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Let $H$ be the set of all sequences of real numbers $x={X_n} $ such that $|X_n|\leq1 $ for all $n$ belonging to Natural Number set $\mathbb{N}.$

Consider the function $d: H X H \to R $ given by $d(x,y) = \sum[(|X_n-Y_n|)/Z^n] $, where $x={X_n} $ and $y={Y_n}$ belong to $H. $ Prove that $(H,d) $ is a metric space.

P.S. Here $Z$ in $d(x,y) $ is not clarified in the question itself so we need to guess what it might be :(

My approach: I could prove this mostly but I have confusion over two points 1. What can $Z $ represent here and how to tackle $ Z$ in $d(x,y) $ in proving triangle inequality axiom. 2. What is the use of $ |X_n|\leq 1$ in this problem?

Also TIA to anyone who formats my question for these symbols :)

user159888
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Atul Singh
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    My guess is they might have hand-written $2^{n}$ which has been mis -read as $Z^{n}$. The condition $|x_n| \leq 1$ is to make the series defining the metric convergent. – Kavi Rama Murthy Apr 12 '20 at 00:06
  • Here's a MathJax tutorial :) – Shaun Apr 12 '20 at 00:13
  • That makes sense. It must be what you are suggesting. Out of curiousity can we have a metric space where H consists of non convergent sequences as well? – Atul Singh Apr 12 '20 at 00:17
  • $H$ has nonconvergent sequences, e.g. $(-1)^n$. – Berci Apr 12 '20 at 00:33
  • Oh yeah true! What do u think might be the reason of the condition |Xn| <=1 and also can we make a metric space out of non convergent sequence ? – Atul Singh Apr 12 '20 at 00:38
  • For bounded sequences, the formula for $d(x,y)$ is a convergent series (assuming $Z=2$, nevertheless $Z>1$ suffices), so it gives a finite number. While without assuming boundedness, it's not guaranteed any more, meaning that $d(x,y)=\infty$ for certain $x, y$, which is not allowed for a metric. – Berci Apr 12 '20 at 09:22

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