1. I want to compare two topologies on $\mathbf{R}$: $\tau_1$=finite complement topology and $\tau_2$=the topology having all sets $(-\infty,a)=\{x|x<a\}$ as basis. If I pick out the element $\mathbf{1}$, take $B_1=\mathbf{R}$-$\{0\}\in \tau_1$ and $B_2=(-\infty,2) \in \tau_2$ I see that $\mathbf{1}\in B_1$, $B_2$ but $B_2\not\subset B_1$ and vice-versa, so I say that the two topologies cannot be compared.
Now if I take another case where I change $B_1=\mathbf{R}$-$\{3\}\in \tau_1$ and keep the rest unchanged, again $\mathbf{1}\in B_1$, $B_2$ but now $B_2\subset B_1$. They not uncomparable anymore. So what do I conclude?

2. Originally Posted by Arenholt
I want to compare two topologies on $\mathbf{R}$: $\tau_1$=finite complement topology and $\tau_2$=the topology having all sets $(-\infty,a)=\{x|x<a\}$ as basis. If I pick out the element $\mathbf{1}$, take $B_1=\mathbf{R}$-$\{0\}\in \tau_1$ and $B_2=(-\infty,2) \in \tau_2$ I see that $\mathbf{1}\in B_1$, $B_2$ but $B_2\not\subset B_1$ and vice-versa, so I say that the two topologies cannot be compared.
Now if I take another case where I change $B_1=\mathbf{R}$-$\{3\}\in \tau_1$ and keep the rest unchanged, again $\mathbf{1}\in B_1$, $B_2$ but now $B_2\subset B_1$. They not uncomparable anymore. So what do I conclude?
It seems you have got the definition of comparison of two topologies wrong. We say that two topologies $\tau_1$ and $\tau_2$ on a set $X$ are comparable if $\tau_1\subseteq \tau_2$ or $\tau_2\subseteq \tau_1$.

Do you see your mistake now?

Originally Posted by caffeinemachine
It seems you have got the definition of comparison of two topologies wrong. We say that two topologies $\tau_1$ and $\tau_2$ on a set $X$ are comparable if $\tau_1\subseteq \tau_2$ or $\tau_2\subseteq \tau_1$.

Do you see your mistake now?
According to a Lemma in Munkres, I can also compare the basis of the two topologies to find their comparability. I am trying to do that way. That's why I have chosen those basis elements or else in this case how can I compare the two topologies directly?

4. Originally Posted by Arenholt
According to a Lemma in Munkres, I can also compare the basis of the two topologies to find their comparability. I am trying to do that way. That's why I have chosen those basis elements or else in this case how can I compare the two topologies directly?
Can you state the exact Lemma you want to use?

Originally Posted by caffeinemachine
Can you state the exact Lemma you want to use?
The Lemma is as follows:
The topology generated by a basis $\mathbf{B'}$ is finer than the topology generated by another basis $\mathbf{B}$ iff for every Bâˆˆ $\mathbf{B}$ and xâˆˆB, there is a B' $\in\mathbf{B'}$ such that xâˆˆBâ€²âŠ‚B.

6. Originally Posted by Arenholt
The Lemma is as follows:
Lemma. The topology generated by a basis $\mathbf{B'}$ is finer than the topology generated by another basis $\mathbf{B}$ iff for every Bâˆˆ $\mathbf{B}$ and xâˆˆB, there is a B' $\in\mathbf{B'}$ such that xâˆˆBâ€²âŠ‚B.
Originally Posted by Arenholt
Now if I take another case where I change $B_1=\mathbf{R}$-$\{3\}\in \tau_1$ and keep the rest unchanged, again $\mathbf{1}\in B_1$, $B_2$ but now $B_2\subset B_1$. They not uncomparable anymore. So what do I conclude?
Above you have found a $B_1$ in the basis for $\tau_1$ and a single $x\in B_1$, namely $x=1$, such that there is $B_2$ in the basis for $\tau_2$ such that $x_1\in B_2\subseteq B_1$. This does not mean that the topologies are comparable.

Originally Posted by caffeinemachine
Above you have found a $B_1$ in the basis for $\tau_1$ and a single $x\in B_1$, namely $x=1$, such that there is $B_2$ in the basis for $\tau_2$ such that $x_1\in B_2\subseteq B_1$. This does not mean that the topologies are comparable.
According to the lemma, I found the required basis for an element x=1. The lemma itself doesn't say that it has to be true for "all the elements of the basis".

8. Originally Posted by Arenholt
I found the required basis for an element x=1.
The bases have nothing to do with a chosen point in $\mathbf R$. The bases are already given.

The bases have nothing to do with a chosen point in $\mathbf R$. The bases are already given.