0
$\begingroup$

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.

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

$\endgroup$
  • 1
    $\begingroup$ which definition? $\endgroup$ – Hagen von Eitzen May 9 '18 at 3:29
  • 1
    $\begingroup$ @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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.