# $\mathbb{Q}=\mathbb{R}$ in usual topology?

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 $\mathbb{R}$?

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Your formula is false. (If every set contained its closure, all sets would be closed, and they are not...) – Mariano Suárez-Alvarez Dec 31 '11 at 17:51
A set ic closed if and only if it contains its closure. Not every set is closed. – Michael Greinecker Dec 31 '11 at 17:53
My source for the foprmula is page 7 on math.cornell.edu/~hatcher/Top/TopNotes.pdf – tourist_in_math Dec 31 '11 at 17:54
@MarianoSuárez-Alvarez It says boundary, not its closure – tourist_in_math Dec 31 '11 at 18:00
The problem is on p.7 of the pdf file, labelled p.6. It gets stated that $A$ and $\bar{A}$ are equal to the interior union the boundary, a couple of lines apart in the para after the enumerated list. The second statement just looks like a typo, it should be $\bar{A}= \operatorname{int}(A) \cup \partial A$. – m_t_ Dec 31 '11 at 18:08

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....