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Let $c$ be the set of all convergent sequences $x=(x_n)$, then $c$ is a metric space with metric

$$d(x,y)=\text{sup}_{1\leq n<\infty}|x_n-y_n|.$$

I have to check all the conditions of metric space.

  • $0=|x_n-x_n|\geq \text{sup}_{1\leq n<\infty}|x_n-x_n|$, if $\text{sup}_{1\leq n<\infty}|x_n-x_n|<0$, then the distance function $d$ will be negative, which is not possible thus $d(x,x)=\text{sup}_{1\leq n<\infty}|x_n-x_n|=0.$

If this is correct, then I can proceed the rest of conditions.

Thanks!

Unknown
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  • what are you trying to show here, that $\sup_n |x_n-x_n|=0$? isn't that obvious since $|x_n-x_n|=0$ for all $n$ and you are just taking the supremum of zeros? – hgmath Feb 19 '21 at 16:34
  • @hgmath Yes I know $sup$ of $0$ is zero. The way I showed right too...isn't it. – Unknown Feb 19 '21 at 17:49

1 Answers1

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We will check the conditions one by one.

  1. (Distance zero iff the convergent sequences are exactly same) If the sequences are exactly same for all $n$, $| x_n-y_n| = 0$, hence the supremum is also $0$. On the other hand, if the supremum is zero, all points in the set must be $0$ as the set contains only points which are non-negative (due to the modulus). Hence, it follows that the sequences are same.

  2. (Distance between two distinct points is positive) If the sequences are different, at least one term will be different, for that particular $n$, $| x_n-y_n| > 0$, so the $sup>0$.

  3. (Symmetry) Follows from the symmetry of the modulus function.

  4. (Triangle Inequality) Hint: Use the triangle inequality for modulus for every $n$ seperately.

Agile_Eagle
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