Cauchy sequence and limitpoint

Aug 2018
44
0
Denmark
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
Aug 2006
22,470
8,640
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?
 
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