Let $X$ be our topological space, and $A\subset C \subset X$.
If $A$ is dense in $X$, then we would have that $\bar A = X$.
What does it mean to say $A$ is dense in $C$?
Is it $\bar A \cap C = C$, or $\bar A = C$?
Asked
Active
Viewed 59 times
2
-
1Let me answer with another question: Is "$\bar{A}$" the closure of $A$ in $X$, or the closure of $A$ in $C$ (where $C$ is a subspace of $X$)? – Neal Jun 05 '19 at 15:47
-
1Follow-up comment: Consider your question in the context of $A = {q\in\mathbb{Q}\ |\ 0<q<1}$, $C = (0,1)$, and $X = \mathbb{R}$. – Neal Jun 05 '19 at 15:47
-
@Neal thanks for the comments. I'm using the closure of A in X. I get your point now. ;) – An old man in the sea. Jun 05 '19 at 16:00
-
Related. – Cameron Buie Jun 05 '19 at 16:33
-
A useful tool is that if $A\subset X$ and $A$ is dense in $X$ then $A\cap U$ is dense in $U$ for every open $U\subset X.$ – DanielWainfleet Jun 06 '19 at 04:32
1 Answers
2
It is $\overline{A}^{(C)}=C$, where the closure is taken in the subspace $C$.
But it is well known that $\overline{A}^{(C)}=\overline{A} \cap C$ where the right hand closure is taken in $X$. And so it is equivalent too $C \subseteq \overline{A}$.
Henno Brandsma
- 242,131
-
-
Henno, by the way, do we also have $int_C(A)=int(A)\cap C$ and $\partial_C(A) = (\partial A )\cap C$? – An old man in the sea. Jul 08 '19 at 14:23
-
@Anoldmaninthesea. No to interior as witnessed by the rationals inside the reals, e.g. So boundary also fails. – Henno Brandsma Jul 08 '19 at 15:32