I Convergence not defined by any metric

[elias001]

##

Question:

For the following Exercise:, (assuming the two exercises in the Assumed Exercises: under the Background below:

Exercise: Let $X$ be the space of all infinite sequences $\{x_n\}$ of real numbers such that $x_{n}=0$ for all but a finite number of $n.$ Define a convergence in $X$ as follows:

A sequence $a_n =(\alpha_{n,1},\alpha_{n,2},\alpha_{n,3}, \ldots )$ converges to $a=(\alpha_1,\alpha_2,\alpha_3, \ldots)$ if the following two conditions are satisfied:

(1) $|\alpha_{n,k} - \alpha_{k}|\rightarrow 0$ as $n \rightarrow \infty$ for every $k \in N,$

(2) there exists $k_{0} \in N$ such that $\alpha_{n,k}=0$ for all $n\in N$ and all $k \geq k_0.$

Prove that this convergence cannot be defined by a metric.

I thought I am ask to show that if I was able to define a metric on the convergence satisfying properties (1) and (2), that the convergence would violate the following property of metric space:

$d(x,y)\leq d(x,z)+d(z,y)$? In the context of the exercise question, would that translate into $d(x_{n_1,p_{n_1}}, x_{n_2,p_{n_2}})>d(x_{n_1,p_{n_1}}, x_{n_3,p_{n_3}})+d(x_{n_3,p_{n_3}}, x_{n_2,p_{n_2}})$?

I emailed the author inquiring how I should approach the exercise, and his reply was

"If you show that the defined convergence does not have a property that every convergence define by a metric has, you are done. Consider the so-called diagonal property of convergence in metric spaces."

The diagonal property the author is referring to is question $2.$ in the Assumed exercise section below. The thing is since the exercise has two properties, I am not sure how to translate the exercise it into a question about diagonal property of $2$ in assumed exercises.?


Background

Assumed exercises;

$1.$ Consider the sequence of sequences $x_n=(x_{n,1},x_{n,2},x_{n,3},\ldots)$ where

$$x_{n,k}=\begin{cases}\frac{1}{k}&\text{if }k\leq n\\0&\text{if }k>n\end{cases}$$

Show that $\{x_n\}$ is convergent in $l^2,$ but divergent in $l^1.$ Is $\{x_n\}$ convergent in $l^{\infty}$?

$2.$ Let $(X,d)$ be a metric space and let $x_{n,k} \in X$ for $k,n \in N$. Prove that if

$$x_{1,1},x_{1,2},x_{1,3}, \ldots \rightarrow x$$

$$x_{2,1},x_{2,2},x_{2,3}, \ldots \rightarrow x$$

$$x_{3,1},x_{3,2},x_{3,3}, \ldots \rightarrow x$$
$$\vdots$$
$$x_{n,1},x_{n,2},x_{n,3}, \ldots \rightarrow x$$
$$\vdots$$

then there exists an increasing sequence of natural numbers $p_n$ such that $x_{n,p_n} \rightarrow x.$

Thank you in advance.

 

[WWGD]

Net convergence, as in that used to determine/define convergence of Riemann Sums, isn't defined by/through a metric, but rather by , obviously by the name,Nets:

https://en.wikipedia.org/wiki/Limit_of_a_sequence

https://en.wikipedia.org/wiki/Limit_of_a_sequence

 

[elias001]

@WWGD sorry for my late reply. The wikipedia pages you linked to does not mention anything about Nets?

 

[WWGD]

@WWGD sorry for my late reply. The wikipedia pages you linked to does not mention anything about Nets?

My bad. Give me a few minutes and I'll look for another one.

 

[WWGD]

@WWGD sorry for my late reply. The wikipedia pages you linked to does not mention anything about Nets?

Ok, let's start with the Wiki article. Let me know if this clears things up :https://en.wikipedia.org/wiki/Net_(mathematics)

 

[Office_Shredder]

To do the diagonalization argument the goal is to construct sequences that all converge to the same point (let's make it $(0,0,....,0,...)$ for simplicity) but you cannot pick one entry from each sequence and have this new sequence converge to 0.

I would recommend looking at very simple sequences and focus on the second property (that there has to be some fixed point after which everything is always 0). You just need to make sure your later sequences use larger indices and you can guarantee this will fail.

 

[elias001]

@WWGD Can I ask what Net has to do with diagonal property for subsequences??

 

[elias001]

@Office_Shredder I have a few questions on when I am asked to show a convergence can not be defined by any metric.

1. I am assuming that the question is basically saying that the convergence in the exercise is not metrizable. Meaning, it is not the same as saying that the convergence defined converges in one norm but not in another norm. I think not defined by any metric does not mean that. Am I correct in my understanding?

2. When it says that the convergence can not be defined by any metric, to show that, I would have to assume that it does, and then to show that it violates one of the axioms of metric space. In this case it would have to violate the triangle inequality: $d(x_{n_1,p_{n_1}}, x_{n_2,p_{n_2}})>d(x_{n_1,p_{n_1}}, x_{n_3,p_{n_3}})+d(x_{n_3,p_{n_3}}, x_{n_2,p_{n_2}})$ somehow?

3. From the row of sequences, each with it converging to the same element: $x_{n,1},x_{n,2},x_{n,3},\ldots ,x_{n,p_n}, \ldots \rightarrow x$,

So for the solution to the second assumed exercise above, we have:

For $n=1$, we have for some $p_1\in \Bbb{N}$ for all $w\geq p_1,\vert x_{p_1}-x\rvert<\epsilon$,

for $n=2$, we take the first $p_2>p_1$ such that for all $w\geq p_2, \vert x_{p_2}-x\rvert<\frac{\epsilon}{2}$,

for $n=3$, we take the first $p_3>p_2$ such that for all $w\geq p_3, \vert x_{p_3}-x\rvert<\frac{\epsilon}{2^2}$,

$\vdots$

for $n=k$, we take the first $p_k>p_{k-1}$ such that for all $w\geq p_k, \vert x_{p_k}-x\rvert<\frac{\epsilon}{2^{k-1}}$.

So we have an increasing sequence $p_1<p_2<\ldots < p_{k-1}< p_{k}$ where for any $\epsilon>0,$ there exists a $p_{k}\in \Bbb{N}, \forall w\geq p_k, \vert x_{p_k}-x\rvert<\frac{\epsilon}{2^{k-1}}<\epsilon$.

I am not sure what am I suppose to do from here?

 

[Office_Shredder]

You very rarely want to try to show you violate an axiom directly. You already know that if a metric space can give the convergence definition then it must have the diagonal property mentioned. It is fairly easy to just construct a bunch of converging sequences which don't satisfy the diagonal property. In theory you can trace this back to violating the triangle inequality but that's annoying and not really educational compared to just finding the diagonal property counterexample.

 

[elias001]

@Office_Shredder the way I constructed the diagonal property is the solution to the assumed exercise. So I am suppose to construct a subsequence that doesn't satisfy the diagonal property based on property (2) of the exercise? Also what does it mean for the subsequence to not satisfy the diagonal property? I mean am I going by the negation of assumed exercise (2) above, or am I constructing a diagonal subsequence that violates property (2) of the exercise?

 

[Office_Shredder]

You want to construct a bunch of convergent sequences that all converge to the same value such that any diagonal you try to pull out is guaranteed to not converge under the definition of convergence.

Here's an example. Suppose we look at functions $\mathbb{R}\to \mathbb{R}$ and define convergence of functions $f_n\to f$ if there exists $x_0$ such that for all $\epsilon>0$, there is some $N$ such that for all $n>N$, if $|x-x_0|>\epsilon$ then $f_n(x)=f(x)$, and also $f_n(x_0)=f(x_0)$.

An example: $f_n(x)=1$ if$|x|>1/n$, $1$ if $x=0$, and $0$ otherwise converges to the constant function 1 with $x_0=0$ and $N=1/\epsilon$.

I claim you can't make a metric that makes exactly these sequences converge to exactly these limits. Proof. Consider for each $m$ $f_n^m$ which is 1 if x is m, or $|x-m|>1/n$ and zero otherwise. For each $m$, $f_n^m$ converges to the constant function 1.

But you should see a problem here. Intuitively if there's a metric then e.g. $f_n^2$ converges to 1 because it's always 1 except for a small hole around 2, and $f_n^3$ is close to 1 because it equals 1 except for a small hole around 3, etc. And in fact if you pick any sequence $f_{a_m}^m$ and try to say as $m\to \infty$ this converges to 1, you're going to have a real problem picking which $x_0$ you're supposed to be using. In fact it's impossible, for any fixed choice of $x_0$ eventually $m> x_0+1$ and all future $f^m_{a_m}$ are zero for some choices of $x$ that are at least 1 away from $x_0$. Hence you can't pick a diagonal sequence that converges to the constant function 1.

 

[elias001]

@Office_Shredder Thank you so much for giving an illustration. my apologies for my late reply. I was not feeling well and it took me sometime to gather my questions as a reply.

I have five things I want to ask you. I will number it in square brackets and highlighted the number in bold for ease of reference.

$\textbf{[1]}$

When one says a sequence of functions/sequence does not satisfy the diagonal subsequence property, does it mean the following:

Let $(X,d)$ be a metric space and let $x_{n,k} \in X$ for $k,n \in N$. Prove that if

$$x_{1,1},x_{1,2},x_{1,3}, \ldots \rightarrow x$$

$$x_{2,1},x_{2,2},x_{2,3}, \ldots \rightarrow x$$
$$x_{3,1},x_{3,2},x_{3,3}, \ldots \rightarrow x$$
$$\vdots$$
$$x_{n,1},x_{n,2},x_{n,3}, \ldots \rightarrow x$$
$$\vdots$$

then for any increasing sequence of natural numbers $p_n$ such that $x_{n,p_n} \not\rightarrow x.$

How does showing that for any increasing sequence....$x_{n,p_n} \not\rightarrow x,$ will show that the convergence is not satisfied for any metric? I am not following the logic in the sense of how does that different/compare with for any increasing sequence...not converging to some norm.

$\textbf{[2]}$

The sequence of functions in $\textbf{Assumed Exercises}$ question 1.:

$x_{n,k}=\begin{cases}\frac{1}{k}&\text{ if }k\leq n\\0&\text{ if }k>n.\end{cases},$ defined in a piecewise manner, seem to satisfy the two properties of the Exercise in the Question: section,
where $x_{n,k}=\frac{1}{k},\text{ if }k\leq n,$ corresponds to: (1) $|\alpha_{n,k} - \alpha_{k}|\rightarrow 0$ as $n \rightarrow \infty$ for every $k \in N,$

and

$x_{n,k}=0, \text{ if }k>n,$ corresponds to: (2) there exists $k_{0} \in N$ such that $\alpha_{n,k}=0$ for all $n\in N$ and all $k \geq k_0.$

Matching the definition of $x_{n,k}$ to that of $\alpha_{n,k},$ we have for the first portion, $\alpha_{n,k}=x_{n,k}, \alpha_{k}=\frac{1}{k}, n \rightarrow \infty$ for every $k \in N,$

For the other portion of the piecewise definition of $x_{n,k},$ how do I find the $k_{0} \in N$ such that $\alpha_{n,k}=0$ for all $n\in N$ and all $k \geq k_0$? The example you gave requires some sort of input argument being at least some distance away from some index value, i mean you require that $x-x_0>\epsilon.$ Is my understanding correct?

$\textbf{[3]}$

I tried solving the question 1. in the Assumed exercise. Since I can't translate to a case where $x\not\in B(x_0,\epsilon),$ then can I rewrite the sequence of functions $x_{n,k}$ as
$x_{n,k}(k)=\begin{cases}\frac{1}{k}&\text{ if }k\leq n\\0&\text{ if }k>n\end{cases}.$ Then for any $\epsilon > 0, x_{n,k}(k)=\frac{1}{k}, \text{ if } k\leq n, \text{ then we can let } \epsilon=\frac{1}{n}, \text{ implying if } \frac{1}{k}\geq \frac{1}{n}=\epsilon,\text{ then }x_{n,k}(k)=0, \text{ if }k>n=\frac{1}{\epsilon}.$ But at this point I don't know how to show that the sequence $x_{n,k}$ does not have the diagonal subsequence property. I don't know how to show $x_{n,{n+1}}-\frac{1}{n+1}\geq\epsilon$.

$\textbf{[4]}$


I also tried to solved the Exercise section. I tried the following: according to the two properties in the Exercise question:

(1) $|\alpha_{n,k} - \alpha_{k}|\rightarrow 0$ as $n \rightarrow \infty$ for every $k \in N,$ which translates to:

$$\text{For any } \epsilon>0,\text{ for every }n>M, k\in N, |\alpha_{n,k} - \alpha_{k}|<|\alpha_{n,k} + \alpha_{k}|<|\alpha_{n,k} |+| \alpha_{k}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}<\epsilon$$

or should it be:

$$|\alpha_{n,k} - \alpha_{k}|<|\alpha_{n,k} + \alpha_{k}|<|\alpha_{n,k} |+| \alpha_{k}|=0<\epsilon?$$

(2) there exists $k_{0} \in N$ such that $\alpha_{n,k}=0$ for all $n\in N$ and all $k \geq k_0;$ which translates to:

There exists $k_{0} \in N$ such that $\alpha_{n,k}<\epsilon$ for all $n\in N, k \geq k_0$.

Then we need to find an $\epsilon_0>0$ and a specific $k_{0} \in N$ in terms of $\epsilon_0$ such that $|\alpha_{k,k_0} - \alpha_{k_0}|>\epsilon.$ Here $\alpha_{k_0,k}$ are the terms of the diagonal subsequence. At this point similar to the issue i run into in $\textbf{[3]}$, I can let $\epsilon_0=\frac{1}{k_0},$, then $k\geq\frac{1}{\frac{1}{k_0}}=\frac{1}{\epsilon_0}.$ From here again, I don't know how to proceed from here.

$\textbf{[5]}$

I tried to rewrite from second paragraph on in terms of even formal mathematical notations for the proof you gave. Can you check if it reflects accurately what you wrote please.

Consider the function from $\Bbb{R}\to \Bbb{R}$ and define convergence of functions $f_n\to f,$ if there exists $x_0$ such that for all $\epsilon>0,$ there exists $N$ such that $\forall n>N,$ if $|x-x_0|>\epsilon,$then
$f_n(x)=\begin{cases}f(x)&\text{ if }x\neq x_0\\f(x_0)&\text{ if }x=x_0\end{cases}.(*)$

As an example: $f_n(x)=\begin{cases}1&\text{ if }|x|>\frac{1}{n}\\1&\text{ if }x=0\\0&\text{ otherwise}\end{cases},\quad\quad f_n(x)\to 1,\text{ with } x_0=0,\text{ and } N=\frac{1}{\epsilon}. (**)$

Proof: Consider for each $m,f_n^m(x)=\begin{cases}1&\text{ if }x=m \\1&\text{ if }|x-m|>\frac{1}{n}\\0&\text{ otherwise}\end{cases}.(***)$ For each $m,f_n^m(x)\to f(x)=1.$
There should be a problem here, since intuitively, if there is a metric, then e.g. $f_n^2(x)\to 1,\text{ for }x\in \Bbb{R}\setminus B(2,\epsilon), f_n^3(x)\to 1,\text{ for }x\in \Bbb{R}\setminus B(3,\epsilon),\text{etc.}$ Continue on, if one tries to pick any sequence $f_{a_m}^m(x),$ and try to say as $m\to \infty,$ that $f_{a_m}^m(x)\to 1,$ one will run into difficulty trying to decide which $x_0$ one is suppose to pick. As a matter of fact, it is impossible for any choice of $x_0,$ eventually $m>x_0+1,$ and all future $f_{a_m}^m(x)=0$ for some choices of $x$ such that $|x-x_0|\geq 1.$ Hence you can't pick a diagonal sequence that converges to the constant function $1.$

I am not sure of the following in my translation of your proof. I think it would be a good practice for me just in case if this is for a test/exam, I might be asked to provide further justification in terms of writing it out in even more formal math notations.

1. when you stated: "if there's a metric then e.g. $f_n^2$converges to $1$ because it's always $1$ except for a small hole around $2$" that I translate: "if there is a metric, then e.g. $f_n^2(x)\to 1,\text{ for }x\in \Bbb{R}\setminus B(2,\epsilon)$" is accurate.

2. When you stated: "In fact it's impossible, for any fixed choice of $x_0$ eventually $m>x_0+1$", did you mean in terms of convergence of sequences when you used the word "eventually" as in "$x$ is eventually in a neighborhood of...."

3. The part where you stated: "...all future $f_{a_m}^m$ are zero for some choices of $x$ that are at least $1$ away from $x_0$." My translation: "all future $f_{a_m}^m(x)=0$ for some choices of $x$ such that $|x-x_0|\geq 1.$" is accurate?

4. The functions $f_n^m(x)$ in $(***)$ is the function $f_n(x)$ in $(**)$ when constructing the diagonal sequence with $\{f_n^m\}_{n=1}^{\infty}$?

5. For 4., how do I write out the algebra to show that if given a supposed diagonal subsequnce, that it does not converge to $1$ using an $\epsilon-\delta$ argument?

Thank you in advance.

 

[Svein]

I seem to remember that metric is a strong topological property. For weaker topologies continuity is defined as that the inverse funcion of open sets are open. Whether this is applicable in this case I cannot say on the top of my head.

 

[elias001]

@Svein Can I ask if you mean that the diagonal subsequence property doesn't have strong topological property? Actually I don't know if you mean by coarse or weak topology or how or why metric space has strong topological property is relevant to diagonal subsequence property. If you have any guesses, I would like to hear about them.

 

[Svein]

Copyied from "Real Analysis" (H.L. Royden):
These are topological spaces.
T_{1}: Given two distinct points x and y, there is an open set which contains y but not x.
T_{2}: Given two distinct points x and y, there are disjoint open sets O_{1} and O_{2} such that x\in O_{1} and y\in O_{2}.
T_{3}: In addition to T_{1}:, given a closed set F and a point x not in F there are disjoint open sets O_{1} and O_{2} such that x\in O_{1} and F\subset O_{2}.
T_{4}: In addition to T_{1}:, given two disjoint closed sets F_{1} and F_{2}, there are disjoint open sets O_{1} and O_{2} such that F_{1}\subset O_{1} and F_{2}\subset O_{2}

Spaces that satisfy T_{2} are called Hausdorff spaces
Spaces that satisfy T_{3} are called regular spaces
Spaces that satisfy T_{4} are called normal spaces

 

[elias001]

@Svein ah kk. thank you for the reference.

 

[Office_Shredder]

@elias001 I'm sorry your post is long and I'm not sure what all your questions are but I'm going to start with yes, you correctly understand the diagonal property and for assumed exercise 1 (which you it in bullet 3). This isn't asking you anything fancy. Fix n. Let $x_{\infty} =(1/k)$ be the obvious thing it's supposed to converge to. In $l_1,$ $d(x_n,x_{\infty}) = \sum_{k\geq n} \frac{1}{k}$. In $l_2$, $d(x_n,x_{\infty})=\sum_{k\geq n} \frac{1}{k^2}$. The first one is infinity, the second one goes to zero. In fact $x_{\infty}$ isn't then in $l_1$.

 

[elias001]

@Office_Shredder thank you for the reply. I want to know if the way

1 . if I rewrote your proof correctly?

2. Using an epsilon delta argument if there is a diagonal subsequence, then how to show that it does not converge to what is not suppose to converge; in the case of your example not converging to $1$, and in exercise question since there is two properties, is it $|\alpha_{n,k}-\alpha_k|\not\to 0$ or $\alpha_k>\epsilon$. Where as in the assumed exercise 2, if I have a diagonal subsequence, $x_{n,p_n}$, how do I show that $x_{n,p_n}\not\to x$?

As a result,

3. how does showing that it does not converge demonstrate that it can not be defined by any metric.

I think I am having trouble because I am dealing with a sequence or sequence of functions that is defined in a piece wise manner and I don't know how to work with epsilon delta argument by taking into account what the terms of sequence equals to based on the piecewise conditions.

My apologizes for the long replied post.