Are Open Balls Open?

You might be tempted to say, "Yes, because it's in the name!" But that's not a proof, or even a good explanation.

Sets are open when they have no boundary points. And a point from a set is on the boundary when not all of its $\varepsilon$-balls also contain points in the complement.

So let's verify this. Below is a diagram with the unit circle around the origin. The unit ball is the points on the interior of the circle (i.e. less than 1 away from the origin). We need to find an $\varepsilon$ small enough so that the blue circle stays inside the unit circle no matter where we drag the point $P$. The circle seems small enough now, but you'll notice that as you drag $P$ towards the edge this choice for $\varepsilon$ is not small enough.

When you're ready to change the rule for $\varepsilon$, open up the left pannel and change the value of $r$ (the radius $\varepsilon$ of the small circle) from the 0.23 it's currently set at to a dynamic rule that will work for any point.

Figure 1: The Open Unit Ball

Can you prove that your rule works? If you're stuck, try using the triangle inequality.