Why does the boundary formula give empty set for (a,b)?

[Silviu]

Hello! This is more of a set theory question I guess, but I have that the definition of the boundary of a subset A of a topological space X is $\partial A = \bar A \cap \bar B$, with $B = X - A$ (I didn't manage to put the bar over X-A, this is why I used B). I think I have a wrong understanding of the complement of a set because if I take (a,b) on the real axis, the boundary should be {a,b}, but $\bar A = (- \infty, a] \cup [b, \infty)$ while $B=R-A = (- \infty, a] \cup [b, \infty)$ so $\bar B = (a,b)$ and $\bar A \cap \bar B = \emptyset$. So where exactly I got it wrong? Thank you

 

[fresh_42]

What was $\overline{A} = \overline{(a,b)}$ again? It's not the complement though!

 

[Silviu]

What was $\overline{A} = \overline{(a,b)}$ again? It's not the complement though!

Doesn't $\bar A$ means all elements not in A? Which in this case is $(-\infty,a] \cup [b,\infty)$?

 

[fresh_42]

No, here it means the closure of $A$. That is the reason, why the bar isn't a good choice for complements in topology. Some write $\mathbb{R}-A= A^C$ which I find ugly. I prefer to write complements as $\mathbb{R}-A= \mathbb{R}\backslash A$. In any case, it's a matter of taste, but $\overline{A}$ as the topological closure of $A$ is pretty usual.

 

[Silviu]

No, here it means the closure of $A$. That is the reason, why the bar isn't a good choice for complements in topology. Some write $\mathbb{R}-A= A^C$ which I find ugly. I prefer to write complements as $\mathbb{R}-A= \mathbb{R}\backslash A$. In any case, it's a matter of taste, but $\overline{A}$ as the topological closure of $A$ is pretty usual.

Oh ok makes sense now. Thank you!