Topologies coinciding at a point or a set. Consider a set equiped with two topologies. What does it mean to say that the two topologies coincide at a point in the set? Is it meaningful to talk about this concept in general. Is it meaningful in the context of metric spaces?
p.s. I know what the relative topology is for any subset of a topological space and this concept is a different one, as is apparent from this problem.
 A: I would guess that it means the every open neighborhood of $x$ in $\tau_1$ contains an open neighborhood of $x$ in $\tau_2$, and visa versa. This essentially means that "continuity at $x$" is the same in each topology.
It's possible, though, that the stronger statement is intended: the neighborhoods of $x$ are the same in each topology.
There's a lovely way to define the "local topology" at a point. Let $(X,\tau)$ be a topology and $x\in X$. Define a new topology, $\tau_x$ which has as a basis the singletons $\{y\}$ where $y\neq x$ and $U\in\tau$ such that $x\in U$.  So the open sets are any set not containing $x$, or any set contain a $\tau$-neighborhood of $x$. It is not hard to see that this is a topology.
In this case, my first definition would be equivalent to saying, for topologies $\tau$ and $\rho$, they coincide at $x$ if and only if $\tau_x=\rho_x$.
Now, in the complete lattice of topologies on $X$, $$\bigcap_{x\in X}\tau_x = \tau$$  So we can recover $\tau$ if we know each $\tau_x$. Then, for a topological space, $Y$, we can define a function $f:X\to Y$ to be $\tau$-continuous at $x$ if it is continuous on $(X,\tau_x)$, and it turns out, due to the lattice meet property above, the function $f$ is continuous on $(X,\tau)$ if and only if it is continuous at each $(X,\tau_x)$.  
There's also a simple "local range topology", $\tau^x$ defined as just the $\tau$-open sets containing $x$, and the empty set. Then you can say that $f:Y\to X$ is continuous "to" $x$ if it is continuous when using the topology $\tau^x$. Again, we can recover $\tau$ if we know ecah $\tau^x$. Indeed, $\tau = \bigcup \tau^x$.
The stronger definition above would happen exactly when $\tau^x=\rho^x$. It's true that if $\tau^x=\rho^x\implies\tau_x=\rho_x$, though so maybe this stronger definition is reasonable.
