# How to show $\tau=\{(-\infty,b)|b\in \Bbb R\}\cup\{R,\emptyset\}$ is topology and second countable?

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

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

• 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. – Martin Sleziak Jun 19 '14 at 6:01

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?
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.