3
$\begingroup$

Let $(X, d)$ be a metric space where the set $X$ is infinite. Show that there exists an open set $O$ such that $O$ and $X \setminus O$ are both infinite and then show that there exists an infinite subset $Y \subseteq X$ such that $(Y, d)$ is a metric space where every subset $A \subseteq Y$ is open.

I'm quite new to the whole concept of metric spaces and I know the definitions of open set in a metric space. How do I find the set $O$ and how do I show it is infinite? Additionally, how do I find the infinite set $Y$?

$\endgroup$

1 Answer 1

5
$\begingroup$

First part: Consider the set $A\subset X$ of all open points. A point is open if the is a ball with center at the point that only contains the point itself and no other. If this set is infinite then it is easy. $O$ can be any infinite subset of $A$ such that $X\setminus O$ is infinite.

If the set of open points is finite, then let $x_0$ be a point that is not open. Therefore for every $n$ there are points other than $x_0$ inside the ball $\{x:\ d(x_0,x)<1/n\}$. In particular all these balls contain infinitely many elements. If for all $n$ the complement the ball $\{x:\ d(x_0,x)<1/n\}$ is finite, then the set of open points would be infinite. This is because those finitely many points in the complement of any of those balls are open points. Therefore, for some $n$ both the ball is infinite and its complement is infinite.

Second part: Again, we work under the assumption that the set of open points is not infinite. If it were, we can just take that set to be $Y$. If it is not, we already constructed one infinite open ball $O_0=O$ such that $X\setminus O_0$ is finite. Take a point $x_0\in O_0$ and put it in $Y$. Now consider $X\setminus O_0$. This set is infinite. So, as in the first part we can construct an infinite open ball $O_1$ such that $(X\setminus O_0)\setminus O_1$ is infinite. We can also pick it to be disjoint from the previous ball, by choosing small enough radius. Pick a point $x_1\in O_1$ and put it in $Y$. Continue in this way producing points $x_n$. The union of all $x_n$ satisfies the condition. This is because each $x_n$ has a ball separating it from all the other points that we put in $Y$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.