Proof of A Closed Interval is Sequentially Compact in the Real Line

Throughout, $\mathbb{N}$ denotes the natural numbers with the order $\le$ and addition $+$. The order $\le$ on $\mathbb{R}$ is in particular a total order, so it is reflexive, transitive and antisymmetric and any two real numbers are comparable; addition, additive inverses and the additive identity $0$ are those of the underlying field, $s-t$ abbreviates $s+(-t)$, and $s<t$ means $s\le t$ and $s\ne t$. The absolute value is written $|\cdot|$, so that $d_{\mathbb{R}}(s,t)=|s-t|$.

Let $(x_m)_{m\in\mathbb{N}}$ be a sequence in $\mathbb{R}$ with $x_m\in[a,b]$ for every $m\in\mathbb{N}$. By the definition of the closed interval,

$$a\le x_m\quad\text{and}\quad x_m\le b\qquad\text{for every } m\in\mathbb{N}.$$

We must produce a point of $[a,b]$ that is the limit of a subsequence.

Step 1: an auxiliary set and its supremum. Let

$$E=\bigl\{t\in\mathbb{R}\ :\ \text{for every } N\in\mathbb{N}\text{ there is } m\in\mathbb{N}\text{ with } N\le m\text{ and } t\le x_m\bigr\}.$$

The set $E$ is nonempty: given $N\in\mathbb{N}$, statement 1 of Properties of the Order on the Natural Numbers gives $N\le N$, and $a\le x_N$, so the defining condition holds for $t=a$ with the choice $m=N$; hence $a\in E$.

The real number $b$ is an upper bound for $E$: let $t\in E$. Applying the defining condition with $N=1$, which lies in $\mathbb{N}$ by Natural Numbers, produces $m\in\mathbb{N}$ with $t\le x_m$, and $x_m\le b$, so $t\le b$ by transitivity.

Thus $E$ is nonempty and bounded above, so by the least upper bound property stated in The Real Numbers it has a least upper bound in $\mathbb{R}$, which is unique by Uniqueness of the Supremum and of the Infimum. Write $c=\sup E$.

Step 2: the supremum lies in $[a,b]$. Since $a\in E$ and $c$ is an upper bound for $E$, we have $a\le c$. Since $b$ is an upper bound for $E$ and $c$ is the least upper bound, we have $c\le b$. Hence $c\in[a,b]$.

Step 3: $c$ is a cluster point of $(x_m)_{m\in\mathbb{N}}$. Let $\varepsilon\in\mathbb{R}$ satisfy $0<\varepsilon$, and let $N\in\mathbb{N}$.

(a) A point of $E$ just below $c$. By statement 3 of Approximation Property of the Supremum and the Infimum in $\mathbb{R}$, applied to the nonempty set $E$ which is bounded above, there is $t\in E$ with $c-\varepsilon<t$.

(b) A tail lying strictly below $c+\varepsilon$. From $0<\varepsilon$, statement 1 of Elementary Order Arithmetic in an Ordered Field gives $0+c<\varepsilon+c$; by Additive Cancellation and Elementary Additive Identities in a Field we have $0+c=c$, and addition in a field is commutative, so $c<c+\varepsilon$. If $c+\varepsilon$ belonged to $E$ then, $c$ being an upper bound for $E$, we would have $c+\varepsilon\le c$; together with $c\le c+\varepsilon$ antisymmetry would give $c+\varepsilon=c$, contradicting $c\ne c+\varepsilon$. Hence $c+\varepsilon\notin E$.

Negating the condition defining $E$ at the point $c+\varepsilon$, there is $N_0\in\mathbb{N}$ such that for every $m\in\mathbb{N}$ with $N_0\le m$ the inequality $c+\varepsilon\le x_m$ fails. For such an $m$, comparability gives $x_m\le c+\varepsilon$, and $x_m\ne c+\varepsilon$ because otherwise reflexivity would give $c+\varepsilon\le x_m$; hence

$$x_m<c+\varepsilon\qquad\text{for every } m\in\mathbb{N}\text{ with } N_0\le m.$$

(c) A single index past both thresholds. Put $N_1=N+N_0$. Statement 6 of Properties of the Order on the Natural Numbers gives $N<N+N_0$ and $N_0<N_0+N$, and addition on $\mathbb{N}$ is commutative by statement 4 of Arithmetic of Addition on the Natural Numbers, so $N_0+N=N_1$. By statement 1 of Properties of the Order on the Natural Numbers we get $N\le N_1$ and $N_0\le N_1$. Applying the defining condition of $E$ to $t$ with the threshold $N_1$ produces $m\in\mathbb{N}$ with $N_1\le m$ and $t\le x_m$; transitivity then gives $N\le m$ and $N_0\le m$.

(d) The estimate. From $c-\varepsilon<t$ and $t\le x_m$, mixed transitivity, statement 2 of Elementary Order Arithmetic in an Ordered Field, gives $c-\varepsilon<x_m$; and $N_0\le m$ gives $x_m<c+\varepsilon$ by part (b). Adding $-c$ to both sides of each inequality, which preserves strict inequality by statement 1 of Elementary Order Arithmetic in an Ordered Field, and simplifying $(c-\varepsilon)+(-c)=-\varepsilon$ and $(c+\varepsilon)+(-c)=\varepsilon$ using Additive Cancellation and Elementary Additive Identities in a Field together with the commutativity and associativity of field addition, we obtain

$$-\varepsilon<x_m-c\quad\text{and}\quad x_m-c<\varepsilon .$$

Statement 9 of Properties of the Absolute Value in an Ordered Field therefore gives $|x_m-c|<\varepsilon$, that is, $d_{\mathbb{R}}(x_m,c)<\varepsilon$.

We have produced, for the given $\varepsilon$ and $N$, an index $m\in\mathbb{N}$ with $N\le m$ and $d_{\mathbb{R}}(x_m,c)<\varepsilon$. Since $\varepsilon$ and $N$ were arbitrary, $c$ is a cluster point of $(x_m)_{m\in\mathbb{N}}$ in $(\mathbb{R},d_{\mathbb{R}})$.

Step 4: extraction of a convergent subsequence. By Existence of a Sequence of Positive Real Numbers with Limit Zero there is a sequence $(\varepsilon_k)_{k\in\mathbb{N}}$ in $\mathbb{R}$ with $0<\varepsilon_k$ for every $k\in\mathbb{N}$ and with limit $0$. Statement 1 of A Cluster Point of a Sequence in a Metric Space is the Limit of a Subsequence yields a strictly increasing sequence $(n_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$, in the sense of Subsequence of a Sequence in a Set, with $d_{\mathbb{R}}(x_{n_k},c)<\varepsilon_k$ for every $k\in\mathbb{N}$, and statement 2 of the same result shows that the subsequence $(x_{n_k})_{k\in\mathbb{N}}$ converges to $c$ in $(\mathbb{R},d_{\mathbb{R}})$.

Since $c\in[a,b]$ by Step 2, the sequence $(x_m)_{m\in\mathbb{N}}$ has a subsequence converging to a point of $[a,b]$. As $(x_m)_{m\in\mathbb{N}}$ was an arbitrary sequence with all terms in $[a,b]$, the set $[a,b]$ is sequentially compact in $(\mathbb{R},d_{\mathbb{R}})$.