TU BEd — Bachelor Level (4 Years), Second Year

Group A: Attempt any ONE question [7 marks]

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Option 1: Axioms of Order and the Closure Property

Definition of Axioms of Order in $\mathbb{R}$

The axioms of order are the fundamental properties that describe the ordering relation "$<$" on the set of real numbers $\mathbb{R}$. For any $a, b, c \in \mathbb{R}$, the following axioms hold:

(O1) Trichotomy Law: For any two real numbers $a$ and $b$, exactly one of the following is true: $a < b, \quad a = b, \quad \text{or} \quad a > b$

(O2) Transitivity: If $a < b$ and $b < c$, then $a < c$.

(O3) Additive Compatibility: If $a < b$, then $a + c < b + c$ for every $c \in \mathbb{R}$.

(O4) Multiplicative Compatibility: If $a < b$ and $c > 0$, then $ac < bc$. If $a < b$ and $c < 0$, then $ac > bc$.

These four axioms together make $\mathbb{R}$ a totally ordered field.

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To Show: $(a + b) \in \mathbb{R}$, but $ab \in \mathbb{R}^+$

Given: $a, b \in \mathbb{R}$

Part 1: Show that $(a + b) \in \mathbb{R}$

By the closure axiom of addition in $\mathbb{R}$, the sum of any two real numbers is again a real number. Since $a \in \mathbb{R}$ and $b \in \mathbb{R}$, it follows directly that: $(a + b) \in \mathbb{R}$

Part 2: Show that $ab \in \mathbb{R}^+$

Here, the statement requires careful interpretation. The product $ab$ belongs to $\mathbb{R}^+$ (the set of positive real numbers) if and only if $a$ and $b$ have the same sign. That is:

• If $a > 0$ and $b > 0$, then by axiom (O4), $ab > 0$, so $ab \in \mathbb{R}^+$.
• If $a < 0$ and $b < 0$, then $-a > 0$ and $-b > 0$. By axiom (O4), $(-a)(-b) > 0$, i.e., $ab > 0$, so $ab \in \mathbb{R}^+$.

In both cases, when $a$ and $b$ have the same sign, the product $ab$ is positive, hence $ab \in \mathbb{R}^+$.

Note: If $a$ and $b$ have opposite signs, then $ab < 0$, so $ab \in \mathbb{R}^-$ (negative reals). If either $a = 0$ or $b = 0$, then $ab = 0 \notin \mathbb{R}^+$.

Conclusion: $(a + b) \in \mathbb{R}$ always (by closure of addition), while $ab \in \mathbb{R}^+$ when $a$ and $b$ have the same sign.

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Option 2: Completeness Axiom and Incompleteness of $\mathbb{Q}$

Definition of Completeness Axiom in $\mathbb{R}$

Completeness Axiom (Least Upper Bound Property): Every non-empty subset $S$ of $\mathbb{R}$ that is bounded above has a least upper bound (supremum) in $\mathbb{R}$.

In symbols: If $S \subseteq \mathbb{R}$, $S \neq \phi$, and $S$ is bounded above, then there exists $\sup S \in \mathbb{R}$ such that:

1. $s \leq \sup S$ for all $s \in S$ (upper bound)
2. If $M$ is any upper bound of $S$, then $\sup S \leq M$ (least upper bound)

This axiom distinguishes $\mathbb{R}$ from $\mathbb{Q}$ and is the foundation of real analysis.

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To Show: The Set of Rational Numbers $\mathbb{Q}$ is Not Complete

Proof by Counterexample:

Consider the set $S = \{x \in \mathbb{Q} : x^2 < 2\}$

Step 1: $S$ is non-empty and bounded above.

Clearly $1 \in S$ since $1^2 = 1 < 2$. Also, every element of $S$ satisfies $x < 2$ (since $x^2 < 2$ implies $x < 2$ for positive $x$, and negative rationals are certainly less than $2$). So $S$ is bounded above by $2$.

Step 2: $S$ has no rational supremum.

We know that $\sqrt{2} \notin \mathbb{Q}$ (since $\sqrt{2}$ is irrational). The only possible candidate for $\sup S$ in $\mathbb{Q}$ would be $\sqrt{2}$, but $\sqrt{2} \notin \mathbb{Q}$.

More rigorously, suppose $\sup S = u \in \mathbb{Q}$. Then by trichotomy, either $u^2 < 2$, $u^2 = 2$, or $u^2 > 2$.

• If $u^2 < 2$, then $u \in S$. But we can always find a rational $r > u$ with $r^2 < 2$ (by density of $\mathbb{Q}$), contradicting that $u$ is an upper bound.
• If $u^2 = 2$, then $u = \sqrt{2} \notin \mathbb{Q}$, a contradiction.
• If $u^2 > 2$, then $u$ is an upper bound, but we can find a smaller rational upper bound $v < u$ with $v^2 > 2$, contradicting that $u$ is the least upper bound.

In all cases, we reach a contradiction. Hence $S$ has no least upper bound in $\mathbb{Q}$.

Step 3: Conclusion.

There exists a non-empty subset $S$ of $\mathbb{Q}$ that is bounded above but has no supremum in $\mathbb{Q}$. Therefore, the completeness axiom fails in $\mathbb{Q}$.

$\boxed{\text{Hence } \mathbb{Q} \text{ is not complete.}}$

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Note: In contrast, in $\mathbb{R}$, the same set $S = \{x \in \mathbb{R} : x^2 < 2\}$ has $\sup S = \sqrt{2} \in \mathbb{R}$, illustrating that $\mathbb{R}$ satisfies the completeness axiom while $\mathbb{Q}$ does not.

Let (X) be a metric space (or a topological space). A subset (F\subseteq X) is closed iff its complement (X\setminus F) is open.

Theorem

If (F_1,F_2,\dots,F_n) are closed subsets of (X), then
[ F=\bigcup_{i=1}^{n}F_i ] is closed.

Proof

Since each (F_i) is closed, each complement [ G_i=X\setminus F_i ] is open.

By De Morgan’s law, [ X\setminus F

X\setminus \left(\bigcup_{i=1}^{n}F_i\right)

\bigcap_{i=1}^{n}(X\setminus F_i)

\bigcap_{i=1}^{n}G_i. ]

Now we show that the finite intersection of open sets is open. Let [ x\in \bigcap_{i=1}^{n}G_i. ] Then (x\in G_i) for every (i). Since each (G_i) is open, for each (i) there exists (\varepsilon_i>0) such that [ B(x,\varepsilon_i)\subseteq G_i. ] Let [ \varepsilon=\min{\varepsilon_1,\varepsilon_2,\dots,\varepsilon_n}. ] Because the number of sets is finite, (\varepsilon>0). Hence [ B(x,\varepsilon)\subseteq B(x,\varepsilon_i)\subseteq G_i ] for every (i). Therefore [ B(x,\varepsilon)\subseteq \bigcap_{i=1}^{n}G_i. ] So (\bigcap_{i=1}^{n}G_i) is open.

Thus (X\setminus F) is open. Hence (F=\bigcup_{i=1}^{n}F_i) is closed. (\square)

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Counterexample for an arbitrary family

The theorem does not hold for the union of an arbitrary family of closed sets.

Consider (\mathbb R) with the usual topology. For each (n\in\mathbb N), let [ F_n=\left{\frac1n\right}. ] Each (F_n) is a singleton, hence closed in (\mathbb R).

But [ \bigcup_{n=1}^{\infty}F_n

\left{1,\frac12,\frac13,\frac14,\dots\right}. ] This union is not closed, because (0) is a limit point of the set: [ \frac1n\to 0 \quad \text{as } n\to\infty, ] but [ 0\notin \left{1,\frac12,\frac13,\dots\right}. ] A closed set must contain all its limit points. Since this union does not contain its limit point (0), it is not closed.

Therefore, although the union of finitely many closed sets is always closed, the union of infinitely many closed sets need not be closed.

Let

$u_n=\left(\frac{3n-1}{n+2}\right)\left(\frac{n^4-n}{n^3+1}\right).$

First factor:

$\frac{3n-1}{n+2}=\frac{3-\frac1n}{1+\frac2n}\to 3 \quad (n\to\infty).$

Second factor:

$\frac{n^4-n}{n^3+1}=n\cdot\frac{1-\frac1{n^3}}{1+\frac1{n^3}}.$

The fraction tends to $1$, so this factor behaves like $n$ and tends to $\infty$.

Bounding $u_n$. For $n\ge 2$, $\dfrac{3n-1}{n+2}\ge\dfrac54>1$ and $\dfrac{n^4-n}{n^3+1}\ge\dfrac{7n}{9}$. Hence

$u_n\ge \frac{35}{36}\,n \quad (n\ge 2),$

so $u_n\to+\infty$.

Conclusion: A convergent sequence must be bounded and have a finite limit. Here $\langle u_n\rangle$ is unbounded above and tends to $+\infty$, so it has no finite limit. The sequence is therefore not convergent; it diverges (properly divergent to $+\infty$).

We show by counterexamples that, when $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both fail to exist, the limits of $f\pm g$ and $fg$ may or may not exist. Take $a=0$.

1. Sum/difference may exist. Let $f(x)=\dfrac1x$, $g(x)=-\dfrac1x$. Neither limit exists at $0$, but $f(x)+g(x)=0$ for $x\ne0$, so $\lim_{x\to0}(f+g)=0$ exists. Similarly, for $f(x)=g(x)=\dfrac1x$ we get $f-g=0$, so $\lim_{x\to0}(f-g)=0$ exists.

2. Sum/difference may fail to exist. For $f(x)=g(x)=\dfrac1x$, $f+g=\dfrac2x$, which has no limit at $0$. For $f(x)=\dfrac1x$, $g(x)=-\dfrac1x$, $f-g=\dfrac2x$, which has no limit at $0$.

3. Product may exist. Let $f(x)=\begin{cases}1,&x>0\\-1,&x<0\end{cases},\qquad g(x)=\begin{cases}-1,&x>0\\1,&x<0.\end{cases}$ The left and right limits of each differ, so neither $\lim f$ nor $\lim g$ exists. But $f(x)g(x)=-1$ for $x\ne0$, so $\lim_{x\to0}f(x)g(x)=-1$ exists.

4. Product may fail to exist. For $f(x)=g(x)=\dfrac1x$, $f(x)g(x)=\dfrac1{x^2}\to\infty$, so the limit does not exist (finitely).

Conclusion: $\lim_{x\to a}(f\pm g)$ and $\lim_{x\to a}fg$ may or may not exist when $\lim f$ and $\lim g$ do not exist. $\blacksquare$

Let $f$ be continuous and one-one on an interval $I$, and let $J=f(I)$.

Existence and one-one property. Since $f$ is one-one, each $y\in J$ equals $f(x)$ for exactly one $x\in I$. Define $f^{-1}(y)=x$. So $f^{-1}:J\to I$ exists. If $f^{-1}(y_1)=f^{-1}(y_2)=x$, then $y_1=f(x)=y_2$. Hence $f^{-1}$ is one-one.

Step 1: $f$ is strictly monotonic on $I$. Suppose not. Then there exist $a<b<c$ in $I$ with $f(b)$ not between $f(a)$ and $f(c)$. Since $f$ is one-one the values are distinct, so say $f(b)>\max\{f(a),f(c)\}$ (the other case is similar). Choose $k$ with $\max\{f(a),f(c)\}<k<f(b)$. By the intermediate value theorem on $[a,b]$ and on $[b,c]$ there are $x_1\in(a,b)$ and $x_2\in(b,c)$ with $f(x_1)=f(x_2)=k$, and $x_1\ne x_2$. This contradicts that $f$ is one-one. So for every $a<b<c$ the value $f(b)$ lies between $f(a)$ and $f(c)$, which forces $f$ to be strictly increasing or strictly decreasing on $I$.

Step 2: $J$ is an interval. By the intermediate value theorem a continuous function maps an interval onto an interval.

Step 3: $f^{-1}$ is continuous. Let $f$ be strictly increasing (the decreasing case is similar), and write $g=f^{-1}$, which is also strictly increasing. Take $y_0\in J$, $x_0=g(y_0)$ and $\varepsilon>0$.

If $x_0$ is an interior point of $I$, take $\varepsilon$ small enough that $[x_0-\varepsilon,x_0+\varepsilon]\subset I$. Then $f(x_0-\varepsilon)<y_0<f(x_0+\varepsilon).$ Let $\delta=\min\{y_0-f(x_0-\varepsilon),\;f(x_0+\varepsilon)-y_0\}>0$. If $y\in J$ and $|y-y_0|<\delta$, then $f(x_0-\varepsilon)<y<f(x_0+\varepsilon)$, and since $g$ is increasing, $x_0-\varepsilon<g(y)<x_0+\varepsilon,\quad\text{i.e. }|g(y)-g(y_0)|<\varepsilon.$

If $x_0$ is an end point of $I$, the same argument works with only the one side that lies in $I$. Hence $g=f^{-1}$ is continuous at every point of $J$.

Therefore $f^{-1}$ exists, and is one-one and continuous on $f(I)$. $\blacksquare$

Statement: If a function $f$ is derivable (differentiable) at a point $x = a$, then $f$ is continuous at that point.

Proof:

Let $f$ be derivable at $x = a$. Then the derivative

$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$

exists (is finite).

We must show that $\lim_{x \to a} f(x) = f(a)$, i.e. $f$ is continuous at $x = a$.

Now, for $x \neq a$,

$f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a)$

Taking limits as $x \to a$,

$\lim_{x \to a} \big[f(x) - f(a)\big] = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \cdot \lim_{x \to a} (x - a)$

$= f'(a) \cdot 0 = 0$

Therefore,

$\lim_{x \to a} f(x) - f(a) = 0 \quad \Rightarrow \quad \lim_{x \to a} f(x) = f(a)$

Hence $f$ is continuous at $x = a$. $\blacksquare$

Converse need not be true (counter-example):

Consider the function

$f(x) = |x| = \begin{cases} x, & x \geq 0 \\ -x, & x < 0 \end{cases}$

Continuity at $x = 0$:

$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-x) = 0, \qquad \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$

Also $f(0) = 0$. Since both one-sided limits equal $f(0)$, the function is continuous at $x = 0$.

Derivability at $x = 0$:

Left-hand derivative:

$f'(0^-) = \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^-} \frac{-x}{x} = -1$

Right-hand derivative:

$f'(0^+) = \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0^+} \frac{x}{x} = 1$

Since $f'(0^-) \neq f'(0^+)$, the derivative at $x = 0$ does not exist. Hence $f$ is not derivable at $x = 0$, although it is continuous there.

Conclusion: Derivability at a point implies continuity at that point, but continuity does not imply derivability. Thus the converse of the given statement is not true.

Monotonic function is integrable. Let $f$ be monotonically increasing on $[a,b]$ (the decreasing case is similar). Then $f$ is bounded, since $f(a)\le f(x)\le f(b)$.

Constant case. If $f(b)=f(a)$, then $f$ is constant, and $U(P,f)-L(P,f)=0$ for every partition, so $f\in\Re[a,b]$ trivially. Now assume $f(b)>f(a)$.

Take the partition $P=\{a=x_0<x_1<\dots<x_n=b\}$. On $[x_{r-1},x_r]$, $m_r=f(x_{r-1})$ and $M_r=f(x_r)$. Therefore $U(P,f)-L(P,f)=\sum_{r=1}^{n}\big[f(x_r)-f(x_{r-1})\big](x_r-x_{r-1}).$

Let $\|P\|=\max_r(x_r-x_{r-1})$. Since each bracket is $\ge0$ and the sum telescopes, $U(P,f)-L(P,f)\le\|P\|\sum_{r=1}^{n}\big[f(x_r)-f(x_{r-1})\big]=\|P\|\,[f(b)-f(a)].$

Given $\varepsilon>0$, take the partition into $n$ equal parts with $\dfrac{b-a}{n}<\dfrac{\varepsilon}{f(b)-f(a)}$. Then $U(P,f)-L(P,f)<\varepsilon$. By the Riemann criterion, $f\in\Re[a,b]$. $\blacksquare$

OR

Constant function. Let $k(x)=c$ on $[a,b]$. For any partition $P$, every subinterval has $m_r=M_r=c$, so $U(P,k)=L(P,k)=c\sum_{r=1}^{n}(x_r-x_{r-1})=c(b-a).$

Hence the upper integral $\inf_P U(P,k)$ and the lower integral $\sup_P L(P,k)$ are both $c(b-a)$. They are equal, so $k\in\Re[a,b]$ and $\int_a^b k\,dx=k(b-a). \qquad\blacksquare$

First alternative: Test the convergence of

[ \frac{2^2}{3^2}x+\frac{2^2 4^2}{3^2 5^2}x^2+\frac{2^2 4^2 6^2}{3^2 5^2 7^2}x^3+\cdots ]

Let the general term be
[ u_n=\left(\frac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdots(2n+1)}\right)^2 x^n. ]

Using the ratio test, [ \frac{u_{n+1}}{u_n}

\left(\frac{2n+2}{2n+3}\right)^2 x. ] Therefore, [ \lim_{n\to\infty}\left|\frac{u_{n+1}}{u_n}\right|

\lim_{n\to\infty}\left(\frac{2n+2}{2n+3}\right)^2 x

x. ]

By d’Alembert’s ratio test:

• If (0<x<1), the series converges.
• If (x>1), the series diverges.

Now test the boundary case (x=1). Then [ u_n=\left(\frac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdots(2n+1)}\right)^2. ] For each (k), [ \left(\frac{2k}{2k+1}\right)^2>\frac{2k-1}{2k+1}, ] because [ 4k^2>4k^2-1. ] Hence [ \frac{2k}{2k+1}>\sqrt{\frac{2k-1}{2k+1}}. ] Multiplying from (k=1) to (n), [ \frac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdots(2n+1)}

\sqrt{\frac{1}{2n+1}}. ] Squaring, [ u_n>\frac{1}{2n+1}. ] But (\sum \frac{1}{2n+1}) diverges. Therefore, by the comparison test, (\sum u_n) diverges at (x=1).

Hence the given series converges for
[ \boxed{0<x<1} ] and diverges for
[ \boxed{x\ge 1}. ]

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OR: Cauchy’s general principle of convergence for a series

Let
[ S_n=u_1+u_2+\cdots+u_n. ] The series (\sum u_n) converges if and only if the sequence ((S_n)) converges.

By Cauchy’s criterion for sequences, ((S_n)) converges if and only if for every (\varepsilon>0), there exists (m\in\mathbb N) such that [ |S_{n+p}-S_n|<\varepsilon ] for all (n\ge m) and (p\ge 0).

But [ S_{n+p}-S_n

u_{n+1}+u_{n+2}+\cdots+u_{n+p}. ] Therefore, [ |S_{n+p}-S_n|

|u_{n+1}+u_{n+2}+\cdots+u_{n+p}|. ]

Thus (\sum u_n) converges if and only if for every (\varepsilon>0), there exists (m\in\mathbb N) such that [ \boxed{ |u_{n+1}+u_{n+2}+\cdots+u_{n+p}|<\varepsilon } ] for all (n\ge m) and (p\ge 0).

This is Cauchy’s general principle of convergence for an infinite series.

Riemann Integrability on Subintervals and Additivity of the Integral

Statement

Let $f$ be a bounded function on $[a,b]$ and let $c \in (a,b)$. Then:

$f \in \Re[a,b] \iff f \in \Re[a,c] \text{ and } f \in \Re[c,b]$

Moreover, in either case,

$\int_a^b f = \int_a^c f + \int_c^b f$

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Part I: If $f \in \Re[a,b]$, then $f \in \Re[a,c]$ and $f \in \Re[c,b]$

Proof

Since $f \in \Re[a,b]$, by the Riemann criterion, for every $\varepsilon > 0$ there exists a partition $P$ of $[a,b]$ such that

$U(P,f) - L(P,f) < \varepsilon$

Let $P = \{a = x_0, x_1, \dots, x_n = b\}$. Since $c \in (a,b)$, we may refine $P$ by inserting the point $c$ (refinement does not increase $U - L$). So assume $c = x_k$ for some $k$, $1 \le k \le n-1$.

Now define partitions of the subintervals:

$P_1 = \{a = x_0, x_1, \dots, x_k = c\} \quad \text{of } [a,c]$ $P_2 = \{c = x_k, x_{k+1}, \dots, x_n = b\} \quad \text{of } [c,b]$

Then

$U(P,f) - L(P,f) = \big[U(P_1,f) - L(P_1,f)\big] + \big[U(P_2,f) - L(P_2,f)\big]$

Since both terms on the right are non-negative,

$U(P_1,f) - L(P_1,f) \le U(P,f) - L(P,f) < \varepsilon$

and

$U(P_2,f) - L(P_2,f) \le U(P,f) - L(P,f) < \varepsilon$

Since $\varepsilon > 0$ is arbitrary, by the Riemann criterion,

$f \in \Re[a,c] \quad \text{and} \quad f \in \Re[c,b]$

$\blacksquare$

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Part II: Conversely, if $f \in \Re[a,c]$ and $f \in \Re[c,b]$, then $f \in \Re[a,b]$

Proof

Let $\varepsilon > 0$ be given. Since $f \in \Re[a,c]$, there exists a partition $P_1$ of $[a,c]$ such that

$U(P_1,f) - L(P_1,f) < \frac{\varepsilon}{2}$

Since $f \in \Re[c,b]$, there exists a partition $P_2$ of $[c,b]$ such that

$U(P_2,f) - L(P_2,f) < \frac{\varepsilon}{2}$

Now let $P = P_1 \cup P_2$, which is a partition of $[a,b]$. Then

$U(P,f) - L(P,f) = \big[U(P_1,f) - L(P_1,f)\big] + \big[U(P_2,f) - L(P_2,f)\big]$

$< \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

Since $\varepsilon > 0$ is arbitrary, by the Riemann criterion,

$f \in \Re[a,b]$

$\blacksquare$

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Part III: Additivity of the Integral

Proof

Assume $f \in \Re[a,b]$ (equivalently, $f \in \Re[a,c]$ and $f \in \Re[c,b]$).

Let $P_1$ be any partition of $[a,c]$ and $P_2$ any partition of $[c,b]$. Then $P = P_1 \cup P_2$ is a partition of $[a,b]$, and we have the identity:

$U(P,f) = U(P_1,f) + U(P_2,f)$

$L(P,f) = L(P_1,f) + L(P_2,f)$

Taking the infimum over all partitions $P_1$ of $[a,c]$ and all partitions $P_2$ of $[c,b]$:

$\overline{\int_a^b} f = \overline{\int_a^c} f + \overline{\int_c^b} f$

Similarly, taking the supremum over all such partitions:

$\underline{\int_a^b} f = \underline{\int_a^c} f + \underline{\int_c^b} f$

Since $f \in \Re[a,b]$, we have $\overline{\int_a^b} f = \underline{\int_a^b} f = \int_a^b f$. Also, since $f \in \Re[a,c]$ and $f \in \Re[c,b]$, we have

$\overline{\int_a^c} f = \underline{\int_a^c} f = \int_a^c f \quad \text{and} \quad \overline{\int_c^b} f = \underline{\int_c^b} f = \int_c^b f$

Therefore,

$\int_a^b f = \int_a^c f + \int_c^b f$

$\blacksquare$

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Summary

Thus Riemann integrability on a closed interval is equivalent to Riemann integrability on each of its subintervals, and the integral is additive over adjacent intervals.

Convergent sequence.
A sequence $\langle u_n\rangle$ is said to be convergent if there exists a real number $l$ such that for every $\varepsilon>0$, there is a natural number $m$ satisfying
$|u_n-l|<\varepsilon \quad \text{for all } n\ge m.$
Then $l$ is called the limit of the sequence, and we write
$\lim_{n\to\infty}u_n=l \quad \text{or} \quad u_n\to l.$

Example.
Let
$u_n=\frac{1}{n}.$
For any $\varepsilon>0$, choose $m\in\mathbb N$ such that $m>\frac1\varepsilon$. Then for all $n\ge m$,
$\left|\frac1n-0\right|=\frac1n\le \frac1m<\varepsilon.$
Hence $\frac1n\to 0$, so $\langle u_n\rangle$ is convergent.

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Theorem (Cauchy’s criterion for convergence).
A sequence $\langle u_n\rangle$ is convergent if and only if for every $\varepsilon>0$, there exists $m\in\mathbb N$ such that
$|u_{n+p}-u_n|<\varepsilon \quad \text{for all } n\ge m \text{ and } p\ge0.$

Proof.

1. Necessary part

Suppose $\langle u_n\rangle$ is convergent, say
$\lim_{n\to\infty}u_n=l.$
Let $\varepsilon>0$ be given. Then there exists $m\in\mathbb N$ such that
$|u_n-l|<\frac{\varepsilon}{2} \quad \text{for all } n\ge m.$
Now take any $n\ge m$ and any $p\ge0$. Then $n+p\ge m$, so
$|u_{n+p}-l|<\frac{\varepsilon}{2}.$
Hence
$|u_{n+p}-u_n|\le |u_{n+p}-l|+|l-u_n|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$
Therefore the given condition holds.

2. Sufficient part

Assume that for every $\varepsilon>0$, there exists $m\in\mathbb N$ such that
$|u_{n+p}-u_n|<\varepsilon \quad \text{for all } n\ge m,\ p\ge0.$

First we prove that $\langle u_n\rangle$ is bounded.
Take $\varepsilon=1$. Then there exists $m_0\in\mathbb N$ such that
$|u_{m_0+p}-u_{m_0}|<1 \quad \text{for all } p\ge0.$
So for every $k\ge m_0$,
$|u_k|<|u_{m_0}|+1.$
The finite set $\{u_1,u_2,\dots,u_{m_0-1}\}$ is also bounded. Hence the whole sequence is bounded.

By the Bolzano–Weierstrass theorem, every bounded sequence has a convergent subsequence. Therefore, there exists a subsequence $\langle u_{n_k}\rangle$ such that
$\lim_{k\to\infty}u_{n_k}=l.$

Now we show that the whole sequence converges to $l$.
Let $\varepsilon>0$. From the given Cauchy condition, choose $m\in\mathbb N$ such that
$|u_{n+p}-u_n|<\frac{\varepsilon}{2} \quad \text{for all } n\ge m,\ p\ge0.$
Since $u_{n_k}\to l$, choose $K$ such that for all $k\ge K$,
$|u_{n_k}-l|<\frac{\varepsilon}{2}.$
Also, since $n_k\to\infty$, choose $K'$ such that for all $k\ge K'$,
$n_k\ge m.$
Take $k\ge\max(K,K')$. Then for any $n\ge m$, using the Cauchy condition,
$|u_n-u_{n_k}|<\frac{\varepsilon}{2}.$
Therefore,
$|u_n-l|\le |u_n-u_{n_k}|+|u_{n_k}-l|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$
This is true for all $n\ge m$. Hence
$\lim_{n\to\infty}u_n=l.$
So $\langle u_n\rangle$ is convergent.

Thus a sequence is convergent if and only if the given Cauchy condition is satisfied.