# How to assert a subspace is finer than the other

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.

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I guess you could compare cardinalities of $C(A, B)$ and $C(B, A)$, but it's not a very precise metric for infinite spaces. –  Alexei Averchenko Oct 25 '12 at 17:05

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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Ok, 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 $A$ –  le duc quang Oct 25 '12 at 16:42
Oh, 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? –  le duc quang Oct 25 '12 at 16:52
But what do you mean by "finer"? –  Cameron Buie Oct 25 '12 at 16:53
Just like what you defined above, T1 is finer than T2 if T2⊆T1 ^^ –  le duc quang Oct 25 '12 at 16:54
Okay, 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. –  Cameron Buie Oct 25 '12 at 17:01