# Question regarding convergence not defined by a metric.

I am trying to understand how to do the following question and I am confused about several issues: What does it mean for convergence to not be defined by a metric (not satisfies the three axiom of metric space or something else entirely)?
Am I suppose to show that the convergence defined in the question, even though it can not be defined by a metric, it nonetheless still satisfies conditions (1) and (2)?
Lastly, how is this different than topology of point-wise convergence? I am having trouble trying to have visual representation of this sequence like what is usually shown in an introductory calculus text for uniform convergence of sequences of functions. Also, the question assumes a another question which I have also stated below.

Question: 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.

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

• @ZeroXLR thank you for the edit. I am still relatively new to LaTex and am not familiar with a lot of the syntax. – Seth Mai Aug 11 '19 at 8:12
• No problem. FYI the commands for the dots in my edit are "\ldots" for "$\ldots$" and "\vdots" for "$\vdots$". – 0XLR Aug 11 '19 at 8:15

First of all, the convergence as defined in the question, from now on referred to as Q-convergence, is different from pointwise convergence in the following way. Consider the sequence of sequences $$(x_{n,k})$$ which is given by $$x_{n,k}=\begin{cases}\frac{1}{n}&\text{ if }n=k\\0&\text{else}\end{cases}$$ Note that pointwise the sequence converges to the sequence $$(0)$$, i.e. as the OP noticed the sequence satifies condition $$1$$ of Q-convergence which is equivalent to pointwise convergence, however, the sequence fails to satisfy condition $$2$$ so it is not Q-convergent.

The question ask something similar from you. You have to show that there is no metric on $$X$$ such that an sequence $$(x_{n})$$ converges to $$x\in X$$ with respect to the metric, if and only if $$(x_{n})$$ is Q-convergent to $$x$$.

I hope this helps you on your way. If you are still stuck let me know and I'll expand my answer as necessary.

• thank you for replying. I want to ask how are the two conditions allow for this sequence of sequences to converge? Are the two conditions both necessary? Also how do the assumed result relates to the two conditions? I think the presence of the two conditions (1) and (2) are really confusing because co dituib (1) refers to convergence if a particular sequences within the sequences of sequences. Condition (2) reminds me the differences between the box vs the product topology. – Seth Mai Aug 11 '19 at 8:00
• I'm not sure I entirely understand your question. Whether the two conditions are really necessary is entirely dependent on the context. Why was this particular mode of convergence defined? What kind of objects is it supposed to study? For example metrics are a clear example of an attempt to define some sense of distance on the elements of sets. For this three conditions are necessary. I suspect Q-convergence was made just as an exercise to challenge people to play around with different topologies and how they can be different, so yes it is a bit arbitrary. – Floris Claassens Aug 11 '19 at 8:50
• I am asking is what the question asks of the reader to show similar to the following question: there does not exists a metric $d$ on $C([0,1])$ such that $lim_{{n} \rightarrow \infty} d(f_n, f_0) = 0$ if and only if the sequence $\{f_n\}_{n \in N}$ converges pointwise to $f_0$ – Seth Mai Aug 11 '19 at 10:33

You are not asked to prove that the convergence in this question satisfies properties (1) and (2), you are instead told that convergence in this question is defined by these properties.

The word "convergence" here is not being used in the standard manner, instead it is being used in an abstract manner.

And in fact what you are asked to do is to prove that this "convergence" cannot be defined in the standard metric manner no matter what metric one chooses.