Section 2.5 Regularity of domains
We also need to talk about the regularity of the set and its boundary. (Recall from Notation 2.8 that is a non-empty open set.) Thankfully, we can make do with just a few definitions.
Definition 2.23. Open balls.
A ball in is a set of the form
Definition 2.24. Boundary.
Definition 2.25. Interior and exterior ball properties.
A open set has the interior ball property at a point if there exists a ball such that If this holds for all we simply say that has the interior ball property. The exterior ball property is defined analogously, except that we now require that the ball
Definition 2.26. Connected set.
An open set is called connected if it cannot be written as the disjoint union of two nonempty open sets
Exercises Exercises
1. (PS3) A domain with a corner.
(See comment below about the examinability of this question.) Consider the open set and the point
(a)
Show that satisfies the interior ball property at
Hint.
Look for an interior ball. Putting the centre along the -axis will make things easier.
Solution.
To see that has the interior ball property at the origin, let Then we easily check that Moreover, all points have which in particular implies that
(b)
Show that does not satisfy the exterior ball property at
Hint 1.
Draw some pictures to convince yourself that this is true! Proving it requires a bit more ingenuity than the previous part.
Hint 2.
One relatively simple proof given by a student in previous years can be outlined as follows. Suppose that were an exterior ball at and assume without loss of generality that Then argue that for sufficiently small the point satisfies
Solution.
The following is adapted from a student’s solution in a previous year. Let with and hence Supposing for contradiction that we can assume without loss of generality that in which case We will show that points with lie in which is the desired contradiction. Clearly all such points lie in To see that they lie in we simply expand
Comment 1.
On the exam, you may need to determine whether a given domain satisfies the interior or exterior ball properties, as these are hypotheses of some of our main results. But doing this ‘by eye’ without giving a detailed rigorous proof will be completely acceptable.
Comment 2.
There were several other nice solutions this year, several of which did not involve any calculus.
2. (PS3) Locally constant functions.
Let be an open set and Suppose that is locally constant in the sense that, for all there exists such that on the ball
(a)
Prove that, if is connected, then is globally constant.
Hint.
Fix Supposing that prove that the sets and are both open and nonempty.
Solution.
Fix a value and suppose for the sake of contradiction that on Then where the sets and are disjoint and nonempty. Moreover, since is continuous and is open, a standard argument shows that is open. Since is locally constant, we easily check that is open as well. But then we have written as the disjoint union of two nonempty open sets, contradicting the assumption that was connected.
Comment 1.
If is locally constant, then it is very easy to check that it is automatically continuous. So the requirement is somewhat redundant.
Comment 2.
Often students attempt a more hands-on argument involving chaining together balls on which is constant. While this can be made to work, it requires some serious effort because the definition of ‘locally constant’ gives us no control (from below) on the radii of these balls, and so if we are not careful they could rapidly shrink to zero.
(b)
Give a counterexample to show that the conclusion in the previous part can fail if is not connected.
Solution.
Let be any two nonempty disjoint open sets, and define a continuous function by in and in Then is locally constant in the sense above, but clearly it is not globally constant.