Why dont we say if $f(U)$ is open for every open set $U$ in $A$, then $f$ is countinous? Definition of continuity in topology;
Let $f :A \to B$.
If $f^{-1}(U)$ is open in A for every open set U in B, we say f is countinous.
Why dont we say if $f(U)$ is open for every open set $U$ in $A$, then $f$ is countinous ? 
Does the image of an open set under a continous mapping need to be open ? 
 Thank you for your help.
 A: No. This would make the constant functions ($f:X\to\mathbb R; \ f(x) = c$) discontinuous, since
$$f(U) = \{c\}$$
for $U\ne\emptyset$ open.
A: Very roughly speaking, the idea of continuity is that points that are close to each others are sent to points that are close to each others. A way of phrasing this is that converging sequences are mapped into converging sequences. It is not hard to check that continuity in this sense is equivalent to continuity defined requiring that the preimages of open sets are open.
The other condition (the image of open sets is open) has a different name: a function that satisfies such property is called an open map. There are a number of example of continuous maps that are not open and viceversa. 
Open but not continuous:
Open maps which are not continuous
Continuous (and closed!!!) but not open:
$f \colon \mathbb{R} \to \mathbb{R},\ f(x) = x^2$
A: The preimage of an open set under a continuous function is open, but the image of an open set under a continuous function is not necessarily open.

Let's say $f: X \rightarrow Y$ is continuous and $U \subseteq Y$ is open. 
Here's a proof of why $f^{-1}(U)$ must also be open.
Let $p \in f^{-1}(U).$ Then by definition of a preimage, $f(p) \in U$. Since $U$ is open in $Y$, there exists $\epsilon > 0$ such that $B_{\epsilon}(f(p)) \subseteq U$. Since $f$ is continuous at $p$, there exists $\delta > 0$ such that $q \in B_{\delta}(p) \implies f(q) \in B_{\epsilon}(f(p)).$
But since $B_{\epsilon}(f(p)) \subseteq U$, this says that anything in $B_{\delta}(p)$ is sent to something in $U$, so that all of $B_{\delta}(p)$ is contained in the preimage of $U$. Thus there exists an open ball $B_{\delta}(p) \subseteq f^{-1}(U),$ so $f^{-1}$ is open in $X$.

Think of open/closed sets as properties that map well in the "backwards" direction, i.e. open image to open preimage under a continuous function. Think of the definition of continuity: for any $\epsilon > 0$, we can go "backwards" and get a $\delta > 0$... such that a ball of radius $\delta$ around some $x$ in the preimage gets mapped to a ball of radius $\epsilon$ around $f(x)$ in the image.
You cannot replicate this in the "forwards" direction.
For example, take a constant continuous function. 
$f: (0,1) \rightarrow \mathbb R$ defined by $f(x) = 1$ is continuous but maps an open set $(0,1)$ to a closed set (singletons are closed) $\{1\}$.
A: Consider the metric space $T=\left \{ 0,1 \right \}$ with the discrete metric $d(x,y)=1$ if $x \neq y$ and $d(x,y)=0$ if $x=y$. 
It is easy to show that any subset of $T$ is open. 
Take any metric space $X$ and any discontinuous function $f:X \rightarrow T$ to obtain a counterexample. 
