Franklin's Notes

Metric topology

Every metric space has a topology induced by its metric, in which the open balls are the subsets of $M$ taking the form and the open subsets of $M$ are precisely those containing an open ball about each of their points. That is, $U\subset M$ is said to be open iff for each $u\in U$, there exists some $r>0$ such that $B(u;r)\subset U$.

The set on the left is open, but the set on the right is not open, and the point $u$ is a point in the set for which the set contains no open ball containing that point.

Proposition 1. A subset $B\subset M$ is closed iff $(x_n)\subset B$ and $\lim x_n = x\in M$ implies $x\in B$. In other words, the closed subsets of a metric space are precisely those containing the limit points of each of their sequences.

Proof. First, suppose $B$ is closed, so that $M\backslash B$ is open. Suppose that there existed $(x_n)\subset B$ with $\lim x_n=x\in M\backslash B$. But then there exists an open ball $B(x;r)\subset M\backslash B$ with $r>0$, meaning that $d(x,x_n)\geq r$ for all $n\in\mathbb N$, contradicting the assumption that $\lim x_n = x$. Thus, we have that $x\notin M\backslash B$ by contradiction, and so $x\in B$ as desired.

Conversely, suppose that $B$ is not closed, so that $M\backslash B$ is not open, and there exists a point $x\in M\backslash B$ with no open ball $B(x;r)$ entirely contained in $M\backslash B$. Then we have that each of the intersections $B(x;r)\cap B$ is nonempty, for $r>0$. By the Axiom of Choice , we may let $x_n\in B(x;n^{-1})\cap B$ for each $n\in\mathbb N$, so that $d(x,x_n)\to 0$ and hence $\lim x_n = x$ for some $(x_n)\subset B$ and yet $x\notin B$. $\blacksquare$




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