Equivalent Topologies In a text I am reading, the author defines two norms on a vector space $X$ to be equivalent if they induce the same topology on $X$. The text does not define what it means however for two topologies to be equivalent. The only definition I can think of that seems reasonable is that two topologies $T_1$ and $T_2$ are equivalent if an open set in $T_1$ is an open set in $T_2$ and conversely. Since open sets in a normed space are determined by the norm-relative open balls, this would mean that any open ball about a point $x$ with respect to one norm not only contains an open ball about $x$ with respect to the other norm but is also contained within an open ball relative to that other norm. Is this the right way to look at the situation? Is my definition of "equivalent toplogies" appropriate in this context?
 A: 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.
A: 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.
A: Two topologies are called equivalent if any non-empty open set of the first topology contains a non-empty open set of the second topology and, conversely, every non-empty open set of the second topology contains non-empty open sets of the first topology.
This does NOT mean that the two topologies are the same.
So give me my two vote down back! :)
In case of normed spaces, this condition can be written in a quite nice way: two norms $||\cdot||_1$ and $||\cdot||_2$ on $X$ are equivalent if and only if there are two positive numbers $C_1,C_2>0$ such that, for all $x\in X$,
$$
C_1||x||_1\leq||x||_2\leq C_2||x||_1
$$
you can verify by yourself that this condition is equivalent at the one between topological spaces just remembering what is a base of open sets of the topology induced by a norm.
