Proof of Uniqueness of Limits in a Metric Space

The order $\le$ of $\mathbb{R}$ is a total order, $\mathbb{R}$ being an ordered field by The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness §reals and Ordered Fields §ordered-field, so Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders applies to its strict relation $<$.

Suppose, for contradiction, that $x\neq y$. By Metric Space §zero, $d(x,y)=0$ would give $x=y$, so $d(x,y)\neq0$; together with $0\le d(x,y)$ from Metric Space §nonnegative, Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization gives $0<d(x,y)$. Put $\varepsilon=d(x,y)/2$. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §halving, $0<\varepsilon$ and $\varepsilon+\varepsilon=d(x,y)$.

Since $(x_{m})$ converges to $x$, Convergent Sequence in a Metric Space §converges applied to this $\varepsilon$ gives $N_{1}\in\mathbb{N}$ with $d(x_{m},x)<\varepsilon$ for every $m\in\mathbb{N}$ with $m\ge N_{1}$; since $(x_{m})$ converges to $y$, it likewise gives $N_{2}\in\mathbb{N}$ with $d(x_{m},y)<\varepsilon$ for every $m\in\mathbb{N}$ with $m\ge N_{2}$. The order of $\mathbb{N}$ is a total order by Arithmetic and Order of the Natural Numbers §partial-order and Arithmetic and Order of the Natural Numbers §trichotomy; let $N=\max\{N_{1},N_{2}\}\in\mathbb{N}$ be the greatest element of $\{N_{1},N_{2}\}$ given by The Maximum and Minimum of Two Elements of a Total Order §exists, unique by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §least-unique. By The Maximum and Minimum of Two Elements of a Total Order §bounds, $N_{1}\le N$ and $N_{2}\le N$, and $x_{N}\in X$, as $(x_{m})$ is a sequence in $X$; hence

$$d(x_{N},x)<\varepsilon\qquad\text{and}\qquad d(x_{N},y)<\varepsilon .$$

By Metric Space §triangle and Metric Space §symmetric,

$$d(x,y)\le d(x,x_{N})+d(x_{N},y)=d(x_{N},x)+d(x_{N},y).$$

Since $d(x_{N},y)\le\varepsilon$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict, Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §strict-sum applied to $d(x_{N},x)<\varepsilon$ and $d(x_{N},y)\le\varepsilon$ gives $d(x_{N},x)+d(x_{N},y)<\varepsilon+\varepsilon=d(x,y)$. Then Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed gives $d(x,y)<d(x,y)$, which Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive excludes. This contradiction shows $x=y$.