7 The fundamentals of Topology - Pennsylvania State University

[Pages:5]7 The fundamentals of Topology

7.1 Open and Closed Sets

Let (X, d) be a metric space.

Definition 7.1. A subset A of (X, d) is called an open set if for every x A there exists r = rx > 0 such that Brx(x) A.

Example 7.2. The open ball B(x, r) is an open set. Indeed, take any point y B(x, r) and set R := r - d(x, y) > 0. If z B(y, R), then the triangle inequality

d(x, z) d(x, y) + d(y, z) < d(x, y) + R = d(x, y) + [r - d(x, y)] = r

Example 7.3. Consider a nonempty set X with the discrete metric d(x, y) = 1 if x = y and d(x, y) if x = y. Then an open ball Br(x) is equal to {x} if r 1. Consequently, every subset A of X is open.

Note the following proposition.

Proposition 7.4. A subset A of (X, d) is open if and only if either A is empty or A is a union of open balls.

Proof. Assume that A is a nonempty open subset of X. Then for every x A, there is rx > 0 such that Brx (x) A. Consequently, A xA Brx (x) A proving our claim. Conversely, if A = , then A is trivially open. So, assume that A = and A is a union of open balls. Then if x A, there is y and r such that x Br(y) and Br(y) is one of the open balls making up the set A. By the above example there is rx > 0 such that Brx (x) Br(y) A.

Example 7.5. Let Y be a subset of (X, d). Then (Y, d) is a metric subspace of X. If x Y and r > 0, then an open ball BY (x, r) in (Y, d) is equal to BY (x, r) = BX (x, r)Y where BX (x, r) stands for an open ball in X. Now consider a subset A of Y . Then, in view of the above proposition, A is open in (Y, d) if and only if either A is empty or A = BY (x, rx) = Y BX (x, rx) = Y U where we have abbreviated U = BX (x, rx). Since U is open in X, it follows that A Y is open in (Y, d) if and only if A = Y U for some open set U in (X, d).

Example 7.6. two metric spaces (X1, d1) and (X2, d2), and let X = X1 ? X2 and

d((x1, y1), (x2, y2)) = max{d1(x1, x2), d2(y1, y2)}.

Since BX ((x, y), r) = BX1 (x, r) ? BX2 (y, r) and BX ((x, y), r) = ( BX1 (x, r)) ? ( BX2 (y, r), it follows that a subset A1 ? A2 is open in X if and only if A1 is open in X1 and A2 is open in X2.

Proposition 7.7. Let T be the collection of all open subsets of X. Then

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(i) , X T .

(ii) If Ai T , i J, then iJ Ai T .

(iii) If A1, . . . , AnT , then

n k=1

Ak

T

.

Proof. The part (ii) is a consequence of the above proposition. To see (iii), let x

1kn Ak, then there are numbers rk > 0 such that Brk (x) Ak for 1 k n. Set r = min{r1, . . . , rn}. Then Br(x) 1kn Ak.

Recall that the complement of a subset A of X is defined by Ac = X \ A.

Definition 7.8. A subset F of (X, d) is called a closed set if the complement F c of F is an open subset of X.

Example 7.9. The closed ball B(x, r) is a closed set. Indeed, if X = B(x, r) and there is nothing to prove since Xc = is open. If X = , take y X \ B(x, r).

Then = d(x, y) - r > 0. If x B(y, ), then

d(x, z) d(x, y) - d(y, z) > d(x, y) - = r

showing that B(y, c) (B(x, r))c. Since this holds for any y B(x, r), (B(x, r))c is open.

Example 7.10. Let (Y, d) be a metric subset of (X, d) and let K Y . Then K is closed in (Y, d) if Y \ K is open in (Y, d). By Example 7.5, Y \ K = Y U where U is an open subset of (X, d). Hence, using deMorgan's laws, F = Y \ (Y \ F ) = Y \ (Y U ) = (Y \ Y ) (Y \ U ) = Y \ U = Y (X \ U ) = Y F where F = X \ U is closed in (X, d). Consequently, K is closed in (Y, d) if and only if K = Y F .

From deMorgan's laws and Proposition 7.7 we deduce the following proposition.

Proposition 7.11. (i) and X are closed.

(ii) The union of a finite collection of closed sets is closed.

(iii) The intersection of arbitrary collection of closed sets is closed.

Closed sets can be characterized in terms of converging sequences.

Proposition 7.12. A set F X is closed if and only if every convergent sequence (xn) such that xn F converges to a point in F .

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Proof. Assume that F is closed and that (xn) is a sequence of points belonging to F and converging to x X. We claim that x F . Arguing by contradiction, we assume that x F c. Since F c is open, there is rx > 0 such that Brx (x) F c. Then, since xn F , d(xn, x) rx, contradicting that d(xn, x) 0. Conversely, assume that every convergent sequence (xn) such that xn F converges to a point in F but F is not closed. Then F c is not open. Hence there is x F c such that for every r > 0, Br(x) F = . In particular, taking for r numbers 1/n for n 1, we find points xn F such that d(xn, x) < 1/n. Hence the sequence (xn) converges to x and x F , contradiction.

Interior, Closure, and Boundary

Definition 7.13. Let A be a subset of a metric space (X, d) and let x0 X.

? x0 is an interior point of A if there exists rx > 0 such that Brx(x) A,

? x0 is an exterior point of A if x0 is an interior point of Ac, that is, there is rx > 0 such that Brx(x) Ac.

? x0 is an adherent point of A if x0 is not an interior point of Ac. That is Br(x) A = for every r > 0,

? x0 is boundary point of A is it is neither an interior point of A nor an interior point of Ac. That is, Br(x) A = and Br(x) Ac = for every r > 0for every r > 0

The set of all interior points of a set A is called the interior of A, written int (A) or A , the set of all adherent points of A s called the closure of A, written cl (A) or A, the set of all boundary points of A is called the boundary of A, written as A.

Example 7.14. Consider an open ball Br(x0). Then every point x Br(x0) is an interior point of Br(x). Hence Br(x0) = Br(x0). A point x satisfying d(x0, x) > r, is an exterior point of A. Every point x X such that d(x, x0) = r is a boundary point of Br(x0) so that Br(x0) = {x| d(x0, x) = r}. A point x is an adherent point of Br(x0) if either x Br(x0) or d(x0, x) = r. Hence Br(x0) Br(x0). The closed ball Br(x0 is not necessarily equal to Br(x0). To see this consider X = R equipped with the discrete metric d. If r = 1, then B1(0) = {0} and B1(0) = {0} while B1(0) = R.

Proposition 7.15. (i) A A.

(ii) A is open if and only if A = A.

(iii) A A.

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(iv) A is closed if and only if A = A.

(v) A = A X \ A.

Proof. (ii) Assume that A is open. By (i), A A and we have to show that A A. If x A, then Brx (x) A for some rx since A is open. Hence x is an interior point of A, implying that x A. Conversely, assume that A = A. This means that If x A, then x is an interior point of A so that Brx(x) A for some rx. Since x is arbitrary point of A, it follows that A is open. (iv) Let A be closed. Then Ac is open and if x Ac, then there is rx > 0 such that Brx(x) Ac implying that x is not an adherent point of A. Hence A A. By (i), A A so that A = A. Conversely, pick x Ac. Since A = A, the point x is not an adherent point of A and so, Brx (x) A = . Hence Brx (x) Ac and x is an interior point of Ac. Since x is an arbitrary point of Ac, Ac is open and hence A is closed in X.

Continuous Functions

The definition of continuity of functions between metric spaces is the - definition of calculus.

Definition 7.16. Let (X, d) and (Y, ) be metric spaces and let f : X Y be a function. The function f is said to be continuous at the point x0 X if the following holds: for every > 0, there exists > 0 such that for all xX

if d(x, x0) < , then (f (x), f (x0)) < .

The function f is said to be continuous if it is continuous at each point of X.

The following proposition rephrases the definition in terms of open balls.

Proposition 7.17. Let f : X Y be a function between metric spaces (X, d) and (Y, ) and let x0 X. Then f is continuous at x0 if and only if for every > 0 there exists > 0 such that

f (B(x0)) B(f (x0)).

Continuity can be expressed in terms of converging sequences as follows.

Theorem 7.18. Let f : X Y be a function between metric spaces (X, d) and (Y, ) and let x0 X. Then f is continuous at x0 if and only if for every sequence (xn) such that xn x0 in (X, d), f (xn) f (x0) in (Y, ).

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Proof. Suppose that f is continuous at x0 and let xn x0. We will prove that f (xn) f (x0). Let > 0 be given. By the definition of continuity at x0, there exists > 0 such that for all x X,

if d(x, x0) < , then (f (x), f (x0)) < .

(1)

Since xn x0, there exists an integer k such that for all n k,

d(xn, x0) < .

(2)

Combining (1) and (2), we get

(f (xn), f (x0)) < for all n k.

(3)

Hence f (xn) f (x0) as required. Conversely, arguing by contradiction assume that f is not continuous at x0. To obtain a contradiction we will construct a sequence (xn) such that xn x0 but the sequence {f (xn)} does not converge to f (x0). Since f is not continuous at x0, there is > 0 such that for all > 0 there exists x satisfying d(x, x0) < and (f (x), f (x0)) . Then taking = 1/n for n N, on can choose xn so that d(xn, x0) < 1/n and (f (xn), f (x0)) . Hence xn x0 but the sequence {f (xn)} does not converge to f (x0).

Recall that the preimage f -1(U ) is defined as f -1(U ) = {x X|f (x) U }

Theorem 7.19. Let f : X Y be a function from a metric space (X, d) to (Y, ). Then the following are equivalent.

(i) f is continuous.

(ii) f -1(U ) is open in X for every open subset U of Y .

(iii) f -1(F ) is closed in X for every closed subset F of Y .

Proof. (i) = (ii) Suppose f is continuous and U is open in Y . If f -1(U ) = , then the claim follows. If f -1(U ) = , let x f -1(U ). Then f (x) U . Since U is

open in Y and f (x) U , there exists a positive number such that B(f (x)) U .

Since f is continuous, there is > 0 such that f (y) B(f (x)) for every y B(x). Hence f (B(x)) B(f (x)). This implies that B(x) f -1(f (B(x))) f -1(U ) showing that f -1(U ) is open in X. (ii) = (iii) Let F be closed in Y . Then F c is open and f -1(F c) = (f -1(F ))c and since f is continuous it follows that (f -1(F ))c is open in X. Hence f -1(F ) is

closed in X as claimed. (iii) = (i) Let x X and let > 0. Since (B(f (x), ))c is closed in Y and since (ii) holds, the set f -1(B(f (x))c) is closed in X. Since f -1(Br(x)c) = (f -1(Br(x))c, the set f -1(Br(x)) is open. The point x belongs to f -1(B(f (x))) and so there exists > 0 such that B(x) f -1(B(f (x)). This implies that f (B(x))

B(f (x)), and f is continuous at x. Since x is arbitrary, f is continuous on X.

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