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.
[Math] For a finite set in $\mathbb{R}$, the interior is empty and the closure and boundary are the set itself
metric-spacesreal-analysis
Best Answer
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 $S=\partial S$.