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

False. One fixed $\delta$ working for every $\epsilon$ would force $f$ to be constant near $a$; the limit needs $\delta$ to shrink as $\epsilon$ shrinks, so $\delta$ must be chosen after $\epsilon$.

False. The limit only sees the punctured neighbourhood $0<|x-a|$; $f(a)$ can be anything (or undefined). Equality $f(a)=L$ is the extra condition called continuity.

True. If $|x-a|<{\delta }_0$ forces $|f(x)-L|<\epsilon$, then any $\delta \le {\delta }_0$ forces the same, since fewer inputs qualify. $\delta$ is not unique — you only need one that works.

False. $f(x)=1/x$ on $(0,1)$ is continuous but the steepness near $0$ blows up, so no single $\delta (\epsilon )$ works at all points at once — that failure is exactly non-uniform continuity.

True. For any point $x$, the ball $B(x,\frac{1}{2})=\{x\}$ (nothing else is within distance $1$), and $\{x\}\subseteq U$ whenever $x\in U$, so every set contains a ball around each of its points.

False. The codomain of a metric is $[0,\infty )$; non-negativity follows from the axioms (set $x=z$ in the triangle inequality with symmetry). Distance has no sign — there are no "sides."

True. In $\mathbb{R}$, $d(0,2)=2=d(0,1)+d(1,2)$ when $y$ lies "between" $x$ and $z$. Equality is allowed ($\le$), it just means no detour was taken.

False. Proving it for one $\epsilon$ proves nothing; a function wiggling within $0.1$ of $L$ passes but has no limit. You must win for every $\epsilon >0$.

True. They differ numerically but each ball contains a ball of the other type, so "every point has a ball inside" transfers — they generate the same topology even though distances differ.

The error is generalising from the identity. The slope inflates output error: for $f(x)=3x-1$ you need $\delta =\epsilon /3$. You must factor $|f(x)-L|=|x-a|\cdot M$ and use $\delta =\epsilon /M$.

The error is bounding $|x+3|$ globally. First cage it near $x=3$: restrict $|x-3|<1$ so $|x+3|<7$, then $\delta =\min (1,\epsilon /7)$. The cap $\min$ keeps both restrictions true at once.

The error: symmetry is an independent axiom. One can define $d$ that satisfies the triangle inequality yet has $d(x,y)\ne d(y,x)$ (a "quasimetric"). All three axioms are required.

The error: it fails the triangle inequality. Take $x=0,y=1,z=2$: $d(0,2)=4$ but $d(0,1)+d(1,2)=1+1=2<4$. Squaring destroys the additivity distance needs.

The quantifiers are reversed. It must be $\forall \epsilon \exists \delta$: the challenge $\epsilon$ comes first. The stated version is trivially satisfiable and proves nothing.

The error: $0\notin (0,1)$, so there is no point where it explodes inside the domain. At every actual point of $(0,1)$ it is continuous; it merely fails uniform continuity there.

The error: a closed ball's openness depends on the space. In the discrete metric, $\{x:d(x,a)\le \frac{1}{2}\}=\{a\}$ is open. "Closed" and "not open" are not opposites.

Because the limit describes behaviour approaching $a$, not the value at $a$. Including $x=a$ would let a single misplaced point $f(a)$ sabotage a perfectly good limit.

Because a tighter target tube (smaller $\epsilon$) generally demands a narrower input window (smaller $\delta$). Forbidding the dependence collapses the definition to constant functions only.

Because every $\epsilon$-$\delta$ argument only ever used $|x-y|$ as a distance and its three properties. Those axioms are the minimum that keep limits, uniqueness, and continuity working; the rest of $\mathbb{R}$ was unnecessary baggage.

It's derivable: $0=d(x,x)\le d(x,y)+d(y,x)=2d(x,y)$ by triangle inequality and symmetry, so $d(x,y)\ge 0$. Listing it would be redundant.

Because $\Vert x+y{\Vert }^2=\Vert x{\Vert }^2+2x\cdot y+\Vert y{\Vert }^2$, and Cauchy–Schwarz gives $x\cdot y\le \Vert x\Vert \Vert y\Vert$, turning the middle term into $2\Vert x\Vert \Vert y\Vert$ and completing the square $(\Vert x\Vert +\Vert y\Vert {)}^2$.

Because uniform continuity gives one $\delta$ that works everywhere, which is what's needed to prove theorems like "continuous on a compact set ⇒ integrable / extends to the closure." See Uniform Continuity and Compactness.

Because the bound $M$ was only valid after the preliminary caging $|x-a|<1$. Both restrictions must hold simultaneously, and the smaller of the two guarantees that.

The definition is stated only for limit points of $D$; at an isolated point there is no punctured neighbourhood inside $D$, so "${\lim }_{x\to a}$" is not defined there.

Everything: $|f(x)-L|<0$ is impossible, so no $\delta$ could ever work. The definition insists $\epsilon >0$; the game is about arbitrarily small, never exactly zero.

$L=c$, since $d(x_n,c)=d(c,c)=0<\epsilon$ for every $\epsilon$ and all $n$. Any $N$ (even $N=1$) works — the identity-of-indiscernibles axiom does the work.

Only eventually constant ones. To get $d(x_n,L)<\epsilon$ with $\epsilon <1$ you need $d(x_n,L)=0$, i.e. $x_n=L$ for all large $n$; nothing else is close enough.

The empty set. No point satisfies $d(x,a)<0$, and $d(x,a)<0$ is impossible; even $d(x,a)<0$... it's empty. Radii must be positive for a nonempty ball.

No. Left of $0$ the value is $-1$, right of $0$ it is $+1$; any punctured window around $0$ contains both, so no single $L$ can trap all outputs within $\epsilon <1$. The limit does not exist.

No. As $x$ grows the slope $2x$ is unbounded, so for a fixed small gap $\delta$ the output gap can exceed any $\epsilon$ far out. It is uniformly continuous on any bounded closed interval, though — see Uniform Continuity and Compactness.