Lecture 7: point-set topology We first finish properties of Euclidean ...

Lecture 7: point-set topology
We first finish properties of Euclidean space from last time.
Theorem 0.1. Suppose a, b,c ∈ R n and c ∈ R. Then
1. |a| ≥ 0 with |a| = 0 if and only if a = 0.
2. |ca| = |c||a|.
3. (Cauchy-Schwarz inequality) |a · b| ≤ |a|| b|.
4. (Triangle inequality) |a + b| ≤ |a| + | b|.
5. |a − b| ≤ |a − c| + |c − b|.
Proof. The first two follow easily; for instance since a 2 ≥ 0 for all a ∈ R (this is actually
part of the definition of ordered field), we get a 2 1 + · · · + a 2 n ≥ 0 and therefore |a| ≥ 0. If
|a| = 0 then by uniqueness of square roots, a 2 1 + · · · + a 2 n = 0 and so 0 ≥ a 2 i for all i, giving
ai = 0 for all i.
For the third item, we first give a lemma.
Lemma 0.2. If ax 2 + bx + c ≥ 0 for all x ∈ R then b 2 ≤ 4ac.
Proof. If a = 0 then bx ≥ −c for all x. Then we claim b must be zero. If not, then plugging
in either 2c/b or −2c/b will give bx < −c, a contradiction. Therefore is a = 0 we must have
b = 0 and therefore b 2 ≤ 4ac as claimed.
Otherwise a = 0. First assume that a > 0. Plug in x = −b/(2a) to get
−b 2 /(4a) + c ≥ 0
giving b 2 ≤ 4ac. Last, if a < 0 then we have (−a)x 2 + (−b)x + (−c) ≥ 0 and applying what
we have proved already to this polynomial, we find (−b) 2 ≤ 4(−a)(−c), or b 2 ≤ 4ac.
To prove Cauchy-Schwarz, note that for all x ∈ R,
0 ≤ (a1x − b1) 2 + · · · + (anx − bn) 2
= (a 2 1 + · · · + a 2 n)x 2 − 2(a1b1 + · · · + anbn)x + (b 2 1 + · · · + b 2 n)
= |a| 2 x 2 − 2(a · b)x + | b| 2 .
So using the lemma, (a · b) 2 ≤ |a| 2 | b| 2 .
The last two items follow directly from the Cauchy-Schwarz inequality. Indeed,
|a + b| 2 = (a · b)(a · b)
= a · a + 2a · b + b · b
≤ |a| 2 + 2|a|| b| + | b| 2
= (|a| + | b|) 2 .
The last inequality follows by taking a − c and c − b in the previous.
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This leads us to the definition of a metric space.
Definition 0.3. A set X with a function d : X × X → R is a metric space if for all
x, y, z ∈ X,
1. d(x, y) ≥ 0 and equals 0 if and only if x = y and
2. d(x, y) ≤ d(x, z) + d(z, y).
Then we call d a metric.
A useful example of a metric space is R n with metric d(a, b) = |a − b|.
Open and closed sets: beginning chapter 2 of Rudin
Let (X, d) be a metric space. We are interested in the possible subsets of X and in what
ways we can describe these using the metric d. Let’s start with the simplest.
Definition 0.4. Let r > 0. The neighborhood of radius r centered at x ∈ X is the set
For example,
Br(x) = {y ∈ X : d(x, y) < r}
1. in R using the metric d(x, y) = |x − y| we have the open interval
Br(x) = (x − r, x + r) = {y ∈ R : x − r < y < x + r} .
2. In R n using the metric d(x, y) = |x − y| we have the open ball
Br(x) = {(y1, . . . , yn) : (x1 − y1) 2 + · · · + (xn − yn) 2 < r 2 } .
To describe that these sets appear to be open (that is, no point is on the boundary), we
introduce a formal definition of open.
Definition 0.5. Let (X, d) be a metric space. A set Y ⊂ X is open if for each y ∈ Y there
exists r > 0 such that Br(y) ⊂ Y .
For each point y we must be able to fit a (possibly tiny) neighborhood around y so that
it still stays in the set Y . Thinking of Y as, for example, an open ball in R n , as our point
y approaches the boundary of this set, the radius we take for the neighborhood around this
point will have to decrease.
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Proposition 0.6. Any neighborhood is open.
Proof. Let x ∈ X and r > 0. To show that Br(x) is open we must choose y ∈ Br(x) and
show that there exists some s > 0 such that Bs(y) ⊂ Br(x). The radius s will depend on
how close y is to the boundary. Therefore, choose
s = r − d(x, y) .
To show that for this s, we have Bs(y) ⊂ Br(x) we take z ∈ Bs(y). Then
Some more examples:
d(x, z) ≤ d(x, y) + d(y, z) < d(x, y) + s = r .
1. In R, the only intervals that are open are the (surprise!) open intervals. For instance,
let’s consider the half-open interval (0, 1] = {x ∈ R : 0 < x ≤ 1}. If it were open, we
would be able to, given any x ∈ (0, 1], find r > 0 such that Br(x) ⊂ (0, 1]. But clearly
this is false because Br(1) contains 1 + r/2.
2. In R 2 , the set {(x, y) : y > 0} ∪ {(x, y) : y < −1} is open.
3. In R 3 , the set {(x, y, z) : y > 0} ∪ {(0, 0, 0)} is not open.
Proposition 0.7. Let C be a collection of open sets.
1. The union
O∈C O is open.
2. If C is finite then
O∈C O is open. This need not be true if the collection is infinite.
Proof. Let x ∈ ∪O∈CO. Then there exists O ∈ C such that x ∈ O. Since O is open, there
exists r > 0 such that Br(x) ⊂ O. This is also a subset of ∪O∈CO so this set is open.
To show that we cannot allow infinite intersections, consider the sets (−1/n, 1/n) in R.
We have
∩ ∞ n=1(−1/n, 1 + 1/n) = [0, 1] ,
which is not open (under the usual metric of R).
For finite intersections, let O1, . . . , On be the open sets from C and x ∈ ∩ n i=1Oi. Then
for each i, we have x ∈ Oi and therefore there exists ri > 0 such that Bri (x) ⊂ Oi. Letting
r = min{r1, . . . , rn}, we have Br(x) ⊂ Bri (x) for all i and therefore Br(x) ⊂ Oi for all i.
This implies Br(x) is a subset of the intersection and we are done.
Definition 0.8. An interior point of Y ⊂ X is a point y ∈ Y such that there exists r > 0
with Br(y) ⊂ Y . Write Y ◦ for the set of interior points of Y .
Directly by definition, Y is open if and only if Y = Y ◦ .
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Examples:
1. The set of interior points of [0, 1] (under the usual metric) is (0, 1).
2. The set of interior points of
is just {(x, y) : y > 0}.
3. What is the set of interior points of Q?
{(x, y) : y > 0} ∪ {(x, y) : x = −1, y ≥ 0}
4. Define a metric on R 2 by d(x, y) = 1 if x = y and 0 otherwise. This can be shown to
be a metric. Given a set Y ⊂ R 2 , what is Y ◦ ?
Definition 0.9. A set Y ⊂ X is closed if its complement X \ Y is open.
Sets can be both open and closed. Consider ∅, whose complement is clearly open, making
∅ closed. It is also open.
Proposition 0.10. Let C be a collection of closed sets.
1.
C∈C C is closed.
2. If C is finite then
C∈C C is closed.
Proof. Just use X \ [∩C∈CC] = ∪C∈C(X \ C).
There is an alternative characterization of closed sets in terms of limit points
Definition 0.11. Let Y ⊂ X. A point x ∈ x is a limit point of Y if for each r > 0 there
exists y ∈ Y such that y = x and y ∈ Br(x).
Examples:
1. 0 is a limit point of {1, 1/2, 1/3, . . .}.
2. {1, 2, 3} has no limit points.
3. In R 2 , B1(0) ∪ {(0, y) : y ∈ R ∪ {(10, 10)} has limit points
{(x, y) : x 2 + y 2 ≤ 1} ∪ {(0, y) : y ∈ R} .
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