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The metric $d: \mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by

$$d(x, y) = \left\{ \begin{array}{lr} ||x||+||y|| & \mbox{if } x \neq y \\ 0 & \mbox{if } x = y \\ \end{array} \right.$$ is a variation of the SNCF metric. Given that $d$ is a metric, show that if $X = \mathbb{R}^2$, then $(X, d)$ is a complete but not connected metric space.

My attempt:

Complete:

A Metric Space is said to be complete if every Cauchy Sequence converges in the space.

Let $\{x_n\}$ be a Cauchy sequence in $d$. We need to show that for all $\epsilon > 0$, there is an $N \in \mathbb{N}$ so that $d(x_n, x) < \epsilon$ (for some $x \in X$) for all $n \geq N$.

Not very sure how to proceed here.

Not connected:

$(X, d)$ is disconnected if it can be written as the union of two non-empty separated sets. We need to find $\emptyset \neq A, \emptyset \neq B \subseteq X$ so that $X = A \cup B$ where $\overline{A} \cap B = A \cap \overline{B} = \emptyset$.

I cannot find any such subsets here.

Any assistance is much appreciated.

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The most straightforward approach is to figure out what topology on $\Bbb R^2$ is generated by $d$.

HINT: If $x\ne\langle 0,0\rangle$, and $0<r\le\|x\|$, what is $B(x,r)$? And what is $B(\langle 0,0\rangle,r)$? You should try to figure this out on your own, but I left a description, without proof, in a spoiler box in my answer to your previous question about this metric.

Once you have that, showing that the space is not connected is very easy. To show that it is complete, you need to figure out what the Cauchy sequences are.

HINT: Every Cauchy sequence in $\left\langle\Bbb R^2,d\right\rangle$ either converges to the origin or is eventually constant.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Mar 18 at 22:11

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