# Proof of the theorem : Interior, Boundary and Exterior are mutually exclusive and exhaustive in $\mathbb{R}$

I wish to have proof for the theorem:

For any $A \subseteq \mathbb{R}$, we have: $Int(A) \cup Bnd(A) \cup Ext(A) = \mathbb{R}$.

where $Int(A)$ is the interior of $A$, $Bnd(A)$ is the boundary of $A$ and $Ext(A)$ is the exterior of $A$.

How can I start? Can you help me out.

[I know how to prove (from definitions) $Int(A), Bnd(A)$ and $Ext(A)$ are mutually exclusive:

ie.

$Int(A) \cap Bnd(A) = Bnd(A) \cap Ext(A) = Int(A) \cap Ext(A) = \emptyset$]

• which definition? – Hagen von Eitzen May 9 '18 at 3:29
• @HagenvonEitzen definitions of $Int(A), Bnd(A)$ and $Ext(A)$. Should I give the definitions as part of my question? – Vinod May 9 '18 at 4:02
• This is true for any topological space, not just metrizable ones. – Math1000 May 9 '18 at 4:15

Pick $x$.
If there is some $\epsilon>0$ such that $B(x,\epsilon) \subset A$ then $x$ in in the interior.
If there is some $\epsilon>0$ such that $B(x,\epsilon) \subset A^c$ then $x$ in in the exterior.
Otherwise, for all $\epsilon>0$, $B(x,\epsilon)$ contains points in $A$ and $A^c$ and hence $x$ in on the boundary.