Proof check: If $(x_{n})$ is a real Cauchy sequence $(x_{n})$ is convergent.

Question

I've thought of a proof of this but I'm not sure if its valid.

Let $ \varepsilon > 0$ be given and suppose $(x_{n})$ is a real Cauchy sequence. Since all Cauchy sequences are bounded, by the Bolzano-Weierstrass theorem we can find a convergent subsequence $(x_{n_{j}})$ with $x_{n_{j}} \rightarrow x$ as $ n_{j} \rightarrow \infty $. We can find some $N_{1} \in \aleph ,$ with $n_{j} > N_{1}$ giving,

$$ \mid x_{n_{j}}-x \mid < \varepsilon $$

Since $x_{n_{j}} \in x_{n}$, we can set $n = n_{j}$ and the above will hold, such that,

$$ \mid x_{n} - x \mid < \varepsilon \ \ \Box $$

Answer 1

You have found a subsequence that converges. However your proof is incomplete. $x_{k_j}$ is just one subsequence it can not in general imply things about the whole sequence. Take for example: \begin{align} x_n = \begin{cases} n & \text{ if } & n \text{ is even }\\ \frac{1}{n} & \text{ if } & n \text{ is odd } \end{cases} \end{align} Clearly $x_{2n+1}$ converges to zero, but if there is $N$ such that for all $n>N$ we have $$|x_{2n+1}|<\epsilon$$ for some $\epsilon>0$. We cannot change $2n+1$ with $n$ and say $|x_n|<\epsilon$. We need more things for that. In this case we must use the Cauchy property!

To prove it correctly. Let $\epsilon>0$ be given. There is $N_1$ such that for all $k_j>N_1$ we have: \begin{align} |x_{k_j}-x|<\epsilon/2 \end{align} Where $x$ is the limit of $x_{k_j}$. Moreover there is also $N_2$ such that for all $n,m>N_2$ we have: \begin{align} |x_n-x_m|<\epsilon/2 \end{align} Choose $N:=\max\{N_1,N_2\}$ so for all $n,k_j>N$ we have: \begin{align} |x_{n}-x|= |x_n-x_{k_j}+x_{k_j}-x|\leq |x_n-x_{k_j}|+|x_{k_j}-x| <\epsilon \end{align} And that proves the result.

Answer 2

It is incomplete. How do you jump from the fact that $(x_{n_j})_{j\in\mathbb N}$ converges to $x$ to the assertion that $(x_n)_{n\in\mathbb N}$ converges to $x$? And it is not well-written. What is that $\varepsilon$?

Answer 3

Your statement that we can set $ n = n_{j} $ is problematic. What you are saying implies that every n is an $n_{j}$ which is not true.