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I'm having trouble proving that a given subset is open. Let's say I'm asked to prove the following set is open. $$A =\{(x,y) : -1 < x < 1, -1 < y < 1\}$$ let $(x_0,y_0) \in A$, then $ |x_0| < 1\;and\; |y_0| < 1$. Then I define $$ r = min\{1-|x_0|,1-|y_0|\}$$

Now I need to show that $D_r(x_0,y_0)\subset A$

So I let $(x,y) \in D_r(x_0,y_0)$ and then the distance between the two points is: $$\sqrt{(x-x_0)^2 + (y-y_0)^2}$$ Now I'm stuck. What exactly do I have to show? Can I say that this distance is less than $r$? How do I deal with the fact that $r$ can be one of two values? What is the general approach to such problems?

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So what you have to do is show that $D_r(x_0, y_0) \subseteq A$. In other words, if $(x, y) \in D_r(x_0, y_0)$, then $(x, y) \in A$. In other words, if both

$$((x - x_0)^2 + (y-y_0)^2)^{1/2} \lt 1 - |x_0|$$

and

$$((x - x_0)^2 + (y-y_0)^2)^{1/2} \lt 1 - |y_0|$$

then $|x| \lt 1$ and $|y| \lt 1$.

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You chose $(x,y)$ to be in $D_r(x_0, y_0)$, which is (I assume by definition) the disc of radius $r$ around that point, so yes, the distance to that point is certainly less than $r$.

The idea is to now show that $(x,y) \in A$, so that you can conclude that every point of $D_r(x_0, y_0)$ is in $A$ and hence $D_r(x_0, y_0) \subset A$. Noting that $(x_0, y_0)$ was arbitrary, you conclude that every point in the set has an open disc around it still within the set which is the criterion for openness.

How to deal with the fact that $r$ may be the value of one of two expressions: well, here's a nice way of characterising $\min\{x,y\}$ for any $x$ and $y$: \begin{align} z < \min\{x,y\} &\iff z < x \text{ and } z < y\\ min\{x,y\} < z &\iff x < z \text{ or } y < z \end{align} These facts are kind of obvious if you think about it, but it's easy to forget how useful they are. If you want to prove an inequality like those on the left, just prove an inequality or two from the right.

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$A$ is a solid square with the boundary cut out. It is an open set if we can pick any point $(x_0, y_0)\in A$ and draw a little tiny ball around that point that is contained all inside $A.$ This is clearly true intuitively, let the picture guide your proof. You had the right idea when you took $r= \min \{1-|x_0|, 1-|y_0|\}$ because (as a picture indicates) this is precisely the radius you take to ensure the ball of radius $r$ around $(x_0, y_0)$ stays inside $A.$

You now have to show your intuition is true: That any point inside this ball is inside $A.$ If you have a point $(a,b)$ inside your ball then by definition, $$\sqrt{ (x_0-a)^2+ (y_0-b)^2 } < r = \min \{1-|x_0|, 1-|y_0|\}.$$ If $(a,b)$ wasn't inside $A$ then at least one of $a$ or $b$ would have absolute value greater than or equal to $1.$ Suppose $a$ has absolute value $\geq 1.$ Then $$\sqrt{ (x_0-a)^2+ (y_0-b)^2 } \geq |a-x_0| \geq |a|-|x_0|> 1-|x_0| \geq r= \min \{1-|x_0|, 1-|y_0|\}$$ which contradicts the previous equation. Hence $(a,b)$ in inside $A.$

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