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:


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

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

This has a straightforward generalisation to a topological space.


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