0
$\begingroup$

In General Topology (Dixmier, 1984), Theorem 1.1.14 states that

Let $E$ be a set, $d$ and $d'$ metrics on E. Suppose there exist constants $c,c'>0$ such that $$ c d(x,y) \leq d'(x,y) \leq c'd(x,y)$$ for all $x,y \in E$. Then the open subsets of $E$ are the same for $d$ and $d'$.

The given proof is

Let A be a subset of $E$ that is open for $d$. Let $x_0 \in A$. There exists an $\epsilon>0$ such that every point $x$ of $E$ satisfying $d(x_0,x)<\epsilon$ belongs to $A$. If $x \in E$ satisfies $d'(x_0,x)

I don't understand this line:

If $x \in E$ satisfies $d'(x_0,x)

Surely if $d(x_0,x)<\epsilon$ then from the inequality $d'(x_0,x)

  • 0
    $c d(x_0,x) \leqslant d'(x_0,x) < c\epsilon$; cancel $c > 0$. – 2017-01-16
  • 0
    How do we know $d'(x_0,x) – 2017-01-16
  • 0
    That's the assumption we start with, "If $x\in E$ satisfies $d'(x_0,x) < c\epsilon$". – 2017-01-16
  • 0
    Oh yes, I understand the implication, but why does this hold for all $x \in A$? When I thought about proving it myself I thought I was supposed to get $d(x_0,x)<\epsilon \implies d'(x_0,x)<\epsilon$. – 2017-01-16
  • 1
    It doesn't (generally) hold for all $x\in A$. What is being shown is that for a $d$-open set $A$, every $x_0 \in A$ is a $d'$-interior point of $A$. So we need to find a $\rho > 0$ so that the $d'$-ball with centre $x_0$ and radius $\rho$ is contained in $A$. Now, since $A$ is $d$-open, we know that there is an $\epsilon > 0$ such that the $d$-ball with centre $x_0$ and radius $\epsilon$ is contained in $A$. Now if we can find $\rho > 0$ such that $d'(x_0,x) < \rho \implies d(x_0,x) < \epsilon$, we're done. Because that means the $d'$-ball with radius $\rho$ is contained in the $d$-ball – 2017-01-16
  • 1
    with radius $\epsilon$, which we know is contained in $A$. To get an implication $d'(x_0,x) < \rho \implies d(x_0,x) < \epsilon$, we need an inequality $d(y,z) \leqslant K\cdot d'(y,z)$. We have that, with $K = \frac{1}{c}$, just written with the constant on the other side of the inequality, as $c\cdot d(y,z) \leqslant d'(y,z)$. So we have $$\frac{\rho}{c} > \frac{d'(x_0,x)}{c} \geqslant d(x_0,x).$$ And we want $d(x_0,x) < \epsilon$. We get that by choosing $\rho$ so that $\frac{\rho}{c} \leqslant \epsilon$. Pick the largest choice, $\rho = c\epsilon$. – 2017-01-16

0 Answers 0