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 $$\mathbb{R}^n$$) ...

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 $$N$$ such that for elements of the Cauchy sequence ... EVERY one of them after $$N$$ belong to the subsequence ... how do we know that such an $$N$$ 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 $$x_k$$ in the expression

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

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

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

Peter

 

[castor28]

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

it appears that the term $$x_k$$ in the expression

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

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

I have to say that proving the above expression for all $$n \gt N$$ by relying on specially chosen elements $$x_k$$ 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