I have this formula, $A=\operatorname{int}(A)\cup \delta(A)$. It says that a set is equal to its interior union its boundary. What goes wrong here: $A=$ the rationals on the real line. Then $\operatorname{int}(A)=\{\}$. Boundary of $A$ is $\mathbb{R}$, so $A=\mathbb{R}$, but $A$ is not $\mathbb{R}$?
$\mathbb{Q}=\mathbb{R}$ in usual topology?
2
$\begingroup$
general-topology
-
0Seems like a typo yes, i think I get this now, thanks. close topic – 2011-12-31
1 Answers
8
Your formula should be $\overline{A} = \operatorname{Int(A)} \cup \operatorname{Bd}(A)$ for subsets $A$ of a topological space $X$. And for $\mathbb{Q}$ the closure equals the boundary, namely $\mathbb{R}$, so no problem there....