this question says only three of the below are topology.
So,according to me,$\tau_2$ is topology it is satisfying all three definition for topology.
And I believe $\tau$ consisting of $\mathbb{R}$,$\phi$ and every interval $[-r,r]$,for $r$ any positive real number is also a topology.So $\tau_9$ is union of both $\tau$ and $\tau_2$ with $\tau_2 \subset \tau$ so is also a topology.
Similarly $\tau_0$ consisting of $\mathbb{R},\phi$ and every interval $[-n,n]$,for $n$ any positive integer is also topology.So $\tau_{10}$ is union of both the topology with one containing the other so is also a topology.
so i think $\tau_2$,$\tau_9$ and $\tau_{10}$ are the required topology,am i correct?
And I also want to know whether $\tau_7$ and $\tau_8$ are topology or not with justification.
$\endgroup$ 41 Answer
$\begingroup$- $\tau_1$ is not a topology since $(a,b)\cup(c,d) \notin \tau_1$ where $a<b<c<d$.
- $\tau_2$ is a topology ($\checkmark$) since $$\bigcup_{i\in I}(-r_i,r_i) =\begin{cases} (-R,R) & I \text{ is finite} \land R = \max\{r_i \mid i \in I\} \\ \mathbb R & I \text{ is not finite} \end{cases}$$ and $$\bigcap_{i\in J}(-r_i,r_i) = (-r,r)$$ where $r = \min\{r_i \mid i \in J\}$ ($J$ is finite).
- $\tau_3$ is not a topology since for some irrational $a$ we have$$\bigcup_{n=1}^\infty(-r_n,r_n) = (-a,a)\notin \tau_3$$ where $\lim_{n\to\infty}r_n = a$.
- $\tau_4$ is not a topology since $$\bigcup_{n=1}^\infty\left[-r_n+\frac 1n,r_n-\frac 1n\right] = (-r,r)\notin \tau_4$$
- $\tau_5$ is not a topology for a similar reason as 3.
- $\tau_6$ is not a topology for a similar reason as 4 (note that if $a$ is irrational then so is $a + \frac{1}{n}$ for any $n\in \mathbb N$).
- $\tau_7$ is not a topology for a similar reason as 4.
- $\tau_8$ is not a topology for a similar reason as 4.
- $\tau_9$ is a topology ($\checkmark$) since one cannot use the "trick" that was used in 4. (try to show it rigorously)
- $\tau_{10}$ is a topology ($\checkmark$) since one cannot use the "trick" that was used in 4. (try to show it rigorously)
