The question reads: In $\mathbb{R}$ (equipped with its standard topology), determine the interior of the set $E$ := {$\frac{1}{n} | n \in \mathbb{N}$}.
Now, in $\mathbb{R}$, $E \equiv (0,1]$. My book states that by definition, the interior of E, is the union of all open sets which are contained in E.
Clearly, the set $(0,1)$ is open with respect to the standard topology of $\mathbb{R}$ and it is also contained in $E$. So wouldn't int($E$) = ($0,1$)?
The solution my professor posted simply states that int($E$) = $\varnothing$. I don't understand how this can be the case.
Thanks.
Best Answer
As you say the interior is the union of every open set included in $E$. What are the open sets in $\mathbf{R}$? The ones containing an open interval, generically $(a,b)$.
Now, is there any nonempty $(a,b)$ included in $E$? The answer is no: for every $(a,b)$, either it does not cut $E$, or there is a $1/n$ in $(a,b)$, but then any $1/n + \varepsilon$, for $\varepsilon > 0$ small enough, is in $(a,b)$ but not in $E$, hence $(a,b)$ is not included in $E$. At last, no nonempty $(a,b)$ is contained is $E$, that is no non-trivial open set, that is to say $E$ has empty interior.
Graphically, $E$ is purely discrete, it consists in isolated points, you have no open (=balls=intervals) in it: for each point of $E$, its neighbors are not inside.