Prove that $\mathscr B$ is a basis for topology $\mathscr T$ on $\mathbb R$. Is $\mathscr T$ is a usual topology on $\mathbb R$?

Question

Let $A=\{\frac{1}{n}:n\in \mathbb N\}$ and $\mathscr{B}=\{(a,b):a<b\}\cup \{(a,b)\setminus A:a<b\}.$ Prove that $\mathscr B$ is a basis for topology $\mathscr T$ on $\mathbb R$. Is $\mathscr T$ is a usual topology on $\mathbb R$?

Obviously, (1)$\bigcup_{a,b\in \mathbb R}(a,b)=\mathbb R$ For the second condition of basis, (2)Case 1:-

Let $x\in B_1,B_2\in \{(a,b):a<b\}$ , then obviously there exists $x\in B_3$ such that $B_3\subset B_1 \cap B_3$

Case2:-$x\in B_1\in \{(a,b):a<b\}$and $B_2 \in\{(a,b)\setminus A:a<b\}$ I am not able to prove existance of $B_3$ such that $B_3\subset B_1 \cap B_3$

Case 3 $B_1,B_2 \in\{(a,b)\setminus A:a<b\}$. How do I prove formally the existance of $B_3$?

to check whether the given basis generate usual topology or not? any open set can be generated by using $\{(a,b):a<b\}$. Please check my answers.

Answer 1

The condition you have to check is

For all $B_1, B_2 \in \mathcal{B}$, and all $x \in B_1 \cap B_2$, there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$.

(A condition you sloppily reproduce, be more exact there!)

Now there are indeed cases to consider: suppose $B_1 = (a_1,b_1)$ and $B_2 = (a_2,b_2)$, so they are of "standard type". If we have $x \in B_1 \cap B_2$, then you can note that $B_1 \cap B_2 = (\max(a_1,a_2), \min(b_1,b_2))$ (the intersection is non-empty, so $\max(a_1,a_2) < \min(b_1,b_2)$ and we can use this intersection as our $B_3$ for all $x$ in the intersection.

You state "it's obvious" and go on. Your thoughts are probably along these lines: (but do make them explicit next time): $B_1$ and $B_2$ are standard open. So their intersection is standard open and the "standard type" open intervals already form a base for those, so the condition follows from open intervals being a base for the standard topology.

("existance" should be "existence")

If $B_1, B_2$ are of type two, an open interval minus $A$ (this case you skipped): You can use the observation above that the intersection of two open intervals is an interval or empty. So if non-empty, we can use $B_1 \cap B_2$ again for all points in it, as

$$((a,b) \setminus A) \cap ((c,d) \setminus A) = ((a,b) \cap (c,d)) \setminus A$$

If $B_1$ is of standard type $(a,b)$ and $B_2$ is of type two, $B_2 = (c,d)\setminus A$ for some $c< d$, then note that if $x \in B_1 \cap B_2$, there are two subcases: a. $x \neq 0$, then $x \notin \overline{A}$ (closure in the standard topology, this holds as $\overline{A} = A \cup \{0\}$, which is standard closed. In that case we have $x \in (a,b) \cap (c,d) \cap (\mathbb{R}\setminus \overline{A})$ which is standard open (!) so there is some open interval (as these form a base for the standard topology!) $(e,f)$ with $$x \in (e,f) \subseteq (a,b) \cap (c,d) \cap (\mathbb{R}\setminus \overline{A}) \subseteq B_1 \cap B_2$$ as required.

The other subcase is $x = 0$. Then $0 \in (a,b) \cap (c,d)$ which is an interval $I$ containing $0$ again, and then $x \in B_3:= I \setminus A \subseteq B_1 \cap B_2$ again.

So the condition checks out again.

Note that this topology could also have been defined by stating that

$\mathcal{S} = \{(a,b): a < b\} \cup \{\mathbb{R} \setminus A\}$ is a subbase for and then the finite intersections of members of $\mathcal{S}$ is exactly the collection $\mathcal{S}$, which is then a base by virtue of being a set of finite intersections from this family. If the set $\mathbb{R} \setminus A$ is not used in the intersection, we get open intervals, and if it is, we get an open interval minus $A$, i.e. type two.

It's clear that $A$ is not closed in the usual topology, but this topology is constructed by adding it as a new closed set, so it's strictly finer than the usual topology, as $\mathbb{R} \setminus A$ is open in it.

Answer 2

The assertion “there exists $x\in B_3$ such that $B_3\subset B_1 \cap B_3$” doesn't make sense because no $B_3$ has been defined at this point. What you should prove is that for each $x\in B_1\cap B_2$, there exists some $B_3\in\mathcal{B}$ such that $x\in B_3$ and that $B_3\subset B_1\cap B_2$.

And $\mathcal B$ doesn't generate the usual topology, since $A^\complement\in\mathcal B$ and $A^\complement$ doesn't belong to the usual topology. The topology generated by $\mathcal B$ is strictly finer that the usual one.

Answer 3

Let's look at Case 2. We let $x\in B_1\cap B_2$ be an arbitrary element of the intersection, where $B_1=(a,b)$ and $B_2=(c,d)\setminus A$. Note that since $x\in B_2$, we have that $x\not\in A$. Now consider the set $C=(max\{a,c\},min\{b,d\})\setminus A$. Because $a<x<b$, $c<x<d$, and $x\not\in A$, we can conclude that $x\in C$. It is easy to verify that $C\subset B_1\cap B_2$.

Case 3 is similar.

This topology is also introduced as "The K topology on $\mathbb{R}$" in Munkres Topology, and there are many examples using this topology in sections 12 through 19.

Answer 4

Let $\Bbb B_1=\{(a,b):a,b\in \Bbb R\}$ and $\Bbb B_2=\{\beta \setminus A: \beta \in \Bbb B_1\}$ and $\Bbb B=\Bbb B_1\cup \Bbb B_2.$

(I). The condition $$\forall b,b'\in \Bbb B\;\forall x\in b\cap b'\;\exists b''\in \Bbb B\;(x\in b''\subset b\cap b')$$ is satisfied IF $$\forall b,b'\in \Bbb B\;(b\cap b'\in \Bbb B)$$ is satisfied.

If $b,b' \in \Bbb B$ let $b=c \setminus d$ and $b'= c'\setminus d'$ where $c,c'\in \Bbb B_1$ and each of $d,d'$ is either $\emptyset$ or $A.$ Observe that $c\cap c'\in B_1$. And $d\cup d'\in \{\emptyset,A\}$. So $$b\cap b'=(c\cap c')\setminus (d\cup d')\in \{(c\cap c'),\; (c\cap c')\setminus A \}\subset \Bbb B.$$

(II). Let $T_{\Bbb B}$ be the topology generated by the base (basis) $\Bbb B.$ Then $A$ is closed with respect to $T_{\Bbb B}$ because $\Bbb R \setminus A $ is the union of $$(-1,1)\setminus A$$ with $$\bigcup \{(n,n+1): 1\leq n\in \Bbb Z \lor -1\geq n\in \Bbb Z\}$$ so $\Bbb R \setminus A \in T_{\Bbb B}.$

But with respect to the "usual" topology on $\Bbb R$ the set $A$ is not closed.

Footnote. If $B$ is a family of subsets of a set $X$ then the conditions $\cup B=X$ and $\forall b,b'\in B\;(b\cap b'\in B)$ are sufficient for $B$ to be a base (basis) for a topology on $X.$