1
$\begingroup$

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 $(2)$?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Jun 26 '17 at 15:26
  • $\begingroup$ @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 $\overline{A\cap Q}^{\left.\tau\right|_A}=A$ ... $\endgroup$ – 0xbadf00d Jun 26 '17 at 20:06
  • $\begingroup$ Hence $\overline{A\cap Q}^{\tau} \supseteq A$, so … $\endgroup$ – Daniel Fischer Jun 26 '17 at 20:16
  • 1
    $\begingroup$ $$A = \overline{A\cap Q}^{\tau\lvert_A} = \bigl(\overline{A\cap Q}^{\tau}\bigr) \cap A$$ $\endgroup$ – Daniel Fischer Jun 26 '17 at 21:19
  • 1
    $\begingroup$ So $$\overline{B\cap Q}^{\tau\lvert_B} \supseteq \overline{A\cap Q}^{\tau\lvert_B} = B\cap \bigl(\overline{A\cap Q}^{\tau}\bigr) = B.$$ $\endgroup$ – Daniel Fischer Jun 27 '17 at 9:54
1
$\begingroup$

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 $\tau_A \subseteq \tau$.

$\endgroup$
0
$\begingroup$

In general, let $Q$ be a dense open set and $B$ its complement.

$\endgroup$
0
$\begingroup$

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 $A\not \in \tau.$

$\endgroup$
  • $\begingroup$ 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 $U\in t.$ $\endgroup$ – DanielWainfleet Jun 28 '17 at 6:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.