How do I check whether this set is open or not in $\Bbb R^2$? Question is given as $S= \{(x,y) ∈ \Bbb R^2 \mid 4x + 3y > 12\}$
So the way how I approached it is, say $(x_1,y_1) \in S$ which is any arbitrary point in $S$. Now $B((x_1,y_1), r)$ ($r$ is the radius of open ball). So $$B((x_1,y_1),r) = \bigl\{(x,y) \in \Bbb R^n\mid \bigl((x-x_1)^2 + (y-y_1)^2\bigr)^{1/2}  < r\bigr\}$$ so we will prove that $(x,y) \in B((x_1,y_1), r)$.
Consider $|x-x_1| = \left((x-x_1)^2\right)^{1/2} <\left((x-x_1)^2 + (y-y_1)^2\right)^{1/2} < r$, 
 which means that $|x-x_1| < r$ which ultimately means  $x_1 - r < x < x_1 + r$, but after this point I don't know what to do, I don't know what to take as $r$ in this problem.
 A: By the point to line distance formula, the distance from some $(x_0, y_0)$ to $ax + by+c=0$ is $$ \left|\frac{ax_0 + by_0 +c}{\sqrt{a^2+b^2}}\right|$$ and in your case we're interested in the line $4x +3y-12=0$. In particular, then, the distance between $(x_0, y_0)$ and that point is $$ d=\frac{|4x_0+3y_0-12|}{5}$$ and your set is precisely defined to be the $(x_0, y_0)$ so that the numerator of that expression is positive, so the distance is actually just $$d=\frac{4x_0 +3y_0 - 12 }{5}$$ Thus, given any point $(x_0, y_0) \in S$ one can draw an open ball of radius $\epsilon$ and get points only in $S$ if one draws a ball of radius, say $$ \epsilon = \frac{4x_0 +3y_0 - 12 }{10}$$ and that's really just the fact that the shortest distance to any point outsider of $S$ is $d$. So, in particular, $S$ is an open set, because we can draw a ball of the radius $\epsilon$ given above containing only points in $S$ around any point in $S$.
A: Lets define the function $g\colon\mathbb{R}^2\to \mathbb{R}$ given by $g(x,y)=4x+3y$
This function is continous since is the sum of two continous functions (four times the proyection to X and 3 times the proyection to Y) .
Since $V=\langle 0;+\infty \rangle$ is a open set and $g^{-1}(V)=\{(x,y) ∈ \Bbb R^2 \mid 4x + 3y > 12\}$, hence, by continuity of $g$, we have that $\{(x,y) ∈ \Bbb R^2 \mid 4x + 3y > 12\}$ is open.
A: You seem to be looking for an elementary solution, and for this I think you should refer to the other answer provided (so take this "answer" as more of an extended comment that provides an alternative approach to solving).
Define the function $f: \Bbb{R}^2 \to \Bbb{R}$ by $f(x,y) = 4x + 3y$. Now, $f$ is just a (linear) polynomial function hence it is continuous. One key property of continuous functions is that preimages of open sets is open. In other words, if $U$ is an open subset of the target space $\Bbb{R}$, then the preimage of $U$ under $f$,
\begin{align}
f^{-1}[U] := \{(x,y) \in \Bbb{R}^2: f(x,y) \in U\}
\end{align}
is an open subset of the domain $\Bbb{R}^2$.
In your particular case, we have that
\begin{align}
S&:= \{(x,y) \in \Bbb{R}^2 : 4x+3y > 12 \} \\
&= \{(x,y) \in \Bbb{R}^2 : f(x,y) > 12 \} \\
&= f^{-1}\left[ (12,\infty) \right]
\end{align}
Note that $(12,\infty)$ is an open subset of the target space $\Bbb{R}$; now since $f$ is continuous, it follows that $S = f^{-1}[(12,\infty)]$ is open in $\Bbb{R}^2$.

This property gives an easy way of checking whether certain sets are open. For example, the "exterior of a unit ball"
\begin{align}
B := \{x \in \Bbb{R}^n: \lVert x\rVert> 1 \}
\end{align}
is open in $\Bbb{R}^n$ because it can be seen as the preimage of an open set under a continuous mapping. More precisely, define $f$ to be the norm function: $f(x) = \lVert x \rVert$. Then, $B = f^{-1}[(1,\infty)]$ is open.
