# Thread: Cauchy sequence and limitpoint

1. ## Cauchy sequence and limitpoint

Hello,

I am trying to prove that if a sequence $\displaystyle \{x_{n}\}$ in a set $\displaystyle M\in X$ ($\displaystyle X$ is a metric space) is a Cauchy sequence and the sequence has a limit point $\displaystyle x_{0}$ then $\displaystyle x_{n} \rightarrow x_{0}$. I wish to understand the details in such a proof whence I will try to explain my thoughts.

I know that the sequence $\displaystyle \{x_{n}\}_{N\in \mathbb{N}}$ is a Cauchy sequence which means that for all $\displaystyle \epsilon >0$ I can find an $\displaystyle N \in \mathbb{N}$ beyond which the distance between any to elements $\displaystyle x_{n}$ and $\displaystyle x_{m}$ of the sequence will be smaller than $\displaystyle \epsilon$. This means that I can make the distance between any two element of the sequence arbitrarily small.

Furthermore we know that $\displaystyle x_{0}$ is a limit point which means that for every $\displaystyle \epsilon>0$ there is a point $\displaystyle m \in M$ such that $\displaystyle m \in B(x_{0},\epsilon)$.

I wish to argue as follows:

1) I would like to argue that $\displaystyle x_{m}=m \in B(x_{0},\frac{\epsilon}{2})$ (I can freely choose $\displaystyle r=\frac{\epsilon}{2}$ since $\displaystyle x_{0}$ is a limit point).

2) Since $\displaystyle \{x_{n}\}$ is a Cauchy sequence $\displaystyle d(x_{n},x_{m})$ will be smaller than any $\displaystyle \epsilon$ I choose, so I choose $\displaystyle \frac{\epsilon}{2}$. Therefore $\displaystyle d(x_{n},x_{m})<\frac{\epsilon}{2}$.

3) Since $\displaystyle x_{m}=m \in B(x_{0},\frac{\epsilon}{2})$ then $\displaystyle d(x_{m},x_{0})<\frac{\epsilon}{2}$.

4) Using the triangle inequality we have $\displaystyle d(x_{n},x_{0}) \leq d(x_{n},x_{m}) + d(x_{m},x_{0}) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

5) We can conclude that $\displaystyle x_{n} \rightarrow x_{0}$

Can someone please comment on the proof and the arguments presented?
How can I convince myself of 1) ?
How do I convince myself of the choice of epsilon (I start my choice of epsilon from the fact that $\displaystyle x_{0}$ is a limit point and from this I choose my epsilon for the distance between $\displaystyle x_{n}$ and $\displaystyle x_{m}$!!).

Thanks

2. ## Re: Cauchy sequence and limitpoint

Originally Posted by detalosi
I am trying to prove that if a sequence $\displaystyle \{x_{n}\}$ in a set $\displaystyle M\in X$ ($\displaystyle X$ is a metric space) is a Cauchy sequence and the sequence has a limit point $\displaystyle x_{0}$ then $\displaystyle x_{n} \rightarrow x_{0}$. I wish to understand the details in such a proof whence I will try to explain my thoughts.
I know that the sequence $\displaystyle \{x_{n}\}_{N\in \mathbb{N}}$ is a Cauchy sequence which means that for all $\displaystyle \epsilon >0$ I can find an $\displaystyle N \in \mathbb{N}$ beyond which the distance between any to elements $\displaystyle x_{n}$ and $\displaystyle x_{m}$ of the sequence will be smaller than $\displaystyle \epsilon$. This means that I can make the distance between any two element of the sequence arbitrarily small.
Furthermore we know that $\displaystyle x_{0}$ is a limit point which means that for every $\displaystyle \epsilon>0$ there is a point $\displaystyle m \in M$ such that $\displaystyle m \in B(x_{0},\epsilon)$.

1) I would like to argue that $\displaystyle x_{m}=m \in B(x_{0},\frac{\epsilon}{2})$ (I can freely choose $\displaystyle r=\frac{\epsilon}{2}$ since $\displaystyle x_{0}$ is a limit point).

2) Since $\displaystyle \{x_{n}\}$ is a Cauchy sequence $\displaystyle d(x_{n},x_{m})$ will be smaller than any $\displaystyle \epsilon$ I choose, so I choose $\displaystyle \frac{\epsilon}{2}$. Therefore $\displaystyle d(x_{n},x_{m})<\frac{\epsilon}{2}$.

3) Since $\displaystyle x_{m}=m \in B(x_{0},\frac{\epsilon}{2})$ then $\displaystyle d(x_{m},x_{0})<\frac{\epsilon}{2}$.

4) Using the triangle inequality we have $\displaystyle d(x_{n},x_{0}) \leq d(x_{n},x_{m}) + d(x_{m},x_{0}) \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

5) We can conclude that $\displaystyle x_{n} \rightarrow x_{0}$
Can someone please comment on the proof and the arguments presented?
How do I convince myself of the choice of epsilon (I start my choice of epsilon from the fact that $\displaystyle x_{0}$ is a limit point and from this I choose my epsilon for the distance between $\displaystyle x_{n}$ and $\displaystyle x_{m}$!!).
The good news is that I would say that the proof is basically correct BUT difficult to follow.
You should know that $\{x_n\}$ is a Cauchy sequence in $M\subseteq X$ a metric metric space.
Also, you are given that $x_0$ is a limit point of the sequence.
You are to prove that $(x_n)\to x_0$.
Being a limit point of the sequence, every open set containing $x_0$ contains some point of the sequence different from $x_0$
One property of Cauchy sequences us that for $\forall\varepsilon>0$ the ball $\mathscr{B}_{\varepsilon }(x_0)$ will contain almost all terms of the sequence.
Now, almost all has a definite meaning: it means all but an finite collection.
I think you have that idea. Can you tighten the argument?