0
$\begingroup$

$$\tau=\{(-\infty,b)|b\in \Bbb R\}\cup\{R,\emptyset\}$$

a) Show $\tau$ is a topology on X

b) Find point of interior,closure and boundary points of $(-\frac{1}3,0)$

c) Show that $\tau$ is second-countable

a)

$O_1:=\emptyset,\Bbb R\in \tau$

$O_2:= G_1,G_2\in\tau\Rightarrow b,c\in R \quad s.t. \quad G_1=(-\infty,b)$ and $G_2=(-\infty,c)$ so $G_1\cap G_2\in \tau$

$O_3:=let \quad\forall(G_\lambda)_{\lambda\in \Lambda}\subset \tau $ and $G_\lambda=(-\infty,a_\lambda), k=\max\{a_\lambda|\lambda\in\Lambda\}$ then $\bigcup_{\lambda\in \Lambda}G_\lambda=G_a=(-\infty,k)\in\tau \quad (a\in\Lambda)$

**b)**$A=(-\frac13,0)$ so interior of A, $A^{o}=\emptyset$ because there is no open interval $(-\infty,a),a\in R$ in A

closure of A should be $(-\frac13,\infty)$ . for $x\in (-\frac13,\infty), $ intersection of every open set containing x (i.e. $(-\infty,a),a\in R$ and $x\leq a$) and A is not empty set.

boundary of A is same with closure

c) can we say it is s.countable because let base $B:=\{ (a,b)|a,b\in \Bbb Z\}$ is a countable base of $\tau$

is this correct? is there any mistake or missing part

$\endgroup$
  • $\begingroup$ Probably you want to take $\sup\{a_\lambda|\lambda\in\Lambda\}$, not $\max\{a_\lambda|\lambda\in\Lambda\}$. Not every subset of $\mathbb R$ has maximum. If you allow $\pm\infty$, then every subset of $\mathbb R$ has supremum. $\endgroup$ – Martin Sleziak Jun 19 '14 at 6:01
2
$\begingroup$

Your answer for b. is incorrect. The interior is empty, as $(-1/3,0)$ does not contain any open sets besides $\emptyset$.

The closure is $[-1/3,\infty)$ (the closure should be closed, of course, as in the complement of an open set). To see this, let $a\in[-1/3,\infty)$, any open set containing $a$ is of the form $(-\infty,b)$ for some $b>a$ and $(-\infty,b)\cap(-1/3,0)-\{a\}\neq \emptyset$. Then, for any $a <-1/3$ we can find $\epsilon$ such that $a+\epsilon<-1/3$ (we're working in $\mathbf{R}$ here). So that $(-\infty,a+\epsilon)$ is a neighborhood of $a$ but $(-\infty,a+\epsilon)\cap(-1/3,0)=\emptyset$.

The boundary is defined as $\text{cl}(A)\cap\text{cl}(\mathbf{R}-A)$. Now, $\mathbf{R}-A=(-\infty,-1/3]\cup[0,\infty)$. Can you find the closure of this set? I'll leave it to you. Here's a hint, $\text{cl}(A)=\text{int}(A)\cup \text{bd}(A)$.

For c. you might want to look at the collection $\{(-\infty,q): q\in\mathbf{Q}\}$. Your example does not work, since, for $a<b\in\mathbf{R}$, $(a,b)$ is not open in our topology (does it contain any open set?). Are you sure you know what second countable is?

$\endgroup$
2
$\begingroup$

For $(b)$, note that $\frac{-1}{3}$ is also a limit point. Because every open set $(-\infty,a)\ni\frac{-1}{3}$, should have element in $(\frac{-1}{3},0))$.

For $(c)$, please note that the members of your $B$ not at all open in $\tau$. But the family $\{(-\infty,q):q\in\mathbb{Q}\}$ is a countable family of open sets and it is a basis because $\mathbb{Q}$ is dense.

$\endgroup$

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.