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I am working on a problem where I am given the following bases for topologies:

  • $\mathcal{B}_{1} = \{ (a,b)|a
  • $\mathcal{B}_{2} = \{[a,b)|a
  • $\mathcal{B}_{3}=\{ (a,b]|a
  • $\mathcal{B}_{4}=\{(a,\infty)|a \in \mathbb{R}\}$, where $(a, \infty) = \{x | x>a \}$
  • $\mathcal{B}_{5} = \{ (-\infty,a)|a \in R\}$, where $(-\infty,a) = \{x | x

And now I have to compare all of them to each other.

So far, I have shown that the topology generated by $\mathcal{B}_{2}$ is finer than the topology generated by $\mathcal{B}_{1}$, $\mathcal{B}_{3}$ is finer than the topology generated by $\mathcal{B}_{1}$, and that the topologies generated by $\mathcal{B}_{2}$ and $\mathcal{B}_{3}$ are incomparable.

Right now, I am trying to ascertain the relationship between the topology generated by $\mathcal{B}_{1}$ and the topology generated by $\mathcal{B}_{4}$, call them $\tau_{1}$ and $\tau_{4}$, respectively.

Then, given a basis element $(a,b)$ for $\tau_{1}$ and a point $x$ of $(a,b)$, the basis element $(a, \infty)$ of $\tau_{5}$ contains $x$ and lies in $\tau_{5}$, correct? But, going the other way, if I have a basis element $(a, \infty)$ of $\tau_{5}$ containing $y$, then, the basis element $(a,b)$ may or may not contain $y$. For example, say we consider the interval $(3,\infty) \in \tau_{5}$. Then, $(3,\infty)$ contains $17$. However, the interval $(3,5) \in \tau_{1}$ does not. In fact, we can continue to choose arbitrarily large $y$-values such that there does not exist a basis element for $\tau_{1}$ that contains them - because of the archimedean property, perhaps?

Is what I'm saying true, and if so, what's a nicer way to express it? If it's not true, what is the actual relationship between the topology generated by $\mathcal{B}_{1}$ and the topology generated by $\mathcal{B}_{4}$?

I'm assuming that the cases for comparing $\mathcal{B}_{1}$ and $\mathcal{B}_{5}$, $\mathcal{B}_{2}$ and $\mathcal{B}_{4}$ and $\mathcal{B}_{5}$, and $\mathcal{B}_{3}$ and $\mathcal{B}_{4}$ and $\mathcal{B}_{5}$ would work similarly...

For $\mathcal{B}_{4}$ and $\mathcal{B}_{5}$, though, it's almost painfully obvious that they're not comparable...at least I think it is.

I still haven't completely mastered doing these types of problems, so some guidance/explanation would be extremely helpful!

Thank you.,

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    For any real $a$, you have $\bigcup_{b>a}(a,b)=(a,\infty)$, so it is immediate that at least $B_1$ is finer than $B_4$ – 2017-02-21
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    @user160738 That part I was able to figure out. It's the proper method of showing that the other way (that $\mathcal{B}_{4}$ is NOT finer than $\mathcal{B}_{4}$) that is eluding me. – 2017-02-21
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    Well you cannot possibly create a set $(0,1)$ out of unioning $(a,\infty)$ because any such union must contain any sufficiently large reals – 2017-02-21
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    @user160738 so my intuitiion was correct. But how do I say that slickly/mathly? – 2017-02-21
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    Say you somehow made $(0,1)$ out of unioning basis element. Then such a union must at least have a form of inteval $(a,\infty)$ such that $a<1$. Otherwise, the union wouldn't contain any element in $(0,1)$. But once that happens, $(a,\infty)$ is contained in the union, and in particular $2$ (or whatever big number like $10^{10^{10}}$ does not lie in $(0,1)$, we said the union was equal to $(0,1)$ so this is contradictory. But really, I think it's enough to say that its very clear any union of $B_4$ elements cannot make a set that is contained in a interval of finite length – 2017-02-21
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    @user160738 and then the other cases are similar? – 2017-02-21
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    Basically, yes. There is no inclusion relation between $B_4$ and $B_5$ due to similar reasons, and $B_1$ is finer than $B_5$ also by the same reason. – 2017-02-21

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