Statement:
Let $(X,d_1)$ and $(X,d_2)$ be two equivalent metrics. Such that The ball $B_{d_1}(x,r_1)\subseteq B_{d_2}(x,r)$. Similarly, $B_{d_2}(x,r_2)\subseteq B_{d_1}(x,r)$. Show that they have the same compact subset
Approach:
In order for these two to have the same compact subset. We need a $G$ which is closed and bounded wrt to $d_1$ this implies that it $G$ is also closed and bounded wrt to $d_2$. In order for $G$ to be closed wrt to $d_1$ $G^c$ has to be open in $X$, every point of $X\cap G^c$ is an interior point of $X\cap G^c$ in $X$. If we let $G^c$ be open in $X$, then for every $x\in G^c$ we can find a radii $r>0$ such that $y\in G^c$ whenver $y\in X$ and $d(x,y)<r$. Let $B_d(x,r)$ be the open ball in $X$, then
$$G^c=\bigcup_{x\in G^c}B_{d_1}(x,r_1)$$
is open in $X$ and $G=X\cap G^c$. Since $d_1$ and $d_2$ are equivalent , then by our definitoon
$$G^c\subseteq\bigcup_{x\in G^c}B_{d_1}(x,r_1)\subseteq\bigcup_{x\in G^c}B_{d_2}(x,r_2)=G^c$$
Hence, $G^c$ is open in both metrics. For these to both be bounded, $G$ has to be bounded in $X$ wrt to $d_1$ $\Longrightarrow$ $G$ is also bounded wrt $d_2$. Let $M>0$, since $d_1(x,y)\geq 0$, implying that $d_1(x,y) + 1 >0$ and $d_1(x,y) < d_2(x,y)+1$.
Questions:
- Is this a correct approach to show that they have the same compact subset?
- Is my proof that they have the same closed set valid?
- I don't know how to continue my proof for both of them being bounded, how would I continue?
Your approach is not correct since, in general, being compact is not equivalent to being closed and bounded.
Asserting that $K$ is compact with respect to $d_1$ means that, given a family $\{A_\lambda\mid\lambda\in\Lambda\}$ of open subsets of $(X,d_1)$ such that $K\subset\bigcup_{\lambda\in\Lambda}A_\lambda$, there is a finite set $F\subset\Lambda$ such that $K\subset\bigcup_{\lambda\in F}A_\lambda$. But the open sets of $(X,d_1)$ and the open sets of $(X,d_2)$ are the same sets. So, if this occurs in $(X,d_1)$, then it also occurs in $(X,d_2)$ (and vice versa).