Defining Topological Continuity I have seen this definition many times:
Topological Continuity: A function $f:X\rightarrow Y$ is continuous if for all open sets $U \subseteq Y$, the preimage $f^{-1}(U)$ is open in $X$.
I don't understand why the notion of openness is important to topological continuity. Can someone explain this? 
By the way, I understand the definition of analytic continuity just fine, just to clarify it's topological I'm struggling with. 
 A: The basic idea is the a function $f:X\to Y$ is continuous iff for any open set $V$ containing $f(x)$, there is an open set $O\ni x$ such that $f(O)⊆V$. Thus, given any nbhd (i.e. a degree of closeness) $V$ of $f(x)$ we can take points $O$ "close enough" to $x$ so that all their images are "close enough" to $f(x)$. In metric language, this translates to $f(B(x,δ))⊆B(f(x),ε)$ -- this is the analytic definition you speak of. 
Indeed, to say that given any $\varepsilon>0$ there is $\delta >0$ such that whenever $d(x,y)<\delta$ then $d'(f(x),f(y))<\varepsilon$ says precisely that the image of the ball $B(x,\delta)$ is contained in the ball $B(f(x),\varepsilon)$. Since every open set $O$ containing $f(x)$ in a metric space contains a ball $B(f(x),\varepsilon)$, this says that for any open set $O$, $f^{-1}(O)$ is open: take $x\in f^{-1}(O)$, i.e. $x$ such that  $f(x)\in O$ and use the above to get $f^{-1}(O)$ contains a ball $B(x,\delta)$. 
Conversely, given the open set $B(f(x),\varepsilon)$, saying $f^{-1}(B(f(x),\varepsilon))$ is open means in particular (since $x\in f^{-1}(f(x))$) that there is an open ball $B(x,\delta)$ wholly contained in $f^{-1}(B(f(x),\varepsilon))$. This is equivalent to $f(B(x,δ))⊆B(f(x),ε)$, of course.
The intuitive aid of "closeness", perhaps more appropriate to say metric spaces, might break down when we allow arbitrary topologies. Nevertheless, it is usually an appropriate tool to figure out what's going on. 
A: I'm going to show that if $f: X \to Y$ (where $X$ and $Y$ are metric spaces) is continuous in the topological sense (t-continuous), then it's continuous in the metric sense (m-continuous). With that proof at hand, I'll bet that you can do the other half -- if $f$ is m-continuous, then it's also t-continuous. 
Here's the first half:
Suppose $f$ is t-continuous at $x_0 \in X$, and $f(x_0) = y_0$. 
Given $\epsilon > 0$, the set of $y$ for which $ d(y - y_0)  < \epsilon$ is the same as $D(y_0, \epsilon)$, which is an open disk in $Y$. Its preimage $U$ is therefore an open set in $X$, and contains $x_0$. Since $x_0$ is in this open set, there's an open disk about $x_0$, say $D(x_0, \delta)$ that's entirely contained in $U$. Now for $d(x,x_0) < \delta$, we have
$x \in D(x_0, \delta)$, so $f(x) \in D(y_0, \epsilon)$. Hence $f$ is m-continuous at $x_0$. 
A: Not sure how much you know about metric spaces and topological spaces, but basically if you're working with topological spaces in general, then openness is pretty much the only thing you have to work with.  So you have to define continuity in terms of that.
http://en.wikipedia.org/wiki/Topological_space
Metric spaces are an important type of topological space, and in any metric space the epsilon-delta definition is equivalent to the topological definition, as others have mentioned.
http://en.wikipedia.org/wiki/Metric_space
So this topological definition generalizes the epsilon-delta definition to topological spaces.
