0

Let $\delta(x,y) = \bigg | \frac{1}{x}-\frac{1}{y} \bigg|$ and $d$ is the usual euclidean metric. Show that $(x_n)$ converges to $a$ using $\delta$ iff it converges to $a$ using $d$.

My attempt:

$(x_n)$ converges to $a$ using $\delta \Rightarrow \delta (x_n,a) \rightarrow 0 \Rightarrow \bigg| \frac{1}{x_n} - \frac{1}{a}\bigg| \rightarrow 0 \Rightarrow \frac{1}{x_n}\rightarrow \frac{1}{a}$ using $d \Rightarrow x_n\rightarrow a$ using $d$.

Conversely, suppose $d(x_n, a)\rightarrow 0 \Rightarrow |x_n-a| \rightarrow 0 \Rightarrow \bigg|\frac{1}{1/x_n} - \frac{1}{1/a}\bigg|\rightarrow0$ using $d$$\Rightarrow \frac{1}{x_n}\rightarrow \frac{1}{a} $ using $\delta \Rightarrow x_n \rightarrow a$ using delta.

Is this correct or is it circular?

fosho
  • 6,334

1 Answers1

1

Nitpick: you have $\frac{1}{x_n} \to \frac{1}{a}$ using $d$; what if some $x_n = 0$?

Otherwise it looks right to me.

Patrick Stevens
  • 36,135
  • Forgot to mention, here X is the set of strictly positive real numbers. Thanks for the response, much appreciated. – fosho Mar 21 '16 at 13:38