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Consider$d((x_1,y_1),(x_2,y_2))=\max\{d_1(x_1,x_2),d_2(y_1,y_2)\}$ and $e((x_1,y_1),(x_2,y_2))=d_1(x_1,x_2)+d_2(y_1,y_2)$. Prove the metrics are equivalent.

Attempted solution:

$U$ is an open set of $(X,e)\implies\forall a\in U \:\:\exists\epsilon>0$ such that $B_{e}(a,\epsilon)\subseteq U$

So that $B_e(a,\epsilon)=\{x\in X:e(a,x)<r\}$

So there exists $B_d(a,\epsilon)=\{x\in X:d(a,x)<\epsilon\}\subseteq B_d(a,\epsilon)\subseteq U$ once $\forall x\in B_d(a,\epsilon)$ $d(a,x)=\max\{d_1(a_1,x_1),d_2(a_2,x_2)\}\leqslant d_1(a_1,x_1)+d_2(a_2,x_2)=e(a,x)<\epsilon$

Question:

Do I need to show the reverse inclusion? If so. How should I do it?

Thanks in advance!

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Notice that for every $x,y\in X$ you have that

$$d(x,y)\leq e(x,y)\leq 2d(x,y).$$

This guarantees that both metrics are equivalent (see Strong equivalence).

Intuitively, the above equation says that given any open set in one metric, you can put inside another set in the other metric and vice versa. Hence both topologies have the same neighbourhoods and thus the same open sets.

EDIT:

The other implication follows from the above equation. Explicitly: for a point $y\in B_e(x,r/2)$ we get that:

$$d(x,y)\leq e(x,y)<r, $$

showing that $y\in B_d(x,r)$.

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  • $\begingroup$ So my answer would be enough? $\endgroup$ – Pedro Gomes Nov 9 '18 at 19:41
  • $\begingroup$ No. You should prove the other inclusion. Taking an open set on $(X,d)$ and proving that it contains a ball of $e$. $\endgroup$ – Dog_69 Nov 9 '18 at 20:03
  • $\begingroup$ That is what I do not know how to do. Could you provide an answer? $\endgroup$ – Pedro Gomes Nov 9 '18 at 20:15
  • $\begingroup$ I have edited the answer and add and edit. $\endgroup$ – Dog_69 Nov 9 '18 at 21:06

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