Image of open set is not open? I'm confused by the proof that $\epsilon$-$\delta$ continuity is equivalent to open-set continuity. One can prove that a function is $\epsilon$-$\delta$-continuous if and only if the preimage of any open set is open. However, it is not true that the image of any open set is open. My question is, what is the difference between image and preimage? The duality of image and preimage suggests to me that a proof that works for one of them should work for the other, but clearly this is not the case. Why not?
 A: You can even have a continuous bijection that doesn't map open sets to open sets. Consider the function
$$
f: [0,1) \cup [2,3] \to [0,2], \; f(x) = \begin{cases}x , & x \in [0,1) \\ x-1, & x \in [2,3] \end{cases}.
$$
This is a continuous bijection. The set $[2,3]$ is open in $[0,1) \cup [2,3]$. But the image of $[2,3]$ is $[1,2]$ which is not open in $[0,2]$.
What happens with this function is that the domain is not connected, but the range is. So the function "glues" the two parts of the domain together continuously. This means that the inverse function has to rip apart these parts, so it's not continuous.
A: The difference is in the following: 
1) $x$ close to $y$ implies $f(x)$ is close to $f(y)$ : true for continuous functions
2) $f(x)$ close to $f(y)$ implies $x$ is close to $y$: generally false for continuous functions
A: Any conjectured "symmetry" between preimage and image will fail because functions are not necessarily one-to-one.  Notice that in the case of $f(x) = x^2$, the image of $(-1,1)$ is $[0,1)$, which is not open.  The reason this fails is because the map has no inverse.  If we reflect the graph across the line $y=x$, we end up with something that is not a function (it's not single valued).  The preimage of $(-1,1)$ under this relation is $[0,1)$ but the result is not a continuous function because it is not a function.
