" — dono same topology generate karte hain chahe distances alag hon.

Error identity se generalize karna hai. Slope output error ko inflate karta hai: $f(x)=3x-1$ ke liye tumhe $\delta =\epsilon /3$ chahiye. Tumhe $|f(x)-L|=|x-a|\cdot M$ factor karna hoga aur $\delta =\epsilon /M$ use karna hoga.

Error $|x+3|$ ko globally bound karna hai. Pehle use $x=3$ ke paas cage karo: $|x-3|<1$ restrict karo jisse $|x+3|<7$, phir $\delta =\min (1,\epsilon /7)$. Cap $\min$ dono restrictions ko ek saath sach rakhta hai.

Error: symmetry ek independent axiom hai. Ek aisa $d$ define kiya ja sakta hai jo triangle inequality satisfy kare lekin $d(x,y)\ne d(y,x)$ ho (ek "quasimetric"). Teeno axioms required hain.

Error: yeh triangle inequality fail karta hai. $x=0,y=1,z=2$ lo: $d(0,2)=4$ lekin $d(0,1)+d(1,2)=1+1=2<4$. Squaring us additivity ko destroy kar deti hai jo distance ko chahiye.

Quantifiers reversed hain. Hona chahiye $\forall \epsilon \exists \delta$: challenge $\epsilon$ pehle aata hai. Stated version trivially satisfiable hai aur kuch prove nahi karta.

Error: $0\notin (0,1)$, isliye domain ke andar koi point nahi hai jahan yeh explode ho. $(0,1)$ ke har actual point par yeh continuous hai; sirf wahan uniform continuity fail karti hai.

Error: ek closed ball ki openness space par depend karti hai. Discrete metric mein, $\{x:d(x,a)\le \frac{1}{2}\}=\{a\}$ open hai. "Closed" aur "not open" opposites nahi hain.

Kyunki limit $a$ ko approach karne ka behaviour describe karti hai, $a$ par value ka nahi. $x=a$ include karne se ek akela misplaced point $f(a)$ ek perfectly good limit ko sabotage kar sakta hai.

Kyunki ek tighter target tube (chhota $\epsilon$) generally ek narrower input window (chhota $\delta$) demand karta hai. Dependence forbid karne se definition sirf constant functions tak collapse ho jaati hai.

Kyunki har $\epsilon$-$\delta$ argument ne sirf $|x-y|$ ko ek distance ke roop mein aur uski teen properties use kiya. Woh axioms minimum hain jo limits, uniqueness, aur continuity ko kaam karte rakhte hain; $\mathbb{R}$ ka baaki hissa unnecessary baggage tha.

Yeh derivable hai: $0=d(x,x)\le d(x,y)+d(y,x)=2d(x,y)$ triangle inequality aur symmetry se, isliye $d(x,y)\ge 0$. Ise list karna redundant hoga.

Kyunki $\Vert x+y{\Vert }^2=\Vert x{\Vert }^2+2x\cdot y+\Vert y{\Vert }^2$, aur Cauchy–Schwarz $x\cdot y\le \Vert x\Vert \Vert y\Vert$ deta hai, middle term ko $2\Vert x\Vert \Vert y\Vert$ bana deta hai aur perfect square $(\Vert x\Vert +\Vert y\Vert {)}^2$ complete ho jaata hai.

Kyunki uniform continuity ek aisa $\delta$ deti hai jo everywhere kaam kare, aur yahi woh hai jo theorems prove karne ke liye chahiye jaise "compact set par continuous ⇒ integrable / closure tak extend hota hai." Dekho Uniform Continuity and Compactness.

Kyunki bound $M$ sirf preliminary caging $|x-a|<1$ ke baad valid tha. Dono restrictions simultaneously hold karni chahiye, aur dono mein se chhota ensure karta hai ki aisa ho.

Definition sirf $D$ ke limit points ke liye stated hai; ek isolated point par domain ke andar koi punctured neighbourhood nahi hoti, isliye "${\lim }_{x\to a}$" wahan defined nahi hai.

Sab kuch: $|f(x)-L|<0$ impossible hai, isliye koi $\delta$ kabhi kaam nahi kar sakta. Definition insist karti hai $\epsilon >0$; game arbitrarily small ke baare mein hai, exactly zero ke baare mein nahi.

$L=c$, kyunki $d(x_n,c)=d(c,c)=0<\epsilon$ har $\epsilon$ aur sabhi $n$ ke liye. Koi bhi $N$ (even $N=1$) kaam karta hai — identity-of-indiscernibles axiom kaam karta hai.

Sirf eventually constant waali. $d(x_n,L)<\epsilon$ paane ke liye $\epsilon <1$ ke saath tumhe $d(x_n,L)=0$ chahiye, yaani $x_n=L$ sabhi large $n$ ke liye; aur kuch itna close nahi hota.

Empty set. Koi bhi point $d(x,a)<0$ satisfy nahi karta, aur $d(x,a)<0$ impossible hai; even $d(x,a)<0$... yeh empty hai. Nonempty ball ke liye radii positive honi chahiye.

Nahi. $0$ ke left mein value $-1$ hai, right mein $+1$; $0$ ke around koi bhi punctured window dono contain karta hai, isliye koi single $L$ sabhi outputs ko $\epsilon <1$ ke andar trap nahi kar sakta. Limit exist nahi karti.

Nahi. Jaise $x$ badhta hai slope $2x$ unbounded hoti hai, isliye ek fixed chhote gap $\delta$ ke liye output gap kahin bhi kisi bhi $\epsilon$ se zyada ho sakta hai. Yeh kisi bhi bounded closed interval par uniformly continuous zaroor hai — dekho Uniform Continuity and Compactness.