# Cauchy sequence and limitpoint

#### detalosi

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

#### Plato

MHF Helper
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?

• 2 people