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 8.8.1Recall that forx,y∈ ℝ^{n},
(∑n )1∕2
|x − y| = |xi − yi|2 .
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 that B
(x,r)
⊆ U. Moregenerally, if U is any subset of ℝ^{n}, x ∈ U is aninterior point of U if there exists r > 0 such thatx ∈ B
(x,r)
⊆ U. In other words U is an open set exactly when every point of U is an interior point ofU.
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 8.8.2A subset, C, of ℝ^{n}is called aclosed set if ℝ^{n}∖ C is an open set. They symbolℝ^{n}∖ C denotes everything in ℝ^{n}which is not in C. It is also called thecomplement of C. Thesymbol 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.
It is roughly the case that 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 8.8.3Let 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,r1)
⊆ 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
Recall ℝ^{2} consists of ordered pairs
(x,y)
such that x ∈ ℝ and y ∈ ℝ. ℝ^{2} is also written as ℝ × ℝ. In
general, the following definition holds.
Definition 8.8.4The Cartesian product of two sets A × B, means
{(a,b) : a ∈ A,b ∈ B }.
If you have n sets A_{1},A_{2},
⋅⋅⋅
,A_{n}
∏n
Ai = {(x1,x2,⋅⋅⋅,xn ) : each xi ∈ Ai}.
i=1
Now suppose A ⊆ ℝ^{m} and B ⊆ ℝ^{n}. Then if
(x,y)
∈ A×B, x =
(x1,⋅⋅⋅,xm)
and y =
(y1,⋅⋅⋅,yn)
, the
following identification will be made.
(x,y) = (x1,⋅⋅⋅,xm, y1,⋅⋅⋅,yn) ∈ ℝn+m.
Similarly, starting with something in ℝ^{n+m}, you can write it in the form
(x,y)
where x ∈ ℝ^{m} and y ∈ ℝ^{n}.
The following theorem has to do with the Cartesian product of two closed sets or two open sets. Also here
is an important definition.
Definition 8.8.5A set, A ⊆ ℝ^{n}is said to be bounded if thereexist finite intervals,
[ai,bi]
suchthat
∏n
A ⊆ [ai,bi].
i=1
Theorem 8.8.6Let U be an open set in ℝ^{m}and let V be an open set in ℝ^{n}. Then U × V is anopen set in ℝ^{n+m}. If C is a closed set in ℝ^{m}and H is a closed set in ℝ^{n}, then C × H is a closedset in ℝ^{n+m}. If C and H are bounded, then so is C × H.
Proof: Let
(x,y)
∈ U × V . Since U is open, there exists r_{1}> 0 such that B
(x,r1)
⊆ U. Similarly,
there exists r_{2}> 0 such that B
(y,r2)
⊆ V . Now
B ((x,y),δ) ≡
( m n )
{(s,t) ∈ ℝn+m :∑ |x − s |2 + ∑ |y − t |2 < δ2}
( k=1 k k j=1 j j )
Therefore, if δ ≡ min
(r1,r2)
and
(s,t)
∈ B
((x,y),δ)
, then it follows that s ∈ B
(x,r1)
⊆ U
and that t ∈ B
(y,r2)
⊆ V which shows that B
((x,y),δ)
⊆ U × V . Hence U × V is open as
claimed.
Next suppose
(x,y)
∕∈
C × H. It is necessary to show there exists δ > 0 such that
B
((x,y),δ)
⊆ ℝ^{n+m}∖
(C × H )
. Either x
∕∈
C or y
∕∈
H since otherwise
(x,y)
would be a point of C ×H.
Suppose therefore, that x
∕∈
C. Since C is closed, there exists r > 0 such that B
(x,r)
⊆ ℝ^{m}∖C. Consider
B
((x,y),r)
. If
(s,t)
∈ B
((x,y),r)
, it follows that s ∈ B
(x,r)
which is contained in ℝ^{m}∖ C.
Therefore, B
((x,y),r)
⊆ ℝ^{n+m}∖
(C × H)
showing C × H is closed. A similar argument holds if
y