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Supplement 2
Metric Space and Topological Properties of Rn II
A metric space (S, d) admits a topology, a concept which we shall define next.
Topological Space: Given a set S, suppose there is a collection U of sets with the
following properties:
(i) S belongs to U
(ii) The empty set ∅ belongs to U
(iii) The union of any sub-collection of sets from U again belongs to U; that is,
U1 ⊂ U =⇒
[
∈ U.
U∈U1
(iv) If U1 , . . . , Um is any finite collection of sets in U, then
m
\
Uj ∈ U.
j=1
The members of collection U are called open sets. A pair (S, U) is called a topological
space. Any metric (S, d) admits a natural way to define open sets that generalizes the
notion of the open interval on the line: Let (S, d) be a metric space. A subset, U is open
in (S, d) if, given any point x ∈ U , there is some ball of radius r > 0 centered on x such
that
B(x, r) ⊂ U.
It is relatively easy to check conditions (i) − (iv). Item (i) holds because everything is in S;
item (ii) holds vacuously, since there are no points in ∅; and item (iii) holds, since a point
in the union must be in some particular U0 ∈ U and the r that works for that U0 , works
Tm
for the union (B(x, r) ⊂ U0 ⊆ ∪U1 U ). Finally, for item (iv), assume x ∈ j=1 Uj for open
sets U1 , . . . , Uj . Then for each j, j = 1, . . . , m, there is an rj > 0 for which B(x, rj ) ⊂ Uj .
Let rmin = min1≤j≤m rj , then
B(x, rmin ) ⊂ B(x, rj ) ⊂ Uj ,
j = 1, . . . , m =⇒ B(x, rmin ) ⊂
m
\
j=1
1
Uj .
Another commonly used term is that of a neighborhood of a point. A set E is a neighborhood of the point x0 ∈ S, if E contains a ball B(x0 , r) centered on x0 . Clearly, open
sets are neighborhoods of x0 , but so are many other types of sets containing x0 .
Examples: We take our examples in Rn for various n
1. In R, open intervals (a, b) are open sets relative to (R, d), d(x, y) = |x − y|. In Rn ,
open intervals are of the form of “rectangles”
n
Y
n
o
(aj , bj ) := x : aj < xj < bj , j = 1, . . . , n .
j=1
2. Let f be a real-valued function on (a, b). The graph of f ,
Gf := {(x, f (x)) : a < x < b}
is NOT open in (R2 , d). Indeed, any ball B((x0 , f (x0 )), r), a < x0 < b, will contain
the vertical line segment {(x0 , y) : f (x0 ) − r < y < f (x0 ) + r}, and only the point
(x0 , f (x0 )) on that line segment is in Gf (a function takes on a one and only one value
at a point).
3. Any of the balls B(Rn ,d) (x0 , r), B(Rn ,d1 ) (x0 , r), and B(Rn ,d∞ ) (x0 , r), defined using the
metrics from Supplement 1 on Rn are open in (Rn , d), d(x, y) = |x − y|. This follows
from the inequalities relating these metrics.
4. The set {(x1 , x2 ) : x21 + x22 < 1} ∪ {(1, 0)} is NOT open in (R2 , d), d(x, y) = |x − y|.
Here we have appended a point of distance exactly 1 to the ball of radius 1 about the
origin in R2 . Any ball B((1, 0), r), will contain points (1 + x, 0) with 0 < x < r, which
are distance greater than 1 from the origin.
The last example motivates the definition: Given a set E ⊂ S, the interior of E,
denoted E 0 , is defined to be the set
n
x ∈ S : B(x, r) ⊂ E,
o
for some r > 0 .
Another way of looking at this is that
E0 =
[n
o
U ∈U :U ⊂E ,
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the union of all open sets contained in E. (Why?) By its very definition, E 0 is open in
(S, d) and E 0 ⊆ E. If E is open, then clearly E 0 = E.
The notion of “closed” interval also carries over to this setting with the following
definition: A subset F of S is defined to be closed in (S, d) if it is the complement of an
open set. That is,
F
is closed in (S, d) ⇐⇒ F = S\U,
U ∈ U (U is open).
By the basic properties of complementation in set theory, the collection of all closed sets
F in (S, d) satisfies:
(i) S belongs to F
(ii) ∅ belongs to F
(iii) The intersection of the sets from any sub-collection F1 of F belongs to F ; that is
\
F ∈F
if F1 ⊂ F .
F ∈F1
(iv) If F1 , . . . , Fm belong to F , then
Sm
j=1
Fj ∈ F .
The closure of E, denoted E − , is is defined to be the intersection of all closed sets
containing E. The boundary of E, denoted ∂E, is the set E − \E 0 . Points in ∂E are
called the boundary points of E.
Examples: Again, we take examples in Rn :
1. The closed interval [a, b] in R is closed in this sense, as are the intervals (n-cubes,
n-cells) in Rn
n
Y
n
o
[aj , bj ] := x : aj ≤ xj ≤ bj , j = 1, . . . , n .
j=1
The boundary points of [a, b] are the endpoints, and the boundary points of
are the “faces” of the n-cells
n
x∈
n
Y
o
[aj , bj ] : at least one xj equals aj or bj .
j=1
2. The closed balls
B(x, r)− := {y : d(x, y) ≤ r}
3
Qn
j=1 [aj , bj ]
are closed in Rn .
3. Any finite collection of points, E = {x(1) , . . . , x(m) } is closed in in Rn .
4. Let E = (x(k) ), be the points of a convergent sequence in Rn . Then E is neither
open nor closed. E has no interior, i.e., E 0 = ∅. E − = E ∪ {x∗ } where x(k) → x∗ as
k → ∞.
The last example again motivates a definition. Let E be a subset of the metric space
(S, d). A point p is said to be a limit point (cluster point) of E if any neighborhood
of p contains a point from E\{p}. In other words, any ball centered on p with positive
radius, no matter how small, will contain a point from the set E which is not its center.
The following Proposition gives some of the basic relations of limit points, the closure
of a set, and the boundary of a set:
Proposition S.2.1. Let E be a subset of the metric space (S, d).
(a) E is closed if and only if E = E − .
(b) The set E is closed if and only if it contains all of its cluster/limit points.
(c) The set E is closed if and only if it contains the limit of every convergent sequence of
points from E.
(d) If p ∈ E − \E, then p must be a cluster point of E.
(e) A point belongs to E − if and only if it is the limit of some sequence of points from E.
(f) A point is in the boundary of E if and only if it belongs to the closure of both E and
its complement S\E.
Proof. (a) If E is closed, then it is a closed set containing itself and so E − ⊆ E. But the
reverse conclusion holds by definition of intersection (that is, we always have E ⊆ E − ).
On the other hand, since E − is the intersection of closed sets, it is closed. Hence E is
closed if E = E − .
(b) Suppose E is closed. Let p be a cluster point of E. If p 6∈ E, then p ∈ S\E
which is open. Hence, there is a r > 0 such that B(p, r) ⊂ S\E. But, this contradicts the
definition of a cluster point. On the other hand, if E contains all of its cluster points, let
p ∈ S\E. Then, p cannot be a cluster point of E, so there is some ball B(p, r0 ) that does
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not contain any points of E. That means, B(p, r0 ) ⊂ S\E. Thus, S\E is open, and hence
E is closed.
(c) This will follow from (b) and the following fact:
Lemma S.2.2. A point p is a cluster point of E if and only if it is the limit of a sequence
of points from E.
Proof of Lemma. For each k, consider the ball Bk := B(p, 1/k). If p is a cluster point
of E, then there is at least one point from E in Bk \{p}. Choose one and label it x(k) .
Then, clearly (x(k) ) is a sequence in E which must converge to p. On the other hand, if
p is the limit of the sequence x(k) in E, then in any of the balls B(p, ǫ), ǫ > 0, there are
infinitely many points from the sequence (all those with index k > Nǫ ).
(d) If p ∈ E − \E, and p is not a cluster point of E, then there is an r0 so that
B(p, r0 ) contains no point of E. But then S\B(p, r0 ) is a closed set containing E, and so
E − ⊆ S\B(p, r0 ). This contradicts the fact that p ∈ E − .
(e) If p ∈ E, then we take x(k) = p for each k (the constant sequence). If p ∈ E − \E,
then (d) says that p is a cluster point of E and (e) follows from the lemma.
(f) The proof of (f) will be an HW exercise.
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