# 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.

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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
Since $C$ is open in $X$, $C \cap B$ is open in $B$ and $A$ is dense in $B$, so $C \cap B \supseteq A \cap (C \cap B) \neq \emptyset$. – kahen Dec 22 '10 at 11:47

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