Proof of Every Cauchy Sequence of Real Numbers Converges

Let $(a_n)_{n=1}^\infty$ be a Cauchy sequence of real numbers. We prove that $(a_n)$ converges to a real number.

For each positive integer $k$, choose $N_k$ so large that whenever $m,n\ge N_k$,

By increasing the $N_k$ if necessary, we may assume that

For each $k$, define the closed interval

Then every term $a_n$ with $n\ge N_k$ lies in $I_k$, because for such $n$ we have

We claim that the intervals are nested: $I_{k+1}\subseteq I_k$ for every $k$. Let $x\in I_{k+1}$. Then

because $I_{k+1}$ has radius $2^{-(k+2)}$. Also, since $N_{k+1}\ge N_k$, the Cauchy property gives

Therefore, by the triangle inequality,

Hence $x\in I_k$. This proves that $I_{k+1}\subseteq I_k$.

Let the left and right endpoints of $I_k$ be

Then $(\alpha_k)$ is increasing and $(\beta_k)$ is decreasing. By the least upper bound property Least Upper Bound Property of the Real Numbers, the set $\{\alpha_k:k\in\mathbb{N}\}$ has a least upper bound; let

Since every $\beta_k$ is an upper bound for the left endpoints, one has $L\le \beta_k$ for all $k$. Also $\alpha_k\le L$ for all $k$. Thus

for every $k$.

Now let $\varepsilon>0$. Choose $k$ so large that $2^{-k}<\varepsilon$. If $n\ge N_k$, then $a_n\in I_k$ and $L\in I_k$, so

By Limit of a Sequence of Real Numbers, this proves that $a_n\to L$ as $n\to\infty$.

Therefore every Cauchy sequence of real numbers converges to a real number.