Can anyone tell me whether there is any theorem (or your experience) to conclude that in 2 subspaces (say $A$, $B$) of a same toplogy (say $X$), $A$ is finer than $B$ or vice versa??? If A is subset of B, should B be finer than A??? Thanks so much.
How to assert a subspace is finer than the other
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1I guess you could compare cardinalities of $C(A, B)$ and $C(B, A)$, but it's not a very precise metric for infinite spaces. – 2012-10-25
1 Answers
It doesn't seem like this is a meaningful question. Given two topologies (say $\mathcal{T}_1$ and $\mathcal{T}_2$) on the same space, we typically say that $\mathcal{T}_1$ is finer than $\mathcal{T}_2$ if $\mathcal{T}_2\subseteq\mathcal{T}_1$.
The subspace topologies on $A$ and $B$ in your example are incomparable, for the simple fact that $A$ isn't an element of the subspace topology on $B$, and vice versa.
Is there a different meaning of "finer" that you're using, here?
Perhaps you're confusing "topology" with "topological space". A topological space is a pair $(X,\mathcal{T})$, where $X$ is a set, and $\mathcal{T}$ is a collection of subsets of $X$ such that:
(i) $\emptyset,X\in\mathcal{T}$;
(ii) $\bigcup\mathcal{A}\in\mathcal{T}$ whenever $\mathcal{A}\subseteq\mathcal{T}$ ("$\mathcal{T}$ is closed under arbitrary unions");
(iii) $U_1\cap\cdots\cap U_n\in\mathcal{T}$ whenever $U_1,...,U_n\in\mathcal{T}$ ("$\mathcal{T}$ is closed under finite intersections").
Such a collection $\mathcal{T}$ is called a topology on $X$, and its elements are called the open sets of the topological space $(X,\mathcal{T})$. A subspace of a topological space $(X,\mathcal{T})$ is a pair $(Y,\mathcal{T}_Y)$, where $Y\subseteq X$ and $\mathcal{T}_Y=\{U\cap Y:U\in\mathcal{T}\}$ ($\mathcal{T}_Y$ is called the subspace topology on $Y$ induced by $\mathcal{T}$). Does that clear things up at all?
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0Ok, thanks so much. I mean in general, is there any special way to prove that a subspace $A$ is finer than a subspace $B$??? For a first time, I assume if $A$ is subset of $B$, then subspace $B$ is finer than subspace $A$ – 2012-10-25
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0Oh, thanks for your consideration. I don't have any trouble working on normal space, but I'm usually confused while working on subspace. For example, is the subspace [0,1] in R finer than (0,1) in R? – 2012-10-25
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0But what do you mean by "finer"? – 2012-10-25
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0Just like what you defined above, T1 is finer than T2 if T2⊆T1 ^^ – 2012-10-25
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1Okay, but again, we only talk about *topologies* being finer than others, and only if they're topologies **on the same set**. Asking if **(sub)spaces** are finer than others isn't meaningful. Now, are you simply wondering how to tell when $\mathcal{T}_B\subseteq\mathcal{T}_A$ for some $A,B\subseteq X$? We wouldn't actually talk about fineness in such a situation (unless $A=B$, in which case $\mathcal{T}_A=\mathcal{T}_B$), but we *can* actually determine exactly when this happens. – 2012-10-25
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0(cont'd): In particular, $\mathcal{T}_B\subseteq\mathcal{T}_A$ if and only if there is some $V\in\mathcal{T}$ such that $B=A\cap V$. It's a good exercise to prove that. Note that simply having $A\subseteq B$ isn't *sufficient* to tell us that $\mathcal{T}_B\subseteq\mathcal{T}_A$, but it is *necessary*. – 2012-10-25
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0Ok, I guess I got it right now. In subspace, not only topology is different than the original topology, but also the set they are defined, right? So we can't compare 2 different subspace because they don't have the same set on which 2 topologies are defined..... Thanks so much.You are so kind. I'm learning topo by myself, reading Munkres book, so I am confused so much. Really appreciate your help :-D – 2012-10-25
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0Exactly right.${}$ – 2012-10-25