# Are the following subsets open wrt the metric topology?

For each of the following metric spaces $(M_i , d_i)$ and subsets $V_i ⊂ M_i$ decide whether $V_i$ is open with respect to the metric topology

1) $M_1= \Bbb R^2,d_1((a,b),(x,y))=|a-x|+|b-y|,V_1=\{(x,y)\in\Bbb R^2|xy>1\}$

2)$M_2=\Bbb R^\Bbb N, d_2(a,b)=$$\sum_{n=1}^{\infty} \frac{2^{-n} |a_n-b_n|}{1+|a_n-b_n|},V_2=\{a \in M_2| \{n \in N \mid a_n \neq 0\} \text{ finite}\} Here is my attempt (by the way this is not a homework question its from a past exam I'm using to help me study) 1)if V is in open in M \implies \exists B_e(a,b) \subset V so choosing an arbitrary (a,b) \in V we choose (2,1) as this satisfies ab>1 now we have to see if there is an open ball from this point to any other point in V. i.e d((a,b),(x,y))<e d((2,1),(x,y))=|2-x|+|1-y| so we need to see is there a finite epsilon s.t. d<e and x,y satisfy xy<1 |2-x|+|1-y|<e \Rightarrow$$\sqrt{(2-x)^2}+\sqrt{(1-y)^2}<e \Rightarrow (2-x)^2+(1-y)^2<e^2=e$

$4-4a+a^2+1-b+b^2<e \Rightarrow a^2-4a-b+b^2<e \Rightarrow a^2(1-\tfrac{4}{a})+b^2(1-\tfrac{1}{b})<e$

and then as we increased the distance between these two points to the furthest apart they can be this implies that $a^2+b^2<e$ and if we suppose these points are still in V this implies that $a^2>\tfrac{1}{b^2}$ and so $a^2+\tfrac{1}{a^2}<a^2+b^2<e \Rightarrow a^2<e \Rightarrow a<e$ and so we have found that epsilon is greater than any real number and so there is no open ball and hence V is not open in M