Let $(X,d)$ be a metric space, Let $\mathscr T$ be a topology induced by $d$, and let $\mathscr U$ be the finite complement topology on $X$. Prove that $\mathscr T$ is finer than $\mathscr U$.
My attempt:-
Let $A\in \mathscr U$, then $X \setminus A$ is finite. I am able to deduce upto this. How do I prove that $A$ is open with respect to $d$? Please help me. I tried like this. Let $x \in A$. How do I choose open ball $B(x;r)\subset A$?
Best Answer
Let $\{p_1,\ldots,p_r\}=X\setminus A$ and define $r=\min\bigl\{d(x,p_j)\,|\,j\in\{1,2,\ldots,r\}\bigr\}$ (unless $X\setminus A$ is empty, in which case $r$ can be any positive number). Then $B(x;r)\subset A$.