I have problems proving the following statement. Prove that two metrics$\quad d_1,d_2 $ are topologically equivalent if and only if $$\forall x\in E\quad \forall \epsilon>0 \quad\exists \delta>0:$$ $$\forall y\in E\quad d_1(x,y)<\delta \Rightarrow d_2(x,y)<\epsilon$$ $$\quad \quad \quad \quad d_2(x,y)<\delta \Rightarrow d_1(x,y)<\epsilon$$
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(I) what is a basis of a topology? (II) what is a basis of a metric space topology? – Will Jagy Oct 31 '18 at 16:01
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Basis of a topology is $$\beta={B_r(x) \quad x \in E, \quad r>0}$$ – Emerald Oct 31 '18 at 16:09
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how could you show that a basis set for one metric is open in the other metric? – Will Jagy Oct 31 '18 at 16:12
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this answer of mine might be relevant to you. – Henno Brandsma Oct 31 '18 at 17:13
1 Answers
Two topologies say $\mathcal T_1, \mathcal T_2$ on a set $X$ are equivalent, if you have $$\forall x\in X, \forall \mathcal U_1 \in \mathcal T_1 \text{ with } x\in \mathcal U_1,~ \exists \mathcal U_2 \in \mathcal T_2 \text{ such that}\quad x\in \mathcal U_2\subset \mathcal U_1 $$ and $$\forall x\in X, \forall \mathcal U_2 \in \mathcal T_2 \text{ with } x\in \mathcal U_2,~ \exists \mathcal U_1 \in \mathcal T_1 \text{ such that}\quad x\in \mathcal U_1\subset \mathcal U_2. $$
Assume you have the hypothesis you gave above, let $x\in X$ and let $\mathcal U_1 \in \mathcal T_1$ such that $x\in \mathcal U_1$, there exists $r>0$ such that $B_{d_1}(x,r)\subset \mathcal U_1$, let such an $r>0$. By hypothese, there exists $\delta>0$ such that $$ \forall y\in X, d_2(x,y)\leq \delta \Rightarrow d_1(x,y)\leq r.$$ Let such a $\delta>0$, we have : $$ \mathcal U_2 := B_{d_2}(x,\delta)\subset B_{d_1}(x,r)\subset \mathcal U_1 \quad \text{ and } \quad x\in \mathcal U_2$$
You can proceed the same way to prove the other part of the statement.
Now, assume the two topologies are equivalent, Let $x\in X$ and let $\varepsilon>0$, the ball $B_{d_1}(x,\varepsilon) \in \mathcal T_1$ thus there exists a $\mathcal U_2 \in \mathcal T_2$ such that $\mathcal U_2 \subset B_{d_1}(x,\varepsilon)$ and $x\in \mathcal U_2$. Let such an $\mathcal U_2\in \mathcal T_2$. There exists $\delta>0$ such that $B_{d_2}(x,\delta)\subset \mathcal U_2$, let such a $\delta>0$. We have : $$ x\in B_{d_2}(x,\delta) \subset \mathcal U_2 \subset B_{d_1}(x,\varepsilon)$$ thus $$ \forall y\in B_{d_2}(x,\delta), y\in B_{d_1}(x,\varepsilon)$$ which can be stated also : $$ \forall y\in X, d_2(x,y)< \delta \Rightarrow d_1(x,y)\leq \varepsilon$$
The other statement can be proven the same way.

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