Ytukyg

I) Let $b_1,b_2,...,b_n....$ be a Cauchy sequence in a complete metric space $(X,d)$. If $T$ is the set of points in the sequence show that $T$ is a bounded set.

As the space is complete, every Cauchy sequence is convergent and every convergent sequence is bounded. As $Tsubseteq{b_i:iinmathbb{N}}$ then $T$ is bounded.

II) Is I) above still true if we do not insist that $(X,d)$ is complete?

If I) is not true then the sequence is not necessarily convergent in $X$ so it remains to show if every Cauchy sequence is bounded.

I devised the following proof:

As ${b_i:iinmathbb{N}}$ is a Cauchy sequence then $forallepsilon>0exists m,n>Ninmathbb{N}$ such that $d(b_n,b_m)<epsilon$.

$max_{iin mathbb{N}}{d(b_i,b_n)}$ must be smaller than some $Linmathbb{R}$, which implies that the $max_{i,jin mathbb{N}}{d(b_i,b_j)}leqslant max_{iin mathbb{N}}{d(b_i,b_n)}+d(b_n,b_m)<L+epsilon $, proving the Cauchy sequence is bounded.

Question:

Is my proof right? If not. What are the alternatives?

Thanks in advance!

The proof is not correct. For instance, asserting that your sequence is a Cauchy sequence means that$$(forallvarepsilon>0)(exists inmathbb{N})(forall m,ninmathbb{N}):m,ngeqslant Nimplies d(b_m,b_n)<varepsilon;tag1$$that is not what you wrote.

It follows from $(1)$ that there is a natural $N$ such that $m,ngeqslant Nimplies d(b_m,b_n)<1$. Therefore, each $b_n$, with $ngeqslant N$, belongs to the ball $B_1(x_N)$. So, the set ${b_n,|,ngeqslant N}$ is bounded. The set ${b_1,b_2,ldots,b_{N-1}}$ is also bounded, since it is finite. So, your set is bounded, since it is the union of two bounded sets.