Can a subbasis help determine which topology is finer?

[Bashyboy]

Homework Statement

If two different topologies on a given space $X$ are given by a basis, then we have the following criteria for determining which topology is finer:

Let $\mathcal{B}$ and $\mathcal{B}'$ be bases for the for topologies $\tau$ and $\tau'$, respectively, on $X$. Then the following are equivalent:

(1) $\tau \subseteq \tau'$

(2) For each $x \in X$ and each basis element $B \in \mathcal{B}$, there is a basis element $B' \in \mathcal{B}'$ such that $x \in B' \subseteq B$.

Homework Equations

The Attempt at a Solution

I am wondering, do we have a similar criteria if the two topologies are given by a subbasis $\mathcal{S}$ and $\mathcal{S}'$? My motivation for asking this question is that I am trying to show that $\{Y \cap S ~|~ S \in S \}$ forms a subbasis for the subspace topology on $Y \subseteq X$, and I thought that it might help me in proving this.

 

[fresh_42]

Do you mean by subbasis $\mathcal{S}$ a basis of the from $(X,\mathcal{B})$ induced topology of $Y$?

 

[Bashyboy]

Here is the definition of subbasis with which I am working:

A subbasis $\mathcal{S}$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$. The topology generated by the subbasis $\mathcal{S}$ is defined to be the collection $\tau$ of all unions of finite intersections of elements in $\mathcal{S}$.

 

[fresh_42]

I can't see any problems. The topology generated by the sets of $\mathcal{S}$ is the least finest topology which contains these sets. Refinements thus should be possible to define analogously if everything is done with rigor. (But as topology is notoriously strange, I would be delighted to learn something new.)