# Standard topology exercise

Let $\tau$ the topology generated from $\mathcal{B}=\{(-\infty,a):a<0\} \cup \{\{0\}\cup (b,+\infty):b>0\}$.

(i) Find $Int[0,1], Int(0,+\infty), Int(-4,+\infty)$

(ii)Find the closure of $\mathbb{Z}$

(iii)Is $\mathbb{R,\tau}$ a $T_2$ space?

(iv) Is $R,\tau$ compact, connected?

(i) $Int[0,1]=\emptyset$, because every open set is unlimited. The same argument applies to $Int(-4,+\infty)=\emptyset$.

$Int(0,+\infty)=(0,+\infty)$, because it's an open set.

(ii)Since every closed set doesn't contain the singleton $0$, the closure of $\mathbb{Z}$ is $\mathbb{R}$

(iii)$R,\tau$ is not an Hausdorff space.

For example, $1$ is not separable (alla Hausdorff) from $2$. In fact, every open neighborhood $U$, $V$ of $1,2$ are non-disjoint.

(iv)It's not connected since $\mathbb{R}=(-\infty,a) \cup [0,+\infty)$, with $a=sup${$a:a <0$}.

I think it's not compact because this open cover $A=(-\infty,a) \cup$ {0} $(b,+\infty)$, with $a<0, b>0$ doesn't have a finite open subcover for $\mathbb{R}$.

First note that $\mathcal B\cup\{\varnothing\}$ is closed under finite intersection and covers $\mathbb R$. This tells us that $\mathcal B$ is a base of the topology, so that a set is open if and only it is a union of sets in $\mathcal B$.

i) $(0,\infty)$ and $[0,1]$ both contain no base elements as subset hence their interiors (the unions of base elements contained in these sets) are empty.

For $(-4,\infty)$ we find $[0,\infty)$ as interior. It is the union of the sets $\{0\}\cup(b,\infty)$ where $b>0$.

ii) There are closed sets that contain the singleton $\{0\}$. For instance the set $[-1,\infty)$ which is closed as component of open set $(-\infty,-1)$. So your argument is not correct. If $a\notin\mathbb Z$ then every open set that contains $a$ also contain elements of $\mathbb Z$. So every $a\notin\mathbb Z$ is a limit point of $\mathbb Z$. This implies that the closure of $\mathbb Z$ is $\mathbb R$.

iii) It is indeed not Hausdorff. As you state there are no disjoint neighborhoods of $1$ and $2$.

iv) Indeed not compact. As said $\mathcal B$ constitutes an open cover of $\mathbb R$ but has no finite subcover.

It is not connected since the sets $(-\infty,0)$ and $[0,\infty)$ are open, disjoint and cover $\mathbb R$. Your argumentation here has shortcomings. Actually you should prove here that $(-\infty,0)$ is open as a union of open sets.

• Thanks, I've understood now! – VoB Aug 29 '17 at 12:41

(i) Your argument is correct for $[0,1]$, but not for the other two. $(-4,+\infty)$ is unbounded (unlimited) as well, so the argument for $[0,1]$ doesn't work for $(-4,+\infty)$. Actually, notice that $2\in \{0\}\cup(1,+\infty)\subset(-4,+\infty).$ Thus $2\in\text{Int}(-4,+\infty)$ which shows that the interior is not exmpty. Concerning $(0,+\infty)$, it is not an open set in your topology. Indeed, your collection $\mathcal{B}$ is a basis for your topology, so if $U$ is an open set and $x\in U$ then $\exists B\in \mathcal{B},x\in B\subset U$. Setting $U=(0,+\infty)$, you see that this is not true for any element of $U$ because a basis element that has positive numbers contains $0$ as well, while $0\notin U$.

(ii) I can't understand your argument here. There are closed sets that contain $0$ (you shouldn't say singleton $0$, i.e, $\{0\}$). If $a<0$ then $(-\infty,a)$ is an open set, thus its complement, $[a,+\infty)$, is closed, and it contains $0$. I can't understand how you reached the conclusion that the closure of $\mathbb Z$ is $\mathbb R$. You contradicted yourself: you said that no closed set contains $0$, but then the closure of $\mathbb Z$, which is a closed set, is $\mathbb R$ and it contains $0$. However, if I'm not mistaken, it is true that the closure of $\mathbb Z$ in your topology is $\mathbb R$.

(iii) Your argument looks fine.

(iv) Your argument looks fine. A remark: your wrote $\mathbb{R}=(-\infty,a)\cup [0,+\infty)$ with $a=\sup\{a\in\mathbb{R}\mid a<0$. But this sup is $0$, hence you can simply write $\mathbb{R}=(-\infty,0)\cup[0,+\infty)$. These two sets are indeed open.