Calculus of One and Several Variables

15.2 Open And Closed Sets

Eventually, one must consider functions which are defined on subsets of ℝⁿ and their properties. The next definition will end up being quite important. It describe a type of subset of ℝⁿ with the property that if x is in this set, then so is y whenever y is close enough to x.

Definition 15.2.1 Recall that for x,y ∈ ℝⁿ,

( )1∕2 \sumn 2 |x - y| = |xi - yi| . i=1

Also let

B (x,r) \equiv {y \in ℝn : |x - y| < r}

Let U ⊆ ℝⁿ. U is an open set if whenever x ∈ U, there exists r > 0 such that B

⊆ U. More generally, if U is any subset of ℝⁿ, x ∈ U is an interior point of U if there exists r > 0 such that x ∈ B

⊆ U. In other words U is an open set exactly when every point of U is an interior point of U.

If there is something called an open set, surely there should be something called a closed set and here is the definition of one.

Definition 15.2.2 A subset, C, of ℝⁿ is called a closed set if ℝⁿ ∖ C is an open set. They symbol ℝⁿ ∖ C denotes everything in ℝⁿ which is not in C. It is also called the complement of C. The symbol S^C is a short way of writing ℝⁿ ∖S.

To illustrate this definition, consider the following picture.

You see in this picture how the edges are dotted. This is because an open set, can not include the edges or the set would fail to be open. For example, consider what would happen if you picked a point out on the edge of U in the above picture. Every open ball centered at that point would have in it some points which are outside U. Therefore, such a point would violate the above definition. You also see the edges of B

dotted suggesting that B

ought to be an open set. This is intuitively clear but does require a proof. This will be done in the next theorem and will give examples of open sets. Also, you can see that if x is close to the edge of U, you might have to take r to be very small. open sets do not have their skins while closed sets do. Here is a picture of a closed set, C.

Note that x

C and since ℝⁿ ∖ C is open, there exists a ball, B

contained entirely in ℝⁿ ∖ C. If you look at ℝⁿ ∖ C, what would be its skin? It can’t be in ℝⁿ ∖ C and so it must be in C. This is a rough heuristic explanation of what is going on with these definitions. Also note that ℝⁿ and ∅ are both open and closed. Here is why. If x ∈∅, then there must be a ball centered at x which is also contained in ∅. This must be considered to be true because there is nothing in ∅ so there can be no example to show it false¹ . Therefore, from the definition, it follows ∅ is open. It is also closed because if x

∅, then B

is also contained in ℝⁿ ∖∅ = ℝⁿ. Therefore, ∅ is both open and closed. From this, it follows ℝⁿ is also both open and closed.

Theorem 15.2.3 Let x ∈ ℝⁿ and let r ≥ 0. Then B

is an open set. Also,

D (x,r) \equiv {y \in ℝn : |y - x| \leq r}

is a closed set.

Proof: Suppose y ∈ B

. It is necessary to show there exists r₁ > 0 such that B

⊆ B

. Define r₁ ≡ r −

. Then if

< r₁, it follows from the above triangle inequality that

|z- x| = |z- y+ y - x| \leq |z- y|+ |y - x| < r1 + |y- x| = r - |x - y|+ |y - x| = r.

Note that if r = 0 then B

= ∅, the empty set. This is because if y ∈ ℝⁿ,

≥ 0 and so y

. Since ∅ has no points in it, it must be open because every point in it, (There are none.) satisfies the desired property of being an interior point.

Now suppose y

. Then

> r and defining δ ≡

− r, it follows that if z ∈ B

, then by the triangle inequality,

|x- z| \geq |x- y|- |y- z| > |x- y|- δ = |x- y|- (|x - y|- r) = r

and this shows that B

⊆ ℝⁿ ∖ D

. Since y was an arbitrary point in ℝⁿ ∖D

, it follows ℝⁿ ∖D

is an open set which shows, from the definition, that D

is a closed set as claimed. ■

A picture which is descriptive of the conclusion of the above theorem which also implies the manner of proof is the following.

The next theorem includes the main ideas for a set to be closed. It says that closed is to retain all limits of sequences which are contained in A.

Theorem 15.2.4 A nonempty set A is closed if and only if whenever x_k ∈ A and lim_k→∞x_k = x, it follows that x ∈ A. In other words, the set is closed if and only if every convergent sequence of points of A converges to a point of A.

Proof: Suppose A is closed and suppose lim_k→∞x_k = x. Does it follow that x ∈ A? If not, then since A is closed, its complement is open and so there is a ball B

contained in A^C. However, this contradicts the assertion that x is the limit of the sequence. Indeed, x_k must be in B

for all k sufficiently large.

Conversely, suppose A retains all limits of convergent sequences. Is A closed? In other words, is its complement A^C open? Suppose x ∈ A^C. Is B

⊆ A^C for small enough positive r? If not, then B

contains a point of A called x_k for each k = 1,2,

. Thus x is a limit of the sequence

and so x ∈ A after all. Hence A^C must indeed be open and so, by definition, A is closed. ■