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Solutions to Topology Homework #4, due Week 8.

Problems: Munkres Section 17 #4, 7, 10, 14, 16, 19, 20

17.4 Show that if U is open in X and A is closed in X, then U A is openin X, and A U is closed in X.

Proof: Let U be open in X and A be closed in X. Then X A isopen since complements of closed sets are open, and X U is closed sincecomplements of open sets are closed. Furthermore, U A = U

(X A) is a

finite intersection of open sets and hence is open. Also, A U = A

(XU)is the intersection of closed sets and hence is closed.2

17.7 Criticize the following proof that

A

A: if{A} is a collection

of sets in X and ifx

A, then every neighborhood U ofx intersects

A.Thus U must intersect some A, so that x must belong to the closure of someA. Therefore, x

A.

The problem lies in the sentence Thus U must intersect some A, sothat x must belong to the closure of some A. Its true that each U inter-sects some A, but nothing says that every such U intersects the same A.Therefore, there may not be a single A which has x as its limit point. (Asimple example is Ai = {1/i} for i N.)

17.10 Show that every order topology is Hausdorff.Proof: Let X be a topological space with order topology T. Let x, y X

be such that x < y . First suppose that neither of these points is the greatestor least element in X. Now in the case where y is the immediate successorof x, then we have x (a, y) for some a < x, and clearly this interval isan open set not containing y. Furthermore, y (x, b) for some b > y, and(a, y)

(x, b) = since y is the immediate successor of x. On the other

hand, if y is not the immediate successor of x then there is some z Xsuch that x < z < y. In that case, (a, z) contains x and (z, b) contains yand (a, z)

(z, b) is disjoint. We now turn to the case where x is the least

element in X. In that case we follow the same reasoning using [x, z) or [x, y)in place of (a, z) or (a, y). The case in which y is the greatest element inX is similar. In all cases we find that there is an open set containing x but

not y and an open set containing y but not x, such that these open sets aredisjoint. Hence X is Hausdorff.

17.14 In the finite complement topology on R, to what point or points doesthe sequence xn = 1/n converge?

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Since (xn) is a sequence with infinitely many distinct points, any open

set in the finite complement topology will contain some points of (xn). Itfollows that (xn) converges to every point in R.

17.16 Consider the five topologies on R given in Exercise 7 of Section 13.(a) Determine the closure of the set K = {1/n|n Z+}.In T1, the standard topology, K = K

{0}.

In T2, the topology ofRK, notice that K has no limit points, since R Kis an open set which doesnt contain K (and hence there is no limit pointof K outside of K), and also every 1/n in K clearly has a small interval ofradius (1/n) (1/(n + 1)) around it which does not intersect K in any otherpoint. Thus, K = K.

In T3, the finite complement topology, we have just shown in the previousproblem that every point in R is a limit point of K, and hence K = R.

In T4, the upper limit topology, notice that (1, 0] is an open set con-taining 0 but no point of K, hence 0 is not a limit point of K. No otherpoint is a limit point of K for the same reason as in the standard topology,so K = K.

In T5, with basis {(, a)}, notice that every point x greater than orequal to 0 will have the property that any open set containing x will alsocontain some point ofK, and hence K is the set of real numbers greater thanor equal to 0.

(b) Which of these topologies satisfies the Hausdorff axiom? The T1axiom?

Clearly T3 and T5 are not Hausdorff since they contain a sequence thatconverges to more than one limit point. Furthermore, T3 is T1 but T5 isnot since finite point sets have open complements in T3 but not T5. Thestandard topology, is both T1 and Hausdorff since we can find disjoint openintervals separating any two reals. Since T4 and T4 are finer than the standardtopology, they have both properties too.

17.19 If A X, we define the boundary of A by the equation BdA =A

(X A).(a) Show that IntA and BdA are disjoint, and A = IntA

BdA.

Proof: From the definition of boundary, and the fact that A itself mustbe disjoint from its complement in X, we can conclude that the boundary ofA is the set of points which are limit points of A and X A. But if y is in

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IntA then there is some neighborhood ofy contained entirely in A and hence

y is not a limit point of X A. This shows that IntA and BdA are disjoint.Furthermore, for any x A such that x is not in the interior of A, thenevery neighborhood ofx must intersect X A and hence x is a limit point ofX A and hence an element of BdA. This shows that A IntA

BdA. The

reverse inclusion is clear from the fact that both IntA and BdA are subsetsof A. 2

(b) Show that BdA = A is both open and closed.Proof: BdA = means that no point is a limit point of both A and XA.

That is, every point p has a neighborhood Np contained entirely within A orentirely within X A. In that case, A =

pA Np and X A =

pA Np

so A and X A are both open and hence both closed. Conversely, if A and

X A are both open and both closed, then every point in A and every pointin X A are in the interior ofA or X A. This means that the union of theinteriors is the entire space X and consequently by part (A), and the factthat the boundary of A equals the boundary ofX A we can conclude thatthe boundary is empty.2

(c) Show that U is open BdU = U U.Proof: U is open exactly when U is its own interior, in which case by

(a) we have U and BdU disjoint, and U = U

BdU. Since this is a disjointunion, it follows that BdU = U U.2

(d) If U is open, is it true that U =Int(U)? Justify your answer.

This is not necessarily true. Suppose U =R

{0} in the standardtopology on R. Then U is open, and U = R, with IntU also equal to all ofR.

17.20 Find the boundary and the interior of each of the following subsets ofR2:

(a) A = {x y|y = 0}. BdA = A, IntA = .(b) B = {xy|x > 0, y = 0. BdB = {xy|x = 0}

{xy|x > 0, y = 0},

IntB = B.(c) C = A

B. BdC = {x y|x < 0, y = 0}

{x y|x = 0}, IntC =

{x y|x > 0}

(d) D = {x y|x Q}. BdD = R2

. IntD = .(e) E = {x y|0 < x2 y2 1}. BdE = {x y|x2 y2 = 0}

{x y|x2 y2 = 1}. IntE = {x y|0 < x2 y2 < 1}.

(f) F = {x y|x = 0, y 1/x}. BdF = {x y|x = 0}

{x y|y = 1/x}.IntF = {x y|x = 0, y < 1/x}.

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