Cauchy Sequences and Completeness in R^n ....

[Math Amateur]

I am reading "Multidimensional Real Analysis I: Differentiation" by J. J. Duistermaat and J. A. C. Kolk ...

I am focused on Chapter 1: Continuity ... ...

I need help with an aspect of the proof of Theorem 1.6.5 (Completeness of \(\displaystyle \mathbb{R}^n\)) ...

Duistermaat and Kolk"s Theorem 1.6.5 and its proof (including the preceding notes on Cauchy Sequences) read as follows:View attachment 7713
In the above proof of Theorem 1.6.5 we read the following:

" ... ... But then the Cauchy property implies that the whole sequence converges to the same limit. ... ... "D&K don't seem to prove this statement anywhere ... presumably they think the proof is obvious ... but I cannot see exactly why it must be true ...

Can someone please demonstrate formally and rigorously that if a subsequence of a Cauchy sequence converges to a
limit then the whole sequence converges to the same limit. ... ...Peter

 

[castor28]

Hi Peter,

Assume that the Cauchy sequence $(x_k)$ contains a subsequence $Y = (y_i)$ that converges to $a$. Given $\varepsilon>0$, there are integers $N_1$ and $N_2$ such that:

• $k, n>N_1 \Rightarrow \Vert x_k-x_n\Vert<\varepsilon/2$
• $m>N_2,x_m\in Y\Rightarrow\Vert x_m-a\Vert<\varepsilon/2$.

Take $N=\max(N_1,N_2)$, and choose $k>N$ such that $x_k$ belongs to the subsequence $Y$; this implies that $\Vert x_k-a\Vert<\varepsilon/2$. For any $n>N$, we have:

$$\Vert x_n-a\Vert \le \Vert x_n-x_k\Vert + \Vert x_k-a\Vert < \varepsilon$$​

and this shows that the sequence $(x_n)$ converges to $a$.

 

[Math Amateur]

Hi Peter,

Assume that the Cauchy sequence $(x_k)$ contains a subsequence $Y = (y_i)$ that converges to $a$. Given $\varepsilon>0$, there are integers $N_1$ and $N_2$ such that:

• $k, n>N_1 \Rightarrow \Vert x_k-x_n\Vert<\varepsilon/2$
• $m>N_2,x_m\in Y\Rightarrow\Vert x_m-a\Vert<\varepsilon/2$.

Take $N=\max(N_1,N_2)$, and choose $k>N$ such that $x_k$ belongs to the subsequence $Y$; this implies that $\Vert x_k-a\Vert<\varepsilon/2$. For any $n>N$, we have:

$$\Vert x_n-a\Vert \le \Vert x_n-x_k\Vert + \Vert x_k-a\Vert < \varepsilon$$
​

and this shows that the sequence $(x_n)$ converges to $a$.

Hi castor28 ... thanks for the help !

But just a small point of clarification ... ... ,,, you write:

" ... ... Take $N=\max(N_1,N_2)$, and choose $k>N$ such that $x_k$ belongs to the subsequence $Y$ ... ... "This implies that there is an \(\displaystyle N\) such that for elements of the Cauchy sequence ... EVERY one of them after \(\displaystyle N\) belong to the subsequence ... how do we know that such an \(\displaystyle N\) exists ...

Hope you can clarify ...Thanks again for your help ...

Peter***EDIT*** Hmm ... already beginning to doubt that my interpretation of your statements is true ...

it appears that the term \(\displaystyle x_k\) in the expression

\(\displaystyle \Vert x_n-a\Vert \le \Vert x_n-x_k\Vert + \Vert x_k-a\Vert < \varepsilon\)

is just an element of the subsequence for which \(\displaystyle k \gt N\) ... ... Is that right?

I have to say that proving the above expression for all \(\displaystyle n \gt N\) by relying on specially chosen elements \(\displaystyle x_k\) surprised me ...

Peter

 

[castor28]

***EDIT*** Hmm ... already beginning to doubt that my interpretation of your statements is true ...

it appears that the term \(\displaystyle x_k\) in the expression

\(\displaystyle \Vert x_n-a\Vert \le \Vert x_n-x_k\Vert + \Vert x_k-a\Vert < \varepsilon\)

is just an element of the subsequence for which \(\displaystyle k \gt N\) ... ... Is that right?

I have to say that proving the above expression for all \(\displaystyle n \gt N\) by relying on specially chosen elements \(\displaystyle x_k\) surprised me ...

Peter

Hi Peter,

Your last interpretation is correct. $N_1$ and $N_2$ correspond to the fact that $X$ is Cauchy and $Y$ is convergent, respectively. By taking $N=\max(N_1,N_2)$, we ensure that, in the part of the sequence after $x_N$, both inequalities are satisfied; of course, the second inequality is only relevant to elements of the sequence that belong to $Y$.

Now, the terms after $x_N$ contain at least one term of $Y$, because $Y$ is an infinite sequence and the condition $n>N$ only removes finitely many terms. We can therefore choose one such term $x_k\in Y$ with $k>N$. We can then keep $x_k$ fixed and show that all the elements $\{x_n\mid n>N\}$ satisfy $\Vert x_n-a\Vert<\varepsilon$, which means that $X$ converges to $a$.

Intuitively, the idea is quite simple. After a certain point, the elements of $X$ are close to each other. On the other hand, $Y\subset X$ contains infinitely many elements close to $a$. As the other points of $X$ are close to that point, they are also close to $a$.

It is true that the proof only uses one particular $x_k\in Y$. This is not a problem, since any $x_k$ close enough to $a$ will do, and we have an infinite supply of them.

 

[Math Amateur]

Hi Peter,

Your last interpretation is correct. $N_1$ and $N_2$ correspond to the fact that $X$ is Cauchy and $Y$ is convergent, respectively. By taking $N=\max(N_1,N_2)$, we ensure that, in the part of the sequence after $x_N$, both inequalities are satisfied; of course, the second inequality is only relevant to elements of the sequence that belong to $Y$.

Now, the terms after $x_N$ contain at least one term of $Y$, because $Y$ is an infinite sequence and the condition $n>N$ only removes finitely many terms. We can therefore choose one such term $x_k\in Y$ with $k>N$. We can then keep $x_k$ fixed and show that all the elements $\{x_n\mid n>N\}$ satisfy $\Vert x_n-a\Vert<\varepsilon$, which means that $X$ converges to $a$.

Intuitively, the idea is quite simple. After a certain point, the elements of $X$ are close to each other. On the other hand, $Y\subset X$ contains infinitely many elements close to $a$. As the other points of $X$ are close to that point, they are also close to $a$.

It is true that the proof only uses one particular $x_k\in Y$. This is not a problem, since any $x_k$ close enough to $a$ will do, and we have an infinite supply of them.

Hi castor28 ... thanks for the help and the really clear explanation ... really helpful!

Peter