Show that $f:\mathbb{R}^2\to\mathbb{R}$ defined by $f(x,y)=x+y$ is continuous using open sets. The problem statement is,
Show that $f:\mathbb{R}^2\to\mathbb{R}$ defined by $f(x,y)=x+y$ is continuous using open sets.
I know that to show $f$ to be continuous I take an arbitrary open set $O\subset\mathbb{R}$ and show that $f^{-1}(O)$ is open in $\mathbb{R}^2$. We can simplify this by first showing that for any $(a,b)\in\mathbb{R},$  $f^{-1}(a,b)$ is open in $\mathbb{R}^2.$ Now, we have
$$f^{-1}(a,b)=\{(x,y):f(x,y)\in(a,b)\}=\{(x,y):a<x+y<b\}$$
To show that $f^{-1}(a,b)$ is open, we need to show that for every $u\in f^{-1}(a,b)$, there exists a $\delta>0$ such that $B_\delta(u)\subset f^{-1}(a,b).$
One way I thought about showing this was to first fix $y$ in $(x,y)\in f^{-1}(a,b)$ and range over all $x$. Do the same for $y$ and then try to find an appropriate $\delta$. I did find this question (which is exactly mine)
Using the open set definition of continuity to directly prove a function is continuous
Yet, the I didn't completely understand the hints or answers given.
Thanks for any help or feedback.
 A: You'll have to check the details, but I think this would be the outline:
Let $(x,y) \in f^{-1}(a,b)$, then $a < x + y < b$ hence we can find an $\epsilon > 0$ (check this!) such that
$$
a + \epsilon < x + y < b - \epsilon.
$$
Then (for $u = (x, y)$) we have $B_{\epsilon/2}(u) \subset f^{-1}(a,b)$, for let $(x', y') \in B_{\epsilon/2}(u)$, then $|x - x'| < \epsilon/2$ and $|y - y'| < \epsilon/2$, therefore (check this one too!)
$$
a < x' + y' < b.
$$
In conclusion, for every point in $f^{-1}(a,b)$ we can find an open neighbourhood $B_{\epsilon/2}(u)$ such that $B_{\epsilon/2}(u) \subset f^{-1}(a,b)$, proving that $f^{-1}(a,b)$ is open.
A: A picture is helpful:

The diagonal lines are the lines $x+y=a$ and $x+y=b$, so $f^{-1}[(a,b)]$ is the strip lying strictly between them. The point $p$ is any point in that region; you want to find an open ball with centre at $p$ that lies wholly within the strip, as the one in the sketch just barely does. If you take the radius of the ball to be the minimum of the distances from $p$ to the two edges of the strip, you’ll be in business, and calculating that minimum in terms of the coordinates of $p$ is just a bit of easy analytic geometry.
A: You could prove that the squares $(x,x+\frac{b-a}{2})\times(a-x,a-x+\frac{b-a}{2})$ cover $f^{-1}(a,b)$ as $x$ ranges over all real values.
A: Exodd is right! In addition, the pre-image of the open interval in $\mathbb{R}$ is a band region in $\mathbb{R}^2$, which is also open.
A: $$ \{(x,y)\mid a<x+y<b\}=\{(x,y)\mid x+y<b\}\cap \{(x,y)\mid x+y> a\} $$
Intersection of two open half-planes.
