Let $(X,d)$ be a metric space. We know that if $x \in X$ , then $Cl(\{x\})=\{x\}$, which implies that $\{x\}$ is closed. However if that's the case, what would the interior of $\{x\}$ be? I was reading this answer but I am still confused, especially when people say "that depends on what topology you're talking about". What if I'm talking about a metric space? Thanks.

I think I should explain a little bit more. The definition I have for "interior" is "the set of all interior points". And we defined "interior point" as "a point $y$ is an interior point of $X$ if there is $B(y;\epsilon) \subset X$. Now, for the set $\{x\}$, $x$ is the only point that is in this set apparently. But then if we start looking for this $\epsilon$, it seems like we have to know what's going on around the point $x$. That looks kinda weird to me because I assume when we define concepts such as "interior" or "open", we don't really have to know what exactly the set $X$ looks like?


1 Answer 1


Consult your definitions—the interior of a set is its maximal open subset. If your topology is e.g. discrete, then $\{ x \}$ is open, so it is its own interior. But in the standard topology on the reals, it's not open. In this case, the empty set is its only open subset, so that is its interior.

Both of the above cases are metrisable, so the restriction to metric spaces does not simplify much.

Edit: It should not be hard to see that our definitions are equivalent, but that is incidental. The topology is defined solely by the metric we have, so it's not so much that we know anything about $X$ as it is that we know how distances work. We can't pick $\epsilon = 0$, so what we need to know is that we can pick a positive $\epsilon$ without anything being that close to $x$. $\{ x \}$ will be its own interior iff we know it is "isolated" in this way.

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    $\begingroup$ As an addendum, singletons are open if and only if the metric space is induced by a discrete metric, so there's only one case in which you have a nonempty interior of singleton sets. $\endgroup$
    – Alan
    Apr 18, 2015 at 8:32

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