# Closed, open subspaces, dense subsets

Two questions from Dugundji's book (not hw, just practice).

1) Let $Y_{1}, Y_{2}$ be subspaces of $X$ and $A \subset Y_{1} \cap Y_{2}$. Assume that $A$ is open in $Y_{1}$ and open in $Y_{2}$. Prove A is open in $Y_{1} \cup Y_{2}$.

Can you please give a hint for this one?

2) a. Let $D$ be dense in $X$. Give an example to show that $D \cap A$ need not to be dense in $A$.

Can't we take $X = \mathbb{R}$, $D=\mathbb{Q}$ then $D$ in dense in $X$. Now take $A =$ irrationals, since the empty set is closed then it cannot be dense in $A$.

b. If $A$ is dense in $B \subset X$ then $A$ is dense in $\overline{B}$.

Attempt:

Let $V \subset \overline{B}$ be an open set, then by definition of subspace topology we have $V = C \cap \overline{B}$ where $C$ is an open subset of $X$. Now consider $C \cap B$ ,this is an open set in $B$ so $A$ intersects this set and hence $A$ is dense in $\overline{B}$. What bothers me, is how do we know that $C \cap B$ is non-empty?

Thanks.

-
In 1), do you mean $A\cap Y_i$ is open in $Y_i$ for $i=1,2$? Or is $A$ actually contained in $Y_1\cap Y_2$? –  Jonas Meyer Dec 22 '10 at 7:03
1) Write down the definition of the subspace topology. 2) Yes, this is fine. 3) Show that if C cap B is empty then C contains a point not in the closure of B. Note that you have not yet used this condition. –  Qiaochu Yuan Dec 22 '10 at 7:06
@Jonas Meyer: just corrected it. Still stuck. –  Paulo Dec 22 '10 at 7:20
@Paulo: Thank you. I thought that must be what you mean, because in the other interpretation it would be false. –  Jonas Meyer Dec 22 '10 at 7:25
@Qiaochu Yuan: what is the relevance of showing "C contains a point not in the closure of B?$. Confused about this hint. – Paulo Dec 22 '10 at 9:57 ## 1 Answer HINT For the first one Recall the definition of$A$being open in the relative topology: if there exists some$A'$open in$X$such that$A'\cap Y_1 = A$. - Right. So$A = C \cap Y_{1}$where$C$is open in$X$and similarly$A = D \cap Y_{2}$where$D$is open in$X$. We have to show that$A$is of the form$V \cap (Y_{1} \cup Y_{2})$where$V$is some open set in$X$. Now take$V = C \cap D$then$V \cap (Y_{1} \cup Y_{2}) = A$and$V$is open being a union of open sets. Why do we nee the assumption$A \subseteq Y_{1} \cap Y_{2}$? – Paulo Dec 22 '10 at 7:32 @Paulo:$A=C\cap Y_1$implies$A\subseteq Y_1$and similarly$A\subseteq Y_2$. Saying that "$A$is open in$Y_j$" means in particular that it is contained in$Y_j$. If instead all you knew was that the intersection of$A$with$Y_j$is open in$Y_j\$, then you would get counterexamples. –  Jonas Meyer Dec 22 '10 at 7:38
@Jonas: Thank you. What I wrote makes sense? –  Paulo Dec 22 '10 at 7:42
@Paulo: Yes, except you wrote "union" where you meant "intersection". –  Jonas Meyer Dec 22 '10 at 7:45
@Jonas: Doh, right. Thanks again! –  Paulo Dec 22 '10 at 7:47