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

Idea. To negate a statement of the form "for all $\epsilon$, there exists $\delta$, such that (implication)", we flip every quantifier and negate the inside.

The positive statement is $\forall \epsilon >0\exists \delta >0:(0<|x-a|<\delta ⟹|f(x)-L|<\epsilon ).$

Negating: $\forall$ becomes $\exists$, $\exists$ becomes $\forall$, and the implication "$P⟹Q$" fails exactly when "$P$ and not-$Q$" holds. So: $\exists \epsilon >0\forall \delta >0\exists x:(0<|x-a|<\delta \text{ and }|f(x)-L|\ge \epsilon ).$

Plain words: there is one bad challenge $\epsilon$ that you can never answer — no matter how small a $\delta$ you pick, some point $x$ inside the punctured $\delta$-neighbourhood is thrown outside the $\epsilon$-tube.

By definition $B(a,r)=\{x:d(x,a)<r\}$. Here $a=3$, $r=\frac{1}{2}$: $B(3,\frac{1}{2})=\{x:|x-3|<\frac{1}{2}\}=(3-\frac{1}{2},3+\frac{1}{2})=(\frac{5}{2},\frac{7}{2}).$ An open ball on the line is just an open interval centred at the point, radius $=r$. See the figure — the endpoints are hollow (not included).

Goal: make $|(5x+1)-11|<\epsilon$. Factor. $|(5x+1)-11|=|5x-10|=5|x-2|$. The multiplier is the constant slope $M=5$ — no caging needed because it doesn't depend on $x$. Solve. $5|x-2|<\epsilon ⟺|x-2|<\epsilon /5$. Choose $\boxed{\delta =\epsilon /5}$. Then $0<|x-2|<\delta \Rightarrow 5|x-2|<5\cdot \frac{\epsilon }{5}=\epsilon .$ ✅

Goal: $|x^2-1|<\epsilon$. Factor. $|x^2-1|=|x-1||x+1|$. The "stuff" $|x+1|$ is unbounded globally, so we must cage it near $x=1$. Cap. Restrict $|x-1|<1\Rightarrow 0<x<2\Rightarrow |x+1|<3$. Combine. $|x^2-1|<3|x-1|$. Beat $\epsilon$ with $|x-1|<\epsilon /3$. Choose $\boxed{\delta =\min (1,\epsilon /3)}$. The $\min$ forces both the cap and the tolerance simultaneously. ✅

Goal: $|\sqrt{x}-3|<\epsilon$. Expose $|x-9|$. We can't factor $\sqrt{x}-3$ directly, so rationalise: multiply by $\frac{\sqrt{x}+3}{\sqrt{x}+3}$: $|\sqrt{x}-3|=\frac{|(\sqrt{x}-3)(\sqrt{x}+3)|}{\sqrt{x}+3}=\frac{|x-9|}{\sqrt{x}+3}.$ Why rationalise? It converts the ugly $\sqrt{x}-3$ into $x-9$, which $\delta$ controls directly. Bound the denominator from below. For $x>0$, $\sqrt{x}\ge 0$ so $\sqrt{x}+3\ge 3$, hence $\frac{1}{\sqrt{x}+3}\le \frac{1}{3}$. Therefore $|\sqrt{x}-3|\le \frac{|x-9|}{3}.$ Choose $\boxed{\delta =3\epsilon }$ (we also need $x>0$; since we work near $9$, e.g. $\delta \le 9$ keeps $x>0$, but $3\epsilon$ suffices for small $\epsilon$). Then $|x-9|<3\epsilon \Rightarrow |\sqrt{x}-3|<\epsilon$. ✅

Recall the negation (from Exercise 1.1 logic applied to uniform continuity): $\exists \epsilon >0\forall \delta >0\exists x,y:|x-y|<\delta \text{ and }|f(x)-f(y)|\ge \epsilon .$ Choose the bad $\epsilon$. Take $\epsilon =1$. For an arbitrary $\delta >0$, build the pair. Let $n$ be large and set $x=\frac{1}{n},y=\frac{1}{n+1}.$ Then $|x-y|=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}\to 0$, so for $n$ large enough $|x-y|<\delta$. But the outputs stay far apart: $|f(x)-f(y)|=|n-(n+1)|=1\ge \epsilon .$ So the same $\epsilon =1$ can never be answered by any single $\delta$. Geometrically: as $x\to 0^{+}$ the graph gets arbitrarily steep, so equally-close inputs give ever-farther outputs. See the figure — two inputs of fixed separation are compared near the origin and far from it. ✅

Using the parent's table of metrics:

• 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: $d_{\infty }=4\le d_2=5\le d_1=7$. ✅ The general fact (true in ${\mathbb{R}}^n$) is that the max-coordinate distance is smallest, the sum-of-coordinates largest, Euclidean in between.

Goal: $\left|\frac{1}{x}-\frac{1}{3}\right|<\epsilon$. Factor out $|x-3|$. $\left|\frac{1}{x}-\frac{1}{3}\right|=\left|\frac{3-x}{3x}\right|=\frac{|x-3|}{3|x|}.$ The multiplier is $\frac{1}{3|x|}$, which blows up as $x\to 0$ — so we must cap to keep $|x|$ away from $0$. Cap. Restrict $|x-3|<1\Rightarrow 2<x<4\Rightarrow |x|>2$, so $\frac{1}{3|x|}<\frac{1}{6}$. Combine. $\left|\frac{1}{x}-\frac{1}{3}\right|<\frac{|x-3|}{6}$. To beat $\epsilon$: $|x-3|<6\epsilon$. Choose $\boxed{\delta =\min (1,6\epsilon )}$. Then $0<|x-3|<\delta \Rightarrow \left|\frac{1}{x}-\frac{1}{3}\right|<\frac{6\epsilon }{6}=\epsilon .$ ✅

Definition recall: $x_n\to L$ iff $\forall \epsilon >0\exists N:n\ge N\Rightarrow d(x_n,L)<\epsilon$. In the discrete metric $d(x,y)=0$ if $x=y$ and $1$ otherwise.

($\Leftarrow$) Eventually constant $\Rightarrow$ converges. If $x_n=L$ for all $n\ge N_0$, then $d(x_n,L)=0<\epsilon$ for every $\epsilon >0$ once $n\ge N_0$. So $x_n\to L$. ✅

($\Rightarrow$) Converges $\Rightarrow$ eventually constant. Pick the specific challenge $\epsilon =\frac{1}{2}$. Convergence gives an $N$ with $d(x_n,L)<\frac{1}{2}$ for all $n\ge N$. But $d$ only takes values $0$ or $1$, and $1<\frac{1}{2}$ is false, so we must have $d(x_n,L)=0$, i.e. $x_n=L$, for all $n\ge N$.

Moral: the discrete metric makes "close" mean "equal." Convergence collapses to eventual equality — the metric axioms alone force this, no real-line structure needed.

Write $\Vert u\Vert =\sqrt{u\cdot u}=d_2(u,0)$ for a vector $u$, so that $d_2(p,q)=\Vert p-q\Vert$.

Step 1 — reduce to a norm statement. Set $u=x-y$ and $v=y-z$. Then $x-z=u+v$, and the goal $d_2(x,z)\le d_2(x,y)+d_2(y,z)$ becomes exactly $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert .(\star )$ Why this substitution? It turns three points into two vectors and reduces the triangle inequality to the single norm inequality $(\star )$.

Step 2 — expand the squared left side. Using $\Vert w{\Vert }^2=w\cdot w$ and distributivity of the dot product: $\Vert u+v{\Vert }^2=(u+v)\cdot (u+v)=\Vert u{\Vert }^2+2(u\cdot v)+\Vert v{\Vert }^2.$

Step 3 — apply Cauchy–Schwarz. Replace the cross term using $u\cdot v\le \Vert u\Vert \Vert v\Vert$: $\Vert u+v{\Vert }^2\le \Vert u{\Vert }^2+2\Vert u\Vert \Vert v\Vert +\Vert v{\Vert }^2=(\Vert u\Vert +\Vert v\Vert {)}^2.$ This is the only inequality in the whole proof — everything else is algebra.

Step 4 — take square roots. Both sides are $\ge 0$ and $\sqrt{\cdot }$ is increasing, so $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert ,$ which is $(\star )$. Substituting back gives $d_2(x,z)\le d_2(x,y)+d_2(y,z)$. ✅

Where the ingredients live: symmetry of the dot product gave the clean cross term; Cauchy–Schwarz bounded it; non-negativity of norms justified taking square roots. See the figure — geometrically $u+v$ is the third side of a triangle, and it can never exceed the two-leg path.

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