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  1. MHB Apprentice

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    #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?

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    #2
    Quote Originally Posted by Arenholt View Post
    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?

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    #3 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    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?
    Last edited by Arenholt; December 21st, 2016 at 06:32.

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    #4
    Quote Originally Posted by Arenholt View Post
    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?

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    #5 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    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 B∈ $\mathbf{B}$ and x∈B, there is a B' $\in\mathbf{B'}$ such that x∈B′⊂B.

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    #6
    Quote Originally Posted by Arenholt View Post
    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.
    Quote Originally Posted by Arenholt View Post
    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.

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    #7 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    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".

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    #8
    Quote Originally Posted by Arenholt View Post
    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.

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    #9 Thread Author
    Quote Originally Posted by caffeinemachine View Post
    The bases have nothing to do with a chosen point in $\mathbf R$. The bases are already given.
    I got my mistake.

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