Let $(X,d)$ be a metric space and let{$x_n$}and {$y_n$}be arbitrary sequences > in $ X$. Which of the following statements are true?

Question

[NBHM 2015- PhD Screening Test]

Let $(X,d)$ be a metric space and let{$x_n$}and {$y_n$}be arbitrary sequences in $ X$. Which of the following statements are true?

(A). If both {$x_n$} and {$y_n$} are Cauchy sequences, then the sequence of real numbers {$d(x_n,y_n)$} is a Cauchy sequence.

(B). If $d(x_n,x_{n+1}) < \frac{1}{n+1}$, then the sequence {$x_n$} is a Cauchy sequence.

(C). If $d(x_n,x_{n+1}) < \frac{1}{2^{n}}$, then the sequence {$x_n$} is a Cauchy sequence.

How to show (A) and (B)? It is given that (A) is true and (B) is false. I could able to prove (C). How to prove and disprove (B) in very short time?. Please help me.

Answer 1

(A) We will use $|d(x,y)-d(x,z)|\le d(y,z)$.

Let $\varepsilon > 0$ be given. Then there exists $N$ such that for $n,m\ge N$ we have that both $d(x_n,x_{n+m}) < \varepsilon/2$ and $d(y_n,y_{n+m}) < \varepsilon/2$. Hence, \begin{align*} |d(x_n,y_n) - d(x_{n+m},y_{n+m})| &\le |d(x_n,y_n)-d(y_n,x_{n+m})| + |d(y_n,x_{n+m}) - d(x_{n+m},y_{n+m})|\\ &\le d(x_n,x_{n+m}) + d(y_n,y_{n+m}) < \varepsilon. \end{align*}

You prove (B) in giving a counterexample (see Fred's answer)...

Answer 2

$(A)$: We have $|d(x_n,y_n)-d(x_m,y_m)| \le d(x_n,x_m)+d(y_n,y_m)$.

$B)$: Look at $X=\mathbb R$ with $d(x,y)=|x-y| and

$x_n=\frac{1}{2}(1+\frac{1}{2}+....+\frac{1}{n})$.