For a finite set in $\mathbb{R}$, the interior is empty and the closure and boundary are the set itself How do I show explicitly that for a finite set in $\mathbb{R}$ the interior is empty and the closure and boundary are the set itself?
For closure is simple: it is union of boundary and interior.But how to find boundary and interior analitically.

First, recall the definitions: 


*

*Closure: The union of a set and its limit points, or the union of a set and its boundary points, or even more broadly, $x$ is in the closure of $S$ if every open interval containing $x$ intersects $S$

*Boundary point: Every neighborhood of $x$ intersects both $S$ and the complement of $S$

*Interior point: at least one neighborhood of $x$ lies wholly in $S$.


Now, a finite set is going to have $n$ elements in it. Let $\epsilon_{0}$ be less than the shortest distance between any two distinct elements in the set, where $d(a,b)=|a - b|$. 
I say that every element, $k$ of the finite set $S$ is a boundary point: The interval $[k-\epsilon_{0},k+\epsilon_{0}]$ contains obviously contains $k$, and also $k+(\epsilon_{0}/2)$. This latter point cannot be an element of $S$ because of how we defined $\epsilon_{0}$. So each $k$ is a boundary point.
I also claim that the interior of a set $S$ is a subset of $S$: Any neighborhood of $x$ contains $x$. If a neighborhood of $x$ is wholly contained in $S$, then $x\in S$. Therefore, all interior elements are elements of $S$, so the interior of $S$ is a subset of $S$.
Finally, since all points in $S$ are boundary points, the requirements for boundary points are mutually exclusive with the requirements for interior points, and interior points must be in $S$, we can conclude that the interior of $S = \{\}$.
