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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) $

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $x \in ∂X$

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