Proof of Uniqueness of Limits and Boundedness of Convergent Real Sequences

For a real number $x$ we write $|x|$ for its absolute value, that is $x$ if $0\le x$ and $-x$ otherwise; by claim 8 of Properties of Complex Conjugation and Modulus this is the modulus of $x$ regarded as a complex number, so that the following hold: $0\le|x|$ by Modulus of a Complex Number; $|x|=0$ if and only if $x=0$, by claim 3; $|xy|=|x||y|$, by claim 4; $|x+y|\le|x|+|y|$, by claim 7; and $|-x|=|-1|\,|x|=|x|$, since $|-1|=1$ by claim 8. Convergence is as in Limit of a Sequence of Real Numbers.

Claim 1. Suppose $A\neq A'$. Then $A-A'\neq0$, so $|A-A'|\neq0$, and since $0\le|A-A'|$ we have $0<|A-A'|$. Put $\varepsilon=|A-A'|/2$, a positive real number. Choose $N_{1}$ with $|a_{n}-A|<\varepsilon$ for all $n\ge N_{1}$ and $N_{2}$ with $|a_{n}-A'|<\varepsilon$ for all $n\ge N_{2}$, and let $n$ be a natural number at least as large as both. Then

which is impossible, since no real number is strictly less than itself. Hence $A=A'$.

Claim 2. Let $A$ be a real number with $a_{n}\to A$. Applying the definition of convergence with $\varepsilon=1$ gives a natural number $N$ such that $|a_{n}-A|<1$ for all $n\ge N$; for such $n$,

The finitely many real numbers $|a_{1}|,\dots,|a_{N}|$ and $1+|A|$ have a largest element $M$: this follows by applying the principle of induction to the number of entries, using that the order of $\mathbb{R}$ is a total order, so that any two entries are comparable. Then $1+|A|\le M$, and since $0\le|A|$ we get $0<1\le1+|A|\le M$, so $M$ is positive. Finally $|a_{n}|\le M$ for every $n$: for $n\le N$ because $|a_{n}|$ is one of the listed entries, and for $n>N$ because $|a_{n}|<1+|A|\le M$. Hence $(a_{n})$ is bounded.