Metric topology question

Question

Let d be the standard euclidean metric but restricted to the set $[0, 1]$. Also define $d'(x, y) = | \sqrt x − \sqrt y|$, for all $0 ≤ x$, $ y ≤ 1$.

1. Are the metrics $d, d'$ equivalent on $X = [0, 1]$?

It seem the answer is no.

2. Do they have the same metric topologies?

It seems the answer is yes but i can't prove my claims any help much appreciated.

Answer 1

Note that $$|x-y|= | \sqrt x - \sqrt y\ |\le 2| \sqrt x - \sqrt y\ |$$

Also, $$| \sqrt x - \sqrt y\ |\le \frac {|x-y|}{|\sqrt x + \sqrt y|}\le \frac {|x-y|}{2 max (\sqrt x , \sqrt y)}$$

Thus the two metrics are equivalent.

Answer 2

You just need to show that each open ball $B_{r_1}(x)$ of one metric is contained in an open ball $B_{r_2}(x)$ of the other, and vice versa. Finding such a ball shouldn't be too bad. If you are still stuck after a while, try drawing some pictures to calculate what the radii should be.

Answer 3

The set of convergent sequences in a metric topology on a set $X$ will determine which subsets of $X$ are closed, and hence which subsets of $X$ are open. So if two metrics on $X$ have the same set of convergent sequences then the two metrics generate the same topology.

If $(x_n)_n$ is a sequence of members of $[0,1]$ and if $x\in [0,1]$ then, as $n\to \infty $ , we have $$d(x_n,x)\to 0\iff |x-x_n|\to 0 \iff$$ $$\iff |\sqrt x\;-\sqrt {x_n}|\;\to 0 \iff$$ $$\iff d'(x,x_n)\to 0.$$ So $d$ and $d'$ have the same set of convergent sequences.

Remark: "$\to 0$" means converging to $0$ as a sequence of real numbers in the usual sense. I only mention this as it can be confusing when we are considering a "non-usual" metric on a subset of $\Bbb R.$