Topology problem and Confusion about Open and Closed Interval This is an excerpt from Topology without tears. There are 10 statements like this 5 on a problem and the job is to seek precisely 3 topologies from the 10 statements
τ 2 consists of R, Ø, and every interval (-r,r), for r any positive real number;
(iii) τ 3 consists of R, Ø, and every interval (-r,r), for r any positive rational
number;
(iv) τ 4 consists of R, Ø, and every interval [-r,r], for r any positive rational
number;
(v) τ 5 consists of R, Ø, and every interval (-r,r), for r any positive irrational
number;
(vi) τ 6 consists of R, Ø, and every interval [-r,r], for r any positive irrational
number; 
But I think every 5 of this is a topology. For suppose, a is postive, rational or irrational real number. Then (-a,a) is on tau. Let, b be an arbitrarily positive real number. Then (-(a+b),a+b) is in tau. This b can be arbitrarily small or large. And assuming a to be arbitrarily close to zero then any other interval (-r,r) can be expressed as (-(a+b),a+b). Then the union of any number of sets is (-(a+b),a+b) which is in tau. And intersection of a finite number of sets will (-a,a) which is in tau.
This argument also work for close interval(I think). So every 5 of them is a topology. But it's not true. where's the problem.
 A: The first of them, $\tau_2$, is a topology on $\Bbb R$: it contains $\varnothing$ and $\Bbb R$, and it is closed under taking finite intersections and arbitrary unions. None of the others is a topology, because none of them is closed under taking arbitrary unions.


*

*$\tau_3$: Let $A=\{r\in\Bbb Q:0<r<\sqrt2\}$; then $\bigcup_{r\in A}(-r,r)=(-\sqrt2,\sqrt2)$, but $(-\sqrt2,\sqrt2)\notin\tau_3$.

*$\tau_4$: Let $A=\{r\in\Bbb Q:0<r<1\}$; then $\bigcup_{r\in A}[-r,r]=(-1,1)$, but $(-1,1)\notin\tau_4$.

*$\tau_5$: Let $A=\{r\in\Bbb R\setminus\Bbb R:0<r<1\}$; then $\bigcup_{r\in A}(-r,r)=(-1,1)$, but $(-1,1)\notin\tau_5$.

*$\tau_6$: Let $A$ be as for $\tau_5$; then $\bigcup_{r\in A}[-r,r]=(-1,1)$, but $(-1,1)\notin\tau_6$.
A: Remember the definition of a topology: $\tau$ is a topology on $X$ if 


*

*$\emptyset, X$ are each in $\tau,

*any finite intersection of elements of $\tau$ is in $\tau$, and

*any union of arbitrarily many elements of $\tau$ is in $\tau$.
So let's take $\tau_3$ for example, which consists of $\emptyset, X$, and every $(-r, r)$ with $r$ a positive rational real.


*

*Clearly the first requirement is satisfied.

*It's not hard to show that the second is, too: $(-r_1, r_1)\cap (-r_2, r_2)\cap . . . \cap (-r_n, r_n)$ is just $(-s, s)$ where $s=\min\{r_1, r_2, . . . , r_n\}$ (do you see why?). Since this $s$ is one of the $r_i$s, $(-s, s)\in\tau$.

*But the third property fails! For example, let $r_1=3$, $r_2=3.1$, $r_3=3.14$, $r_4=3.141$, and so on. Then $$\bigcup (-r_i, r_i)=(-\pi, \pi)$$ but since $\pi$ is irrational, that's not in $\tau$!

I can't follow the argument you give that each of the $\tau$s is a topology; once you understand the counterexample I've given, it would be a good idea to try to write down in detail a purported proof that $\tau_3$ is a topology; you'll likely see exactly where it breaks down, then.
