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I know that we say that the topology $\mathcal{T}$ on a set $X$ is finer than the topology $\mathcal{T}'$ on $X$ if for any $U\subset\mathcal{T}'$, $U$ is also open in $\mathcal{T}$. And also, $\mathcal{T}$ may have several bases.

When we are showing that $U$ is open in $\mathcal{T}$ we find basis elements $B$ of a given basis $\mathcal{B}$ of $\mathcal{T}$ such that $U=\bigcup_x{B_x}$ for all $x\in U$. So, my question is: If we consider a different basis of $\mathcal{T}$ is it possible that $\mathcal{T}$ is not finer than $\mathcal{T}'$?

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Is my answer acceptable? If not, let me know what you are yet to understand. – Michael Albanese Sep 27 '13 at 18:52
I guess I actually forgot to accept your answer, sorry – Xena Sep 28 '13 at 5:51
No worries, just wanted to check. – Michael Albanese Sep 28 '13 at 6:37
up vote 2 down vote accepted

No. Fineness of topologies has nothing to do with bases for those topologies.

Let $\mathcal{T}$ and $\mathcal{T}'$ be two topologies on a set $X$. We say $\mathcal{T}$ is finer than $\mathcal{T}'$ if $\mathcal{T}' \subseteq \mathcal{T}$. That is, every set which is open in the topology $\mathcal{T}'$ is open in the topology $\mathcal{T}$, but the latter could contain open sets that the former doesn't.

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So if a set is open it has nothing to do with the basis, yes? – Xena Apr 22 '13 at 10:01
I suppose you could say that it has nothing to do with a basis. A set $U$ is open in the topological space $(X, \mathcal{T})$ if $U \in \mathcal{T}$. – Michael Albanese Apr 22 '13 at 10:15
Assume $\mathcal{B}$, $\mathcal{B}'$ are both bases of $\mathcal{T}$ and $U=\bigcup_x{B_x}$ where $B_x\in \mathcal{B}$ for all $x\in U$. Now suppose that $\mathcal{B}'$ does not contain such $B_x$'s. Then how can we express $U$ as a union of basis elements of $\mathcal{B}'$? Why there is no such a $\mathcal{B}'$? – Xena Apr 22 '13 at 10:27
The fact that $\mathcal{B}'$ is a basis for the topology implies $U = \bigcup_xB_x'$ where $B_x' \in \mathcal{B}'$ (just as $\mathcal{B}$ being a basis for the topology implies $U = \bigcup_xB_x$ where $B_x \in \mathcal{B}$). It may be the case that $\{B_x \mid x \in U\}\cap\{B_x' \mid x \in U\} = \emptyset$, but that isn't relevant. For example, $\mathcal{B} = \{(a, b) \mid a, b \in \mathbb{Q}\}$ and $\mathcal{B}' = \{(a, b) \mid a, b \in \mathbb{R}\setminus\mathbb{Q}\}$ are both bases for the standard topology on $\mathbb{R}$, but $\mathcal{B}\cap\mathcal{B}' = \emptyset$. – Michael Albanese Apr 22 '13 at 12:09

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