Prove that y>2x+1 is open? The answer is inserted but what I'm looking for is a heavy breakdown on this. My professor tried to explain to me a harder version but I don't understand it.

Solution:


What my prof does for the problems he has gone over is he first draws out a disk contained in S. Then he finds another point within the disk, moving it some small increment less than the radius of the disk. He tries to find a radius that will work for any point in S to show that it's open.
In this case, he would do something like this ... (as a proof sketch)
Let $z=(x,y) \in S$ with open disc of radius $\delta$, $D_r(z)$. If we take another point in the disc around z, say $z_0$, we would have $z_0 = (x+\alpha,y+\beta)$ for $|\alpha|<\delta$ and $|\beta|<\delta$.
Then $y+\beta > 2(x+\alpha)+1$
$y+\beta > 2x + 2\alpha+1$
$y-2x-1>2\alpha-\beta$
Then he would probably do triangle inequality stuff here to get a value to get the radius $\delta$.
I am not looking for alternative solutions (such as the use of inverses which we haven't covered).
Can somebody please really really break this down for me and justify everything? I would really appreciate it, I've tried so hard to learn it but I just don't get it!
Thank you.
 A: First, you probably missed $1$ when you typed $y+\beta>2(x+a)$ (i.e. the RHS should be $2(x+a)+1$). 
Second, I think your professor was trying to show you the reasoning behind the the formula $\delta=\frac{1}{3}(y-2x-1)$. Basically, you want $\delta$ such that whenever $|\alpha|<\delta$ and $|\beta|<\delta$, then $y-2x-1>2\alpha-\beta$. (Again, the $1$ comes from $y+\beta>2(x+a)+1$.) But note that
$$
2\alpha-\beta\leq|2\alpha-\beta|\leq2|\alpha|+|\beta|<3\delta
$$
so any $0<\delta\leq\frac{1}{3}(y-2x-1)$ works. In particular, you might as well set $\delta=\frac{1}{3}(y-2x-1)$.
I'll say the most potentially confusing things about this are: using 2 symbols $r$ and $\delta$ for the radius (perhaps you miscopied him?) and using different notation than in the textbook: your professor used $x$ and $y$ instead of $a$ and $b$ as in the text.
Edit: A definition (among other equivalent definitions) of an open set is that for any point in the set, there is a positive radius (which may depend on the point) such that the open ball around that ball with that radius is contained entirely in the set.
So that was exactly what your instructor did: consider any point $(x,y)\in S$, construct a suitable radius ($\delta=\frac{1}{3}(y-2x-1)$) and check that the ball $B_\delta((x,y))$ is contained in $A$. 
It's quite straightforward, you just need to be mindful of your definitions.
Edit 2:
So what happened was


*

*Consider any point $(x,y)\in A$

*Construct a radius $\delta=\frac{1}{3}(y-2x-1)>0$

*Show that any point $z_0\in B_\delta(x,y)$ belongs to $A$

*Step 3 is achieved by writing $z_0=(x+\alpha,y+\beta)$ and using
$$
y-2x-1>2\alpha-\beta
$$
to infer that
$$
(y+\beta)-2(x+\alpha)-1=(y-2x-1)-(2\alpha-\beta)>0
$$
which proves $z_0\in A$.

A: Let $f:\mathbb{C}\to\mathbb{R}$ be a continuous function. As $]0,\infty[$ is an open set of $\mathbb{R}$, one easily verify that 
$$\{z\in\mathbb{C}:f(z)>0\} = f^{-1}\big[]0,\infty[\big]$$
is a open set of $\mathbb{C}$.
In your case, set $f(x,y) = y-2x-1$.
