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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.

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4 Answers 4

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Let $S=\{x_1,\ldots,x_n\}\subset \mathbb R$. If $x_j\in S$, then any open ball centered at $x_j$ contains infinitely many points, so clearly it contains a point not in $S$, and so $x_j$ isn't an interior point of $S$. Hence the interior is empty.

A singleton set $\{x\}$ is closed as $x=\bigcap_{n=2}^\infty \left[x-\frac1n,x+\frac1n\right]$ is the intersection of closed intervals. So $S=\bigcup_{j=1}^n \{x_j\}$ is closed as the finite union of closed sets, and hence $S=\overline S$.

Since we have $\overline S=S^\circ\cup\partial S$, $S^\circ=\varnothing$, and $S=\overline S$, it follows that $S=\partial S$.

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First, recall the definitions:

  1. 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 $S$
  2. Boundary point: Every neighborhood of $x$ intersects both $S$ and the complement of $S$
  3. 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 $S = \{\}$.

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None of the points in your finite set $A \subset \mathbb{R}$ are interior points: no open interval around any point $x \in A$ is contained in $\{ x \}.$ This shows that the interior of $A$ is the empty set.

On the other hand the set of boundary points of $A$ consists exactly of $A$: every point $x \in A$ is a boundary point since every open interval around $x$ contains $x$ but also points which are not in $A$; but for any point $x \not \in A$ you can find a small enough open interval around $x$ which contains not point in $A.$

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An open set contains an open interval and thus has infinitely many points. Thus the interior of a finite set is empty. Any point outside the set has a positive minimum distance from points in the set, so there is an open interval around it not intersecting the set. Thus the set is closed because its complement is open. A closed set with empty interior is equal to its boundary.

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