(verification) Cauchy sequence can have at most one cluster points

Question

Suppose Cauchy sequence $x_n$ holds more than $2$ different cluster points - $x_k$ and $x_l$.

Then by definition of the cluster points of the sequence, it means that $x_k$ holds more than two convergent point $x_k$ and $x_l$ for $k,l \gt$ certain $N$.

if $x_k \neq x_l$ then $d(x_k,x_l) = e >0$ and it is contradictory to the Cauchy property which for $k,l>N$ $d(x_k,x_l) \lt e \;\forall e\gt0$

Answer 1

The cluster points need not be terms of the sequence. You have to do some more work.

Suppose $a$ and $b$ are different cluster points, say $a<b$. Set $\varepsilon=(b-a)/4$.

Since the sequence is Cauchy, there exists $N$ such that, for $m,n>N$, $$ |x_m-x_n|<\varepsilon $$ On the other hand, since $a$ and $b$ are cluster points, there exist $k,l>N$ such that $$ |x_k-a|<\varepsilon \qquad |x_l-b|<\varepsilon $$ Now find the contradiction.

$$b-a=|b-x_l+x_l-x_k+x_k-a|\le|b-x_l|+|x_l-x_k|+|x_k-a|\le\frac{3}{4}(b-a)$$