Writing skills: Proof of the relation between $\epsilon - \delta$ and open sets continuity In order to check my math writing skills, I worked on writing the following basic proof.
Theorem:
If a function $f: X \to Y$ is continuous, then $G \subseteq Y$ is open implies that $f^{-1} (G)$ is open.  
Proof:
Assume that the function $f$ is continuous, and assume that $G \subseteq Y$ is open. Let $x^* \in f^{-1} (G)$ be arbitrary. Thus, $f(x^*) \in G$, and there is a $y \in G$ such that $y = f(x^*)$. By the definition of open set, there is a $\epsilon^*$ such that $B_{\epsilon^*} (y) \subseteq G$. Applying this to the definition of continuous function, there is a $\delta^*$, that can depend from $\epsilon^*$, such that for every $x \in X$, if $ |x - x^* | < \delta^*$, then $\lvert f(x) - f(x^*) \rvert < \epsilon^*$. We want to prove that there is a $\epsilon > 0$ such that $B_\epsilon (x^* ) \subseteq f^{-1} (G)$. Let $\epsilon = \delta^*$, and let $\bar{x} \in B_{\delta^*} (x^*)$ be arbitrary. Thus, $|\bar{x} - x^* | < \delta^*$ and, from the definition of continuous function, $|f(\bar{x}) - f(x^*)| < \epsilon^*$, that is $f(\bar{x}) \in B_{\epsilon^*} ( f(x^*)) \subseteq G$.
QED
I do have some doubts. In particular I don't like the switch from impersonal to plural first person. Moreover, I am not sure if I should divide it in paragraphs or not. Actually, I am really looking forward to any feedback.  
Thank you in advance.
PS: Of course, I do hope there are no mathematical mistakes.
 A: Everyone has their own preferences and writing styles but I thought I'd give rewriting it a go with a few comments. (I wouldn't be surprised if a make a typo)

Proof:
Suppose that $f:X \to Y$ is continuous, and that $G \subseteq Y$ is open. Fix $x_0 \in f^{-1}(G)$. Then $f(x_0) \in G$, and since $G$ is open there exists $\varepsilon > 0$ such that $B_{\varepsilon}(f(x_0)) \subseteq G$. Continuity of $f$ implies that there exists $\delta >0$ such that for each $x \in X$ with $|x - x_0| < \delta$ we have $|f(x) - f(x_0)| < \varepsilon$. Observe that if $x \in B_\delta(x_0)$, then $|x - x_0|< \delta$ and so $|f(x) - f(x_0)| < \varepsilon$. In particular $f(x) \in B_{\varepsilon}(f(x_0))$ so that $x \in f^{-1} (B_{\varepsilon}(f(x_0)) \subseteq G$. Hence $B_\delta(x_0)$ is an open set in $f^{-1} (G)$ containing $x_0$, and since $x_0$ was arbitrary $f^{-1} (G)$ is open.

So a few things, some personal:

*

*I find the word "suppose" fits a bit more nicely that assume. Perhaps it's less
demanding?

*$\varepsilon$ is better that $\epsilon$     :)

*Using $^*$ for your variable names can make them hard to read; I prefer $x_0$ for a value that you have fixed at the beginning of your proof. Also there were a lot of $\varepsilon^*$s that could have been avoided. If you can avoid using subscripts/superscripts/*/bars/etc do so, but sometimes you need it. If you do need to use $x$ and $\overline{x}$ for example, try and use the $\overline{x}$ on the variable which occurs the least. Superscripts in subscripts can be especially ugly.

*When you say "Let $x^* \in f^{-1}(G)$ be arbitrary" the fact that $x^*$ is arbitrary is already assumed. Later when you are referring to the fact that it was arbitrary you can state it.

*You only used $y$ for a short while and later you started referring to $B_{\varepsilon}(y)$ as $B_{\varepsilon}(x^*)$ so I don't think you really needed the $y$ in the first place.

*If you want to state what you wish to prove its better to do it right after you have defined $x^*$ rather that half way through.

*Try and avoid using "by the definition of", you can make it flow a bit more smoothly without it.

*You referred to the continuity of $f$ twice to do the same thing, you don't really need to state it the second time.

*The proof sort of ended abruptly. What have we shown by proving that $f(\overline{x}) \in G$.

I hope this was useful! I tried to be fairly thorough.
If you or anyone else have critiques of my writing I'd be glad to hear it. Writing at 1 a.m. isn't my best work.
