If $f^{-1}(G)$ is open in $X$ for every open set $G$ in $Y$, then $f$ is continuous. Question on proof. Let $X,Y$ be metric spaces and $f:X\rightarrow Y$. If $f^{-1}(G)$ is open in $X$ for every open set $G$ in $Y$, then $f$ is continuous. 
The text I am using proves this proposition like so:
Suppose that $f$ is discontinuous at some point $a\in X$. Then there is an $\epsilon >0$ such that for every $\delta>0$, there is an $x\in X$ with $d(x,a)<\delta$ and $d(f(x),f(a))\geq \epsilon.$ It follows that, although $a$ belongs to the inverse image of the open set $B_{\epsilon}(f(a))$ under $f$, the inverse image does not contain $B_{\delta}(a)$ for any $\delta>0$, so it is not open.
I am having trouble understanding how $B_{\delta}(a)$ is not contained in $f^{-1}(B_{\epsilon}(f(a)))$. If $d(x,a)<\delta$ implies  $d(f(x),f(a))\geq \epsilon$.
My questions are:
1) What does the preimage of $f^{-1}(B_{\epsilon}(f(a)))$ look like? Is it a set with radius greater than $\delta$ or less than $\delta$? How could I see this if I were to draw a picture? This answer may clarify questions 2) and 3).
2) If $B_{\delta}(a)$ is not contained in $f^{1}(B_{\epsilon}(f(a)))$ then $B_{\delta}(a)$ is a larger set than $f^{-1}(B_{\epsilon}(f(a)))$. Is this correct? 
3) If so doesn't this mean that $f^{-1}(B_{\epsilon}(f(a)))$ maps points inside the $B_{\delta}(a)$?
Thank you for any help and comments!
 A: 1) $f^{-1}(B_{\epsilon}(f(a)))$ is the set $\{x\in X:|f(x)-f(a)|<\epsilon\}$. It is hard to come up with a picture for this, because the set could be somewhat wild. It does not have to be a ball though. In the case where $f$ is continuous this set will be open and thus it will be the union of open sets (you might have a basis for the topology). An example that shows you that the set need not to be a ball consider the map $f:\mathbb{R}\rightarrow [0,1]$ via: $f(x)=\sin(x)$. Consider the open interval $(-1/2,1/2)=B_{1/2}(f(0))$. 
The preimage of this will be: $(-\pi/6+2n\pi,\pi/6+2n\pi),(5\pi/6+2n\pi,7\pi/6+2n\pi)$. This is an union of a bunch of open interval, so your question "Is it a set with radius greater than $\delta$" is not really applicable, since again, the preimage needs not to be a ball. 
2) It is not larger. Just not contained. An example that would show this is again by above, $\epsilon=1/2$, and $a=0$, and let $\delta=\pi/2$. Then $B_\delta(0)$ is not contained in $f^{-1}(B_\epsilon(f(a)))$, but it does not contain it either. 
3)Is it above from above?
A: The $\epsilon$-$\delta$ definition of continuity, you should know.  It's of the form $f$ is continuous $\iff (\forall A, \exists B : C \implies D)$.  So $f$ is not continuous $\iff \neg (\forall A, \exists B : C \implies D)$.  But 
$$
\neg (\forall A, \exists B : C \implies D) \\
= \exists A : \forall B, C \nRightarrow D
$$
But it's more meaningful to come to the same conclusions in your context of discussion.  $C \nRightarrow D$ amounts to $x\in B_\delta \nRightarrow f(x) \in B_\epsilon$ which is logically equivalent two $\exists x\in B_\delta$ such that $f(x) \notin B_\epsilon$.
But the last part is true if and only if $B_\delta \nsubseteq f^{-1}(B_\epsilon)$.  Prove that.
The inverse image isn't necessarily "circular" in shape, so it doesn't necessarily have a circle's radius per se. 
Once you've proved the statement, you'll get a picture of what continuity is.  Every inverse image of an open set of a continuous map $f$ is open meaning it contains an open ball around each of its points (when we're in a metric space, such that there is a definition of open ball). 
