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 $x_{n,p_n} \rightarrow x.$
Thank you in advance.
 A: 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.
A: 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.
