0
$\begingroup$

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.

  • 1
    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.2012-10-25

1 Answers 1

4

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?

  • 0
    Exactly right.${}$2012-10-25