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Dropping either condition destroys the implication. The increasing sequence $a_n=n$ has a lower bound one but no upper bound, and no finite limit of a sequence. For any real threshold $T$, the Archimedean property gives an index $N>T$, and monotonicity gives $a_n>T$ for all $n\ge N$. Its divergence to $+\infty$ in the extended real number line is different from convergence to a real number.

Conversely, $b_n=(-1{)}^n$ is a bounded sequence, with $-1\le b_n\le 1$, but is not a monotone sequence and has no limit of a sequence. Its subsequence at even integers is constantly one, and its subsequence at odd integers is constantly minus one. These two subsequential limits cannot both equal a single limit of a sequence. ^[ref]

The ambient real numbers also matter. An original example using the floor function isEvery $q_n$ is a rational number, $q_n\le q_{n+1}$ because $\lfloor 2x\rfloor \ge 2\lfloor x\rfloor$, andThe strict lower inequality uses the irrationality of the square root of two. Hence this bounded sequence converges to $\sqrt{2}$ in the real numbers, but has no limit of a sequence in the rational numbers. Convergence to a rational candidate would also be convergence in the real numbers, contradicting uniqueness of the limit of a sequence. The formulas give $q_1=1$, $q_2=5/4$, $q_3=11/8$, $q_4=11/8$: this counterexample even has a plateau. This illustrates why completeness of the real numbers matters: the rational numbers lack the least-upper-bound property needed to contain the required supremum. ^[ref]

Every Cauchy sequence of real numbers converges to a real number. Explicitly, the conditionimplies the existence of $L\in \mathbb{R}$ with $x_n\to L$. The hypothesis compares approximations with one another, without requiring a candidate answer. This makes completeness of the real numbers an existence principle: sufficiently consistent approximations cannot approach a missing destination. ^[ref].

The equivalent least-upper-bound property says that every nonempty set $A\subseteq \mathbb{R}$ bounded above has a supremum $s\in \mathbb{R}$. Precisely, every $x\in A$ satisfies $x\le s$, and for every $\epsilon >0$ some $x\in A$ satisfies $s-\epsilon <x$. The second condition excludes any smaller upper bound. No largest element of $A$ is required.

The rational numbers lack this property. The set $A=\{q\in \mathbb{Q}:q\ge 0,q^2<2\}$ contains one and has upper bound two, but has no supremum in $\mathbb{Q}$. A proposed rational supremum $s$ would have to satisfy $s^2=2$: if $s^2<2$, a sufficiently small rational increase still lies in $A$; if $s^2>2$, a sufficiently small rational decrease remains an upper bound. The irrationality of the square root of two excludes equality. In $\mathbb{R}$, its supremum is $\sqrt{2}$. ^[ref].

Construct this square root by the bisection method, starting with rational bounds $[l_0,u_0]=[1,2]$. At each step square the midpoint and retain the half whose endpoints bracket two after squaring. The first four resulting intervals are $[1,3/2]$, $[5/4,3/2]$, $[11/8,3/2]$, and $[11/8,23/16]$. After $k$ steps, $u_k-l_k=2^{-k}$ and $l_k^2<2<u_k^2$. Both endpoint sequences are Cauchy sequences because every later endpoint remains in the current interval. Completeness of the real numbers gives a common limit of a sequence $L$; continuity of squaring gives $L^2=2$. The midpoint approximates $L$ with error at most $2^{-(k+1)}$. Thus exact rational arithmetic produces certified accuracy while completeness of the real numbers supplies the exact answer.

Richard Dedekind approached completeness of the real numbers by filling gaps in the rational numbers with Dedekind cuts. In the preface to his 1872 Continuity and Irrational Numbers, he traced the problem to teaching differential calculus in 1858 and wanting arithmetic foundations for limits. His construction makes a missing boundary, such as $\sqrt{2}$, into a number defined by the separation it produces among rational numbers. ^[ref].

A Dedekind cut can be represented by a nonempty proper subset $A$ of the rational numbers satisfying two conditions: if $q\in A$ and $r<q$ is rational, then $r\in A$; and $A$ has no maximum. The complementary set contains all the rational values on the other side of the boundary. Using the convention that the lower set never includes its boundary avoids two representations of a rational boundary.

The set $\{q\in \mathbb{Q}:q<0\}$ represents zero. The set $\{q\in \mathbb{Q}:q<0\text{ or }{q}^2<2\}$ represents $\sqrt{2}$, even though no rational number occupies that boundary by the irrationality of the square root of two. Thus a Dedekind cut describes an irrational number entirely using rational comparisons. Ordering these cuts by the subset relation and defining their arithmetic gives a construction of the real numbers. ^[ref].

Set $r=p/q$. If $r\notin [1,2]$, then $|\sqrt{2}-r|\ge \sqrt{2}-1>1/4\ge 1/(4q^2)$. If $1\le r\le 2$, irrationality of the square root of two implies that the integer $2q^2-p^2$ is nonzero. Factoring a difference of squares therefore givesThus the requested bound holds for every integer $p$ and every nonzero integer $q$, without using the Liouville approximation theorem. For completeness, irrationality follows because $p^2=2q^2$ in lowest terms would make $p$ even and then $q$ even, contradicting coprimality.

A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.