Let $([0,1],d)$ be a metric space. A set is said to be open iff for every element in the set there is some epsilon ball, containing the element, that lies withing the set. $[0,1]$ is said to be open within itself;
How can I possibly find an epsilon ball around the points $0$ or $1$? What's wrong here?
Thanks in advance?
Best Answer
We are viewing $[0,1]$ as a subspace of $\mathbb{R}$, so we are looking at the the subspace metric topology, i.e any open set intersected with $[0,1]$ is open in $[0,1]$. So about the point $0$ take $r=1/2$ and $B_r(0)=\{x\in\mathbb{R}:|x-0|<r\}$. Then
$$B_r(0)\cap[0,1] $$
is an open set under the subspace metric topology and contains the point $0$. The same can be done with the point $1$. Also, the whole subspace $[0,1]$ is open in itself since, taking $r=2$, we have
$$[0,1]=B_r(0)\cap[0,1] $$ for example ($r=2$ was arbitrary, I just needed a big enough radius to encompass all of $[0,1]$). Hope that helps!