Showing that $U$ open in $(X, d) \Rightarrow U$ open in $(X, d')$

Question

We have $X$ as a non empty set and $d$ and $d'$ are metrics on $X$.

Suppose there exists $c > 0$ with $d(x,y) \leq cd'(x,y)$ for all $x,y \in X$.

I want to show that $U$open in $(X,d)$ $\Rightarrow U$ open in $(X,d')$ and I am struggling on where to start with this specific problem.

I know the definition of an open set is the following: Let $(X,d)$ be a metric space. A subset $U \subset X$ is an open set if for each $x \in U$ there exists $r > 0$ such that $B_{X}(x,r) \subset U$.

To show that $U$ is open in the metric $(X,d)$ I first assume that $U$ is open for every $u \in U$.

I am not sure whether to use the definition to prove this so that is where I am struggling.

Answer 1

HINT

Assume $U$ is open in $(X,d)$. Let $u \in U$. Then $\exists r > 0$ such that the open ball around $u$ with radius $r$ (call it $B_r(u)$) is also in $U$. Can you find $r' > 0$ such that $B_{r'}(u) \subset U$, where the ball is with respect to the $d'$ metric?

Answer 2

Suppose that $U$ is open in $(X,d)$. Then, for each $u\in U$, there is a $\varepsilon_u>0$ such that $d(x,u)<\varepsilon\implies x\in U$. But then\begin{align}d'(x,u)<\frac{\varepsilon_u}c\implies&c.d'(x,u)<\varepsilon_u\\\implies&d(x,u)<\varepsilon_u\\\implies&x\in U.\end{align}

Answer 3

Suppose that $U$ is open in $(X,d)$ for every $x\in U$, there exists $e>0$ with $B_d(x,e)\subset U$. Consider $B_{d'}({e\over c})$, for every $y\in B_{d'}({e\over c})$, $d'(x,y)<{e\over c}$ this implies that $d(x,y)\leq cd'(x,y)<e$, we deduce that $B_{d'}(x,{e\over c})\subset B_d(x,e)\subset U$.