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(MATH323)2002_s_midall.pdf
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Math 323 Midterm Exam, Spring 2002
(1)
(25 points) Find a CW-structure of the punctured torus with two vertices (0-cells). Then compute its euler number. (Hint: Present your answer in the square with four sides appropriately identi.ed)

(2)
(25 points) Which of the following are closed? No reason needed.

1.
{n.1}. R, with respect to B = {[a, b)}.

2.
{n.1}. R, with respect to B = {(a, b]}.

3.
{f C[0, 1] : f(0) f(1)}. C[0, 1], with L1-topology.

4.
{f C[0, 1] : f(0) f(1)}. C[0, 1], with pointwise convergence topology.

5.
{n by n matrix A : all eigenvalues of A are 1}. Rn 2 , with the usual topology.



(3)
(25 points) Let T = {(.a, 2a) : all a> 0} {., R}, B = {[.a, 2a) : all a> 0}.

1.
Prove that T is a topology and B is a topological basis on R. Explain why B {., R} is not a topology.

2.
Find the limit points and the closures of {0}, {1}, {.1} with respect to T and B. No explanation needed.

3.
Compare T and the topology induced by B. Explain.



(4)
(25 points) Let X be a topological space in which any single point is closed. Prove that (A.). . A. (the limit of limit is limit). Moreover, .nd a counterexample showing the necessity of the condition that any single point is closed.


limit in T limit in B closure in T closure in B
{0}
{1}
{.1}

(Hint: If U is open, then U . x is also open. You should be able to construct a counterexample in which the space X consists of two points.)
Answer to Math 323 midterm, Spring 2002
(1)
The following is a CW structure with 0-cells V , W ; 1-cells a, b, c, d; and one 2-cell (given by the shade). The Euler number is 2 . 4+1= .1.

(2)
The limit points of {n.1} with respect to B = {[a, b)} is {0}, which is not contained in {n.1}. Therefore it


is not closed.
{n.1} has no limit points with respect to B = {(a, b]}. Therefore it is closed.
By making adjustment near 0 and 1, we see that any continuous function is an L1-limit point of {f C[0, 1] :
f(0) f(1)}. Therefore the subset is not L1-closed. {f C[0, 1] : f(0) f(1)} is closed in the pointwise convergence topology because the complement U = {f C[0, 1] : f(0) <f(1)} is open. Speci.cally, if f(0) <f(1), then any function g su.ciently close to f at 0
f(0).f(1)
and 1 (say, within . = ) also satis.es g(0) <g(1). In math symbols, we have
2
f(0) . f(1)
f U =. B(f(0),f(1); 0, 1; ) . U.
2
An n by n matrix A has all eigenvalues equal to 1 means det(A . tI)=(t . 1)n . Note that the n +1 coe.cients of the degree n polynomial det(A . tI) are continuous functions of the entries of A. For example, for 2 by 2 case, we have
A = ab . det(A . tI)=(ad . bc) . (a + d)t + t2 . (A)=(ad . bc, .a . d, 1).
cd
Rn+1
and : Rn 2 is continuous. Thus the subset is .1(coe.cients of (t . 1)n) (for 2 by 2 case, this is .1(1, .2, 1)), and is closed.
(3) . and R have been included in T . The union i(.ai, 2ai)=(.a, 2a), wher