Equivalent Topologies

Solution 1:

Given two topologies $\tau,\tau'$ on a set $X$, we say $\tau$ is coarser (or weaker) than $\tau'$ iff $\tau\subset\tau'$. Accordingly we say $\tau'$ is finer (or stronger) than $\tau$.

Equivalently, this means $\tau$ has more (read:not less) opens, or equivalently $\tau$ has more (read:not less) neighborhoods, or equivalently the identity set map $(X,\tau)\to (X,\tau')$ is continuous, or equivalently for each subset the closure in $\tau$ is included in the closure in $\tau'$.

Thus the following are equivalent ways of saying that $\tau,\tau'$ are the same:

• $\tau=\tau'$

• each of $\tau$ and $\tau'$ is coarser/finer than the other

• $\tau,\tau'$ have exactly the same opens

• $\tau,\tau'$ have exactly the same neighborhoods

• for each subset, its closures in $\tau$ and in $\tau'$ coincide.

• the identity set map $(X,\tau)\to (X,\tau')$ is a homeomorphism.

Solution 2:

Two norms are equivalent if they induce the same topology. So it does not matter whether the author of your book defines the notion of "equivalent topologies", since he does not need it (at least not here). I.e., you work here with equality of topologies. (Which is reflexive, symmetric and transitive, so you may call the same topologies equivalent, if you wish. However, the name "equivalent" would be redundant.)

You also asked whether this:

open set in $T_1$ is an open set in $T_2$ and conversely

is the same as equality of topologies. Indeed, it is. Every set open in $T_1$ being open in $T_2$ is inclusion $T_1\subseteq T_2$. The converse implication gives you the converse inclusion.

BTW I guess all of this was already said in comments.