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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?

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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 $\tau_2\subseteq \tau_1$.

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?

Do you see your mistake now?

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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?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?

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Can you state the exact Lemma you want to use?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?

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The Lemma is as follows:Can you state the exact Lemma you want to use?

The topology generated by a basis $\mathbf{B'}$ is finer than the topology generated by another basis $\mathbf{B}$ iff for every B∈ $\mathbf{B}$ and x∈B, there is a B' $\in\mathbf{B'}$ such that x∈B′⊂B.

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 B∈ $\mathbf{B}$ and x∈B, there is a B' $\in\mathbf{B'}$ such that x∈B′⊂B.

Above you have foundNow 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?

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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".Above you have founda$B_1$ in the basis for $\tau_1$ and asingle$x\in B_1$, 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.

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

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