I'll denote closure of A with $A_C$ because I cant get the bar for some reason. also $Int(A)$ is interior of A, $Bdry(A)$ is the boundary of A and A' the accumulation points. I'm trying to prove the following:

$Int(A) \cup Bdry(A)=A_C$

Is this a valid proof?

Take $x \in Int(A) \cup Bdry(A)$ then $x \in Int(A)$ or $x \in Bdry(A)$

if $x \in Int(A)$ then $x \in A_C$

if $x \in Bdry(A)$ then $x \in A_C \cap (X$ \ $A)_C$ thus $x \in A_C $ so $Int(A) \cup Bdry(A) \subset A_C$

Take $x \in A_C$ then $x \in A' \cup A$ thus $x \in A'$\ $A$ or $x \in Int(A)$

if $x \in A'$\ $A$ then $x \in (X$\ $A)_C $ so $x \in A_C \cap (X$ \ $A)_C$ and $x \in Bdry(A)$ so $x \in Int(A) \cup Bdry(A)$

if $x \in Int(A)$ then $x \in Int(A) \cup Bdry(A)$ so $A_C \subset Int(A) \cup Bdry(A) $

  • $\begingroup$ The fourth line doesn't seem right to me. It leaves out the points in $A'\cap (A-Int(A))$. They belong to $(X-A)_C$ though, so what follows still holds. $\endgroup$ – alex Mar 8 '14 at 8:00
  • 1
    $\begingroup$ fyi, the latex command for the bar is \overline and for the set difference backslash you're trying to do it's \setminus. Like this $\overline{(X\setminus A)}$. While we're at it, $X^{\circ}$ and $\partial X$ for interior and boundary might make things a little easier on the eyes, too. $\endgroup$ – Callus Mar 8 '14 at 8:04

Based on the flaws suggested in the comments this I think (IMHO) this is an easier way to approach some parts of the proof.

To prove the line that $x \in ∂X \implies x \in \overline A $

Suppose $x$ is in the boundary of $A$ and $x$ is not in some closed set $B$ which contains $A$. Then $x \in B^c$ which is open and hence there is a neighbourhood $V_x$of $x$ which entirely avoids $A$ leading to a contradiction since every neighbourhood of $x$ must contains elements in $A$ and $A^c$.

Similar reasoning can be used to show that $x \in \overline A \implies x \in A^{\circ}$ or $x \in ∂X$.

Suppose $x \in \overline A$ and $x$ is an exterior point of $A$. Then there is a neighbourhood of $x$ which entirely avoids $A$. But then there is a closed set which contains $A$ but not $x$. This leads to a contradiction since $x \in \overline A \implies x$ is in every closed set containing $A$. Then $x$ is not an exterior point of $A \implies x$ is either an interior point or a boundary point of $A \implies x \in A^{\circ}$ or $x \in ∂X$


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.