$A\cup A'=A\cup \partial A$ Let $A$ be a subset of $\mathbb{R}$. Suppose $A'$ and $\partial A$ represent the sets of all limit points and of all boundary points of $A$ respectively. 
Then $A\cup A'=A\cup \partial A$
Is  this true or false? I am trying to come up with some example to show that it is false, but till now my all attempts have failed.
 A: This is true. If $x\in A\cup A'$, either $x\in A$ (then $x\in A\cup \partial A$) or $x\in A'$. If $x\in A'\setminus A$, there exists a sequence $x_n\to x$ with $x_n\in A$. There also exists a sequence $y_n\equiv x\to x$ with $y_n\notin A$, so $x\in \partial A$. 
If on the other hand $x\in \partial A$, there exists a sequence in $A$ with $x_n\to x$. If all such sequences are constant, then $x\in A$. If there exists a sequence with $x_n\neq x$ for all $n$, then $x\in A'$. Hence $A\cup \partial A \subset A\cup A'$.
A: It is true. By the definition of the boundary you always have $A' \supset \partial A$, though the other points in $A'$ should already belong to $A$.
A: It's true. Let $A^c$ be the complement of $A.$ We have $A\cup A'=\overline A.$ And we have $$ A\cup \partial A\subset \overline A \cup \partial A=$$ $$=\overline A \cup (\overline A \cap \overline {A^c})=$$ $$=\overline A=$$ $$=A\cup (\overline A \backslash A)=$$ $$=A\cup (\overline A \cap A^c)\subset A\cup (\overline A \cap \overline {A^c})=$$ $$=A\cup \partial A.$$
