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?
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general-topology
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5Your formula is false. (If every set contained its closure, all sets would be closed, and they are not...) – 2011-12-31
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1A set ic closed if and only if it contains its closure. Not every set is closed. – 2011-12-31
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0My source for the foprmula is page 7 on http://www.math.cornell.edu/~hatcher/Top/TopNotes.pdf – 2011-12-31
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0@MarianoSuárez-Alvarez It says boundary, not its closure – 2011-12-31
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0You mean under b) at the top of the page? That is a set about the whole space. If $X=\mathbb{R}$ and $A=\mathbb{Q}$, then $\mathbb{R}=\text{int}(\mathbb{Q})\cup\partial(\mathbb{Q})\cup\text{int}(\mathbb{R}-\mathbb{Q})=\emptyset\cup\mathbb{R}\cup\emptyset$. – 2011-12-31
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0Proposition 1.1 (c) and (d) on page 6 of the lecture notes states what the first two commenters point out. Where exactly do you believe to see a statement to the contrary on page 7? – 2011-12-31
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0sorry, page 6 says... "subsets when we want to emphasize the disjointness, so
" – 2011-12-31 -
1The 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 $\bar{A}= \operatorname{int}(A) \cup \partial A$. – 2011-12-31
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0Seems like a typo yes, i think I get this now, thanks. close topic – 2011-12-31