Is the family topology for $\mathbb{R}$

Question

I have a problem about showing that; is the family r creates topology for $\mathbb{R}$ ? $$r = \{U \subset \mathbb{R} : 0 \notin U \text{ or } \mathbb{R} -\{-1,1\} \subset U\} $$

1. First part is okey. I can show that $\emptyset$ and $\mathbb{R} \in r$.
2. If $0 \notin U$ for all $U_a$, it is okey that the union is also in $r$. However, other parts are not clear for me.
3. If $0 \notin U$ for all $U_a$, it is also okey that the finite intersection is also in $r$. However, other parts are again not clear for me.

Any help will be appreciated. Thank you!

Answer 1

Suppose you have $U_1,\dots,U_n\in r$; if $0\in U_i$, for $i=1,\dots,n$, then $\mathbb{R}-\{-1,1\}\subset U_i$, for all $i$, so $$ \mathbb{R}-\{-1,1\}\subset \bigcap_{i=1}^n U_i $$ Otherwise $0\notin U_i$ for some $i$ and therefore $$ 0\notin \bigcap_{i=1}^n U_i $$ So closure under finite intersections is proved.

Suppose you have a family $(U_\lambda)_{\lambda\in\Lambda}$ of members of $r$. If one of them satisfies $\mathbb{R}-\{-1,1\}\subset U_\lambda$, then $$ \mathbb{R}-\{-1,1\}\subset\bigcup_{\lambda\in\Lambda}U_\lambda $$ If none of them satisfies the condition above, then $0\notin U_\lambda$, for every $\lambda\in\Lambda$ and so $$ 0\notin\bigcup_{\lambda\in\Lambda}U_\lambda $$

Note that you can use the same exact proof for the family $r_{(A,B)}$ defined by $$ r_{(A,B)}=\{U\subset\mathbb{R}: A\not\subset U\text{ or }B\subset U\} $$ for any $A,B\subset\mathbb{R}$, with $A\ne\emptyset$.

Answer 2

HINT: Let $U_0=\Bbb R\setminus\{-1,1\}$; the only $U\subseteq\Bbb R$ such that $U_0\subseteq U$ are $U_0$, $U_0\cup\{-1\}=\Bbb R\setminus\{1\}$, $U_0\cup\{1\}=\Bbb R\setminus\{-1\}$, and $\Bbb R$. Thus,

$$r=\wp\big(\Bbb R\setminus\{0\}\big)\cup\big\{\Bbb R,\Bbb R\setminus\{1\},\Bbb R\setminus\{-1\},\Bbb R\setminus\{-1,1\}\big\}\;.$$

That is, only four of the sets in $r$ contain $0$, and we know exactly what they are. It’s not hard to check that the intersection of any two of these four sets is one of the four, and that the intersection of one of these sets with a set that does not contain $0$ is also in $r$. This shows that $r$ is closed under finite intersections.

The proof that $r$ is closed under arbitrary unions is similar. The union of any collection of sets not containing $0$ is a set not containing $0$, and it’s not hard to check that the union of such a set with one of the other four members of $r$ is one of the four members containing $0$.