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A metric is defined on $X=(0,1)$ by $d(x,y)=\left|\frac 1x -\frac 1y\right|$.

Is this space complete or not?

My attempt: I have shown that there does not exist any $x_n$ that tends to 0 ($x_n\neq 0$ for all $n$). Because if $x_n$ tends to 0 then $1/x_n$ tends to infinity. Therefore $d(x_n,x_m)=\vert 1/x_n - 1/x_m\vert$ is not less than epsilon. Because keeping $m$ fixed and tending $n$ to infinity we can get $d(x_n,x_m)$ greater than epsilon.

Finally I have to show if $x_n$ tends to $x\neq 0$ then x is in $(0,1)$. $d(1/x_n, 1/x) < \epsilon$ implies $|x-x_n| <x \epsilon$. I can't proceed any further.

Noa Even
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Algebra Algebra
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1 Answers1

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Consider the sequence $x_n=1/n$. Then $x_n\in(0,1)$ and $x_n$ is Cauchy, since $d(x_n,x_m)=|1/n-1/m|\le1/n+1/m \le2/N$, for $n,m\ge N\gt0$.

$x_n$ doesn't converge in $X$. Take a prospective limit, say $l\in(0,1)$. Then let $N$ be large enough so that $x_N\lt l/2$. Then $d(x_n,l)\ge l/2$, for all $n\ge N$.

Hence $(X,d)$ is not complete.

  • You should show (or at least mention!) that $(x_n)$ is a Cauchy sequence in the given metric. The fact that $x_n\to 1$ in the Euclidean metric is irrelevant here. – TonyK Mar 28 '20 at 11:32
  • @TonyK my mistake. I should have read more carefully. –  Mar 28 '20 at 11:38
  • But $d(x_n,1)$ is not defined, because $1\notin X$. You have to look at $d(x_n,x_m)$. – TonyK Mar 28 '20 at 12:09
  • @TonyK thanks. I really bogeyed this one. –  Mar 29 '20 at 20:35