I know the usual induced topology using the collection of open balls. But the question arises from the following problem:

Let $(X,d)$ be a metric space, where X is a closed set. I now want to verify, that the metric above induces said topology. One axiom to check if it is a topological space, is to see if $X$ and $\emptyset$ themselves are open. Obviously, $\emptyset$ is, but for $X$ I could pick a point on the edge.

Say for $[0,1]\subseteq \mathbb{R}$ there is no open ball around $1$ that is completely contained in $X$. Hence, X is not open. But it does induce a metric ( for example with the euclidean distance ). This would mean there is a metric space which doesn't induce a topological space with the above topology. I know that this can't be though, so where is the mistake in the argument?

Disclaimer: I know there are tons of questions about metric induced topologies, but to my knowledge none cover this specific problem.


Think of "open" and "closed" as relative terms.

$[0,1]$ is closed relative to $\Bbb R$, but $[0,1]$ is open relativ to itself. Indeed any set is open relativ to itself because that is how topologies are defined.

Sitting in $0\in[0,1]\subset \Bbb R$ you can leave the set by an infinitesimal move to the left. But sitting in $0\in[0,1]$ with not ambient space like $\Bbb R$ there is no way to move left. There is nothing left of $0$ inside of $[0,1]$. You cannot leave the set by an infinitesimal move, hence it is an inner point like all the other points.


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