Show that the sequence $(x_n)$ defined by $x_n=\left(\frac{1}{n-1}, \frac{n-2}{n-3} \right)$ is divergent in discrete metric space

cauchy-sequencesconvergence-divergencemetric-spacessequences-and-seriessolution-verification

Let $(\Bbb R^2, d)$ be a discrete metric space. Let $(x_n)$ be a sequence in $\Bbb R^2$ defined by $x_n=\left(\frac{1}{n-1}, \frac{n-2}{n-3} \right)$ for all $n \in \Bbb N$. Show that $x_n$ is divergent.

Definition. Let $(X,d)$ be a metric space. A sequence $(x_n)$ in $X$ is said to be convergent to $x \in X$ if for any $\varepsilon>0$, there exists $N \in \Bbb N$ such that for all $n \in \Bbb N$ with $n \ge n_0$, we have $d(x_n,x) < \varepsilon.$

Definition. Let $(X,d)$ be a metric space. A sequence $(x_n)$ in $X$ is said to be a Cauchy sequence if for any $\varepsilon > 0$, there exists $N \in \Bbb N$ such that $d(x_n,x_m) < \varepsilon$ for all $n,m > N$.

Lemma. Let $(X,d)$ be a metric space. If $(x_n)$ in $X$ is a convergent sequence, then $(x_n)$ is a Cauchy sequence.

Attempt:

  1. First Approach:
    Notice that
    \begin{align*}
    \text{since} \ x_i \ne x_j \ \text{for any} \ i \ne j, \ \text{we have} \ d(x_i, x_j)=1, \qquad \ldots (*)
    \end{align*}
    by definition of $d$. To show that the sequence $(x_n)$ is divergent, by using the negation of the above Lemma, we want to show that $(x_n)$ is not a Cauchy sequence, that is, we want to show that there exists $\varepsilon_0 > 0$ such that for any $N \in \Bbb N$, there exists at least one $n>N$ and $m>N$ such that $d(x_n,x_m)\ge \varepsilon$.
    To this end, we take $\varepsilon_0=1>0$. Then for any $N\in \Bbb N$, we choose a number $n>N$ and let $m:=n+1$. Clearly, $n \ne n+1$ for any $n \in \Bbb N$. Hence, by $(*)$, we have $d(x_n,x_{n+1})=1 \ge \varepsilon_0$. Thus, there exists $\varepsilon_0=1>0$ such that for any $N \in \Bbb N$, there exists $n>N$ and $m=n+1>N$ such that $d(x_n,x_{n+1})=1\ge \varepsilon_0$. Therefore, by definition, $(x_n)$ is not a Cauchy sequence. Hence, the sequence $(x_n)$ is divergent, as desired. $\qquad \Box$

  2. Second Approach:
    Notice that
    \begin{align*}
    \text{since} \ x_i \ne x_j \ \text{for any} \ i \ne j, \ \text{we have} \ d(x_i, x_j)=1, \qquad \ldots (*)
    \end{align*}
    by definition of $d$. Suppose that the sequence $(x_n)$ is convergent. Let $x_n \to x_j \in \Bbb R^2$ for some $j \in \Bbb N$. Then, for any $\varepsilon>0$, there exists $N \in \Bbb N$ such that for any $n \in \Bbb N$ with $n \ge N$, we have $d(x_n,x_j) < \varepsilon$. In particular, if we take $\varepsilon=\frac{1}{2}>0$, there exists $N \in \Bbb N$ such that for any $n \in \Bbb N$ with $n \ge N$, we have $d(x_n,x_j)<\frac{1}{2}$. Hence, $x_n=x_j$ for any $n \ge N$, by definition of $d$, i.e., $x_N=x_j,x_{N+1}=x_j,x_{N+2}=x_j$, and so on. By definition of $d$, we have $d(x_n,x_j)=0$ for any $n \ge N$. This contradicts the observation in $(*)$. Hence, the sequence $(x_n)$ is divergent, as desired. $\qquad \Box$

Does the both of above approach correct? If not, how to approach it correctly? If only one of them is correct, which one of them can I use to solve the problem? Thanks in advanced.

Best Answer

Both of your approaches are correct, as you realise that $d(x_n, x_m) = 1$ for $n \neq m$, which means the sequence cannot be Cauchy hence it cannot converge. Here I expand my comment into an answer.

For a quicker proof, observe the following fact: a sequence in a discrete metric space is convergent if and only if it is eventually constant. To prove this, note $(\Longleftarrow)$ is true for all metric spaces. For $(\Longrightarrow)$, observe that if $d(x_n, x) < 1$ for all $n \geq N$, then $x_n = x$. $\blacksquare$

In your case, your sequence is not eventually constant, hence it cannot converge in the discrete metric.

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