Eventually, one must consider functions which are defined on subsets of ℝ^{n} and their
properties. The next definition will end up being quite important. It describe a type of
subset of ℝ^{n} with the property that if x is in this set, then so is y whenever y is close
enough to x.
Definition 15.2.1Recall that forx,y∈ ℝ^{n},
( )1∕2
∑n 2
|x − y| = |xi − yi| .
i=1
Also let
B (x,r) ≡ {y ∈ ℝn : |x − y| < r}
Let U ⊆ ℝ^{n}. U is an open setif whenever x ∈ U, there exists r > 0 such thatB
(x,r)
⊆ U. More generally, if U is any subset of ℝ^{n}, x ∈ U is aninterior point of Uif there exists r > 0 such that x ∈ B
(x,r)
⊆ U. In other words U is an open set exactlywhen 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 15.2.2A subset, C, of ℝ^{n}is called aclosed set if ℝ^{n}∖ C isan open set. They symbol ℝ^{n}∖ C denotes everything in ℝ^{n}which is not in C. It isalso called thecomplement of C. The symbol S^{C}is a short way of writing ℝ^{n}∖S.
To illustrate this definition, consider the following picture.
PICT
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
(x,r)
dotted suggesting that B
(x,r)
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.
PICT
Note that x
∕∈
C and since ℝ^{n}∖ C is open, there exists a ball, B
(x,r)
contained
entirely in ℝ^{n}∖ C. If you look at ℝ^{n}∖ C, what would be its skin? It can’t be in ℝ^{n}∖ 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 ℝ^{n} 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
false^{1} .
Therefore, from the definition, it follows ∅ is open. It is also closed because if x
∈∕
∅, then
B
(x,1)
is also contained in ℝ^{n}∖∅ = ℝ^{n}. Therefore, ∅ is both open and closed. From this,
it follows ℝ^{n} is also both open and closed.
PICT
Theorem 15.2.3Let x ∈ ℝ^{n}and let r ≥ 0. Then B
(x,r)
is an open set.Also,
D (x,r) ≡ {y ∈ ℝn : |y − x| ≤ r}
is a closed set.
Proof: Suppose y ∈ B
(x,r)
. It is necessary to show there exists r_{1}> 0 such that
B
(y,r )
1
⊆ B
(x,r)
. Define r_{1}≡ r −
|x− y|
. Then if
|z − y|
< r_{1}, it follows from the
above triangle inequality that
|z− x| = |z− y+ y − x|
≤ |z− y|+ |y − x|
< r1 + |y− x| = r − |x − y|+ |y − x| = r.
Note that if r = 0 then B
(x,r)
= ∅, the empty set. This is because if y ∈ ℝ^{n},
|x − y|
≥ 0
and so y
∕∈
B
(x,0)
. 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.
is an open set which shows, from the definition, that
D
(x,r)
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.
PICT
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 15.2.4A nonempty set A is closed if and only if wheneverx_{k}∈ A and lim_{k→∞}x_{k} = x,it follows that x ∈ A. In other words, the set is closedif 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
(x,r)
contained in A^{C}. However, this contradicts the assertion that x is
the limit of the sequence. Indeed, x_{k} must be in B
(x,r)
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
(x,r)
⊆ A^{C} for
small enough positive r? If not, then B
(x, 1)
k
contains a point of A called x_{k}
for each k = 1,2,
⋅⋅⋅
. Thus x is a limit of the sequence
{x }
k
and so x ∈ A
after all. Hence A^{C} must indeed be open and so, by definition, A is closed.
■