sets with same image under continuous map Suppose I have a continuous function $f : X \rightarrow Y$ of topological spaces $X $ and $Y$. If I have two sets $U$ and $V$ in $X$ such that the image under $f$ of both of these sets is the same $f(U) = f(V) $ and $U$ is open- would $V$ have to be open?
 A: No.
Consider $X=Y=\mathbb{R}$ under the usual topology and $f(x)=x^2$. Then, if $U=(-1,1)$ and $V=[0,1)$, $f(U)=f(V)=[0,1)$, but $U$ is open and $V$ is not.
For the question in the comment, we can always make the image open just by restricting the domain $X$ to $(-1,1)$ and $Y$ to $f(X)=[0,1)$ with the subspace topology.
A: Consider this for a constant function $f$.  (If you want $f(U)$ to be open as well then let the space $Y$ consist of a single element.)
A: No. Let $X = \mathbb R^2$ and $Y = \mathbb R$ taken as the $y$-axis, why not. Let $U$ be the open unit square with corners at $(0,0),(1,0),(1,1),(0,1),$ and let
$f(x,y)=y.$ So the square just maps the square sideways to the vertical segment. But, if we let $V$ be that same vertical segment, $f(V) = V.$ But $V$ is not open as s subset of $X = \mathbb R^2$
NOTE THAT $f(U) = V$ IS OPEN AS A SUBSET OF $Y = \mathbb R.$ 
Well, why not. I believe @AndreNicolas's comment alluded to this example: Let $X=U$ be the standard open unit disk in the plane, and let $Y = V = \{(0,0)\}$ be a single point. Finally, let $f:X \rightarrow Y,$ that is every point in the disk is mapped to the origin. So $f(U) = f(V) = V$ again. But $V,$ while an open subset of $Y,$ is a closed subset of $X,$ furthermore $V$ is not an open subset of $X.$ There are sometimes these sets called clopen, just so you know. There was this bit about Paris and France, this is pretty much it. 
A: Indeed, it is possible that $f$ continuous and $f(V)=f(U) \implies U,V$  are open only if $f(V)=f(U)$ are open in $Y$. You may try proving it. If you need more hints, you may comment on it.
