# If $Q$ is a dense subset of a topological space $(Ω,τ)$ and $A∈τ$, then $A∩Q$ is $\left.τ\right|_A$-dense. Does the same apply to closed $A$?

Let $(\Omega,\tau)$ be a topological space and $Q\subseteq\Omega$ be $\tau$-dense. We can show that if $A\in\tau$, then $A\cap Q$ is $\left.\tau\right|_A$-dense.

This allows us to make the following conclusions:

If $\mathcal E$ denotes the Euclidean topology on $\mathbb R$ and $f:\Omega\to\mathbb R$ is $(\tau,\mathcal E)$-continuous, then $$\sup_Af=\sup_{A\:\cap\:Q}f\;.\tag1$$

Moreover, if $\tau$ is induced by a metric $d$ and $x\in A$, then $$d(x,x_n)\xrightarrow{n\to\infty}0\tag2$$ for some $(x_n)_{n\in\mathbb N}\subseteq A\cap Q$.

Now, I would like to make similar conclusions for $A$ replaced by a closed $B\subseteq\Omega$. In particular, I'm interested in the case $\Omega=\mathbb R$ and $B=[0,\infty)$. Is there something which prevents us to make these conclusions? If not, how can we show $(1)$ and $(2)$?

• For $B = [0,\infty)$, things work, since $B$ is the closure of an open set. For general closed subsets, $B\cap Q$ can be empty. – Daniel Fischer Jun 26 '17 at 15:26
• @DanielFischer So, if $B=\overline A^\tau$ for some $A\in\tau$, then $\overline{B\cap Q}^{\left.\tau\right|_B}=B$, right? How do we prove that? Clearly, we know that $\overline{A\cap Q}^{\left.\tau\right|_A}=A$ ... – 0xbadf00d Jun 26 '17 at 20:06
• Hence $\overline{A\cap Q}^{\tau} \supseteq A$, so … – Daniel Fischer Jun 26 '17 at 20:16
• $$A = \overline{A\cap Q}^{\tau\lvert_A} = \bigl(\overline{A\cap Q}^{\tau}\bigr) \cap A$$ – Daniel Fischer Jun 26 '17 at 21:19
• So $$\overline{B\cap Q}^{\tau\lvert_B} \supseteq \overline{A\cap Q}^{\tau\lvert_B} = B\cap \bigl(\overline{A\cap Q}^{\tau}\bigr) = B.$$ – Daniel Fischer Jun 27 '17 at 9:54

As pointed out in comments, for dense $Q$ and closed $B$ we may have $Q \cap B = \emptyset$. The reason this does not happen for $A \in \tau$ is due to the definition of dense (for all open sets $U\in \tau$, $Q \cap U \neq \emptyset$). Further $Q$ is dense in the subspace topology on $A$ as $A$ open implies $\tau_A \subseteq \tau$.
In general, let $Q$ be a dense open set and $B$ its complement.
For a very simple counter-example consider that with the usual topology on $\mathbb R$ the set $Q=\mathbb R$ \ $\{0\}$ is dense and $A=\{0\}$ is closed and not empty but $A\cap Q$ is empty.
When $(\Omega, \tau )$ is a top'l space and $A\in \tau$ then $\tau|_A= \{B: A\supset B\in \tau\}.$ So if $Q$ is a dense subset of $\Omega$ and if $\phi \ne B\in \tau|_A$ then $\phi \ne B\in \tau,$ so the $\tau$-denseness of $Q$ implies $Q\cap B \ne \phi.$ This may fail if $A\not \in \tau.$
• BTW. Here is an exercise that has some useful consequences: Let $(X,t)$ be a top'l space and $Y\subset X$ where $Y$ is $t$-dense. If $U\in t|_Y$ and $Cl_Y(U)=Cl_X(Cl_Y(U))$ then $Cl_Y(U)=Cl_X(U)$ and $U\in t.$ – DanielWainfleet Jun 28 '17 at 6:58