TITLE: Equivalent Topologies
QUESTION [8 upvotes]: 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?

REPLY [12 votes]: 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.