### WebCT Test Questions

Find the flaw in the following proof that all horses are the same color (you may choose more than one answer - this question will be graded on an all or nothing scale):

We proceed by induction. Let S={n in N | in any set of n horses, all members are the same color}. Since in any set of 1 horse, all members are the same color, we know that 1 is in S. For the induction step, let k be an element of N and assume that k is an element of S. We must show that k+1 is an element of S, ie in any set of k+1 horses, all members are the same color. Let A be a set of k+1 horses, call them H1,...,H(k+1). We will show that they all have the same color. Since {H2,...,H(k+1)} is a set of k horses, and k is in S, we know that H2,...,H(k+1) all have the same color. Since {H1,...,Hk} is a set of k horses, and k is in S, we also know that H1,...,Hk all have the same color. Thus H1 and H2 are the same color, and hence H1,...,H(k+1) are the same color. Therefore, k+1 is an element of S. Therefore S=N, ie all horses are the same color.

 1 The flaw is in the part of the proof showing 1 is an element of S 2 The flaw is in the setup of the induction step 3 The flaw is in the statement Since {H2,...,H(k+1)} is a set of k horses, and k is in S, we know that H2,...,H(k+1) all have the same color. 4 The flaw is in the statement Since {H1,...,Hk} is a set of k horses, and k is in S, we also know that H1,...,Hk all have the same color. 5 The flaw is in the statement Thus H1 and H2 are the same color, and hence H1,...,H(k+1) are the same color. Therefore, k+1 is an element of S.

According to the poster on my office door, can an orange peel be flattened?

 1 No, if you try it at home, you will see that it can not be done. 2 Sometimes it can, depending on what kind of orange you use. 3 Yes it can, if it is cut and stretched.

In geometry, two things are the same if they match up when they are placed on top of each other.
In projective geometry, two things are the same if they look the same from different perspectives.
In topology, two things are the same if they can be continuously deformed into one another.

Which of the following are topologically equivalent to a donut?

 1 mug with one handle 2 mug with two handles 3 iron 4 ball 5 vest

Notice that in the following statements on the left, the type and order of quantifiers matters very much since switching them around results in a statement that is logically very different than the original.

Match the statement to the proof that either proves or disproves it.

 1. For all x in R\{0}, there exists y in R s.t. xy=1 a. Let x be an arbitrary element of R\{0}. We will produce y in R s.t. xy is not 1. Take y=0. Notice that y is in R. Also, xy=x(0)=0 by multiplication by 0. Since 0 is not equal to 1, we have shown xy is not equal to 1, as desired. 2. There exists y in R s.t. for all x in R\{0}, xy=1 b. Let x be an arbitrary element of R\{0}. We will produce y in R s.t. xy=1. Take y=1/x. Notice that 1/x is in R since x is not 0 by def. Also, xy=x(1/x)=x/x=1, as desired. 3. There exists x in R\{0} s.t. for all y in R, xy=1 c. Let y in R be arbitrary. We will produce x in R\{0} s.t. xy is not 1.Case 1 y not 0Take x=2/y. Notice that x is in R\{0} since y is not equal to 0. Also, xy=2/y(y)=2(y/y)=2(1)=2. Since 2 is not equal to 1, we have shown that xy is not equal to 1. Case 2 y=0 Take x=3. Notice that x is in R\{0} since 3 is in R\{0}. Also, xy=3(0)=0. Since 0 is not equal to 1, we have shown that xy is not equal to 1. So, for each y in R, we have produced x in R\{0} so that xy is not equal to 1, as desired. 4. For all y in R, there exists x in R\{0} s.t. xy=1 d. We will produce y in R s.t. for all x in R\{0}, xy is not equal to 1. Take y=0. Notice that y is in R. Let x be in R\{0}. We will show that xy is not equal to 1. Notice that xy=x(0)=0, by multiplication by zero. Since 0 is not equal to 1, we have shown that xy is not equal to 1, as desired.
 1 --> Choose Match abcd 2 --> Choose Match abcd 3 --> Choose Match abcd 4 --> Choose Match abcd

Match the the following definitions

 1. f is a function if for all x, there exists y s.t. f(x)=y, and a. for all x1 and x2, f(x1)=f(x2)--->x1=x2 2. f is one-to-one if b. for all x1 and x2, x1=x2 --->f(x1)=f(x2) 3. f is onto if c. for all y, there exists x s.t. f(x)=y
 1 --> Choose Match abc 2 --> Choose Match abc 3 --> Choose Match abc

I9ve added an eighth bridge to the town of Konigsberg (now Kaliningrad, in Russia) as you can see in the following picture. Which of the following are true? 1 If I take away the bridge that I added (on the far right), then it is impossible to find a continuous path that crosses each bridge exactly once. 2 With the picture as is, I can prove that there exists a continuous path that crosses each bridge exactly once by physically producing such a path. 3 With the picture as is, I can prove that there exists a continuous path that crosses each bridge exactly once without actually physically producing such a path. 4 We can prove that there exists a subset of seven of the eight bridges so that a single continuous path may traverse all seven of those bridges exactly once. 5 We can't prove that there exists a subset of seven of the eight bridges so that a single continuous path may traverse all seven of those bridges exactly once because it is impossible.

What is the negation of the definition of f is continuous at xo

(For all E>0, there exists D>0 s.t. |x-xo|<D ---> |f(x)-f(xo)|<E for all x)?

 1 (For all E>0 there exists D>0 s.t. |x-xo|E for all x) 2 (There exists E>0 s.t. for all D>0, there exists x s.t. |x-xo|E) 3 (There exists E>0 s.t. for all D>0, there exists x s.t. |x-xo|>D and |f(x)-f(xo)|>E) 4 (There exists E>0 s.t. for all D>0, there exists x s.t. |x-xo| |f(x)-f(xo)|>E) 5 (For all E<0, there exits D<0 s.t. |x-xo| |f(x)-f(xo)|>E for all x)

Which of the following are true?

 1 One of the great skills mathematicians develop is to break problems up into smaller pieces that are easier to work on. In topology we do the same thing - lots of times it is hard to prove something about a big space, and so we take a little snapshot of it and analyze things locally (via open sets). 2 While a metric space is a space that has a notion of distance on it, a topological space is more general. 3 While the intersection of 2 topologies is a topology, it is not always true that the union of 2 topologies is a topology. In fact, we can find one example where 2) in the def of topology does not hold for the union of 2 topologies, and we can find another example where 3) in the def of topology does not hold for the union of 2 topologies. 4 The standard euclidean metric and the taxicab metric both give metrics on R^2. In addition, while SAS is true in R^2 with the euclidean metric, it is not true in R^2 with taxicab metric. 5 The metric on the upper half-plane induced as a subspace of R^2 is very different from the metric in the upper half-plane given as the hyperbolic metric on R^2. 6 The topology on the upper half-plane induced as a subspace of R^2 is very different from the topology in the upper half-plane induced from the hyperbolic metric on R^2 (where opens are hyperbolic epsilon balls (ie d < epsilon) ).

Recall that a basis is a shorthand way to write a topology. Once we have a basis, we look at the toplogy generated by this bases by looking at the set of arbitrary unions of basic elements.
{(a,b) | a, b are real and a < b} is a basis for the standard topology on R (call this R).
{[a,b) | a, b are real and a < b} is a basis for the lower limit topology on R (call this R_l).
{[a,b) | a, b are rational and a < b} is a basis for some topology on R (call it R_lQ).
Which of the following are true?

 1 The following argument demonstrates that R_l is not contained in R . Look at 0 in [0,1), which is a basis element of R_l. For any basis element in R, say (a,b), if 0 is in (a,b) then a<0. Hence a/2 is in (a,b), but a/2 < 0, and so a/2 is not in [0,1). Thus (a,b) is not contained in [0,1), and so R_l is not contained in R. 2 The following argument demonstrates that R_l is not contained in R_lQ . Look at Pi in [Pi,4), which is a basis element of R_l . For any basis element in R_lQ , say [a,b), where a and b are rational, if Pi is in (a,b) then a < Pi. Hence (Pi + a)/2 is in (a,b), but (Pi + a)/2 < Pi, and so (Pi + a)/2 is not in [Pi,4). Thus (a,b) is not contained in [Pi,4), and so R_l is not contained in R_lQ . 3 R is (strictly) contained in R_lQ , which is (strictly) contained in R_l . 4 {(a,infinity) | a is real} is a basis for a topology on R, and the topology generated by this basis is equal to the standard topology on R (ie R is contained in it and it is contained in R ). 5 {[a,b) | a, b are real and a < b} Union {(a,b] | a, b are real and a < b} is a subbasis for a topology on R, but not a basis for a topology on R. The topology it generates is formed from taking arbitrary unions of finite intersections of subbasis elements. This topology is the discrete topology. 6 {x | x is real} is a basis for the discrete topology, and so it is also a subbasis for it. 7 While the last two "answers" both are subbases for the same topology on R, the discrete topology, it is not always true that for each subbasis element S1 of one set, and x in S1, I can find a subbasis element S2 of the other set such that x in S2 contained in S1.

Which of the following are true?

 1 An open set is an element of a topology. 2 If (X,T) is a topology with a basis, then an open set U is a set so that for each element x of U, there is a basis element B_x so that x is in B_x which is inside of U. Since the arbitrary union of open basic sets is open, we can think of an open set as the union of all B_x, and in this way think of U as being made up of snapshots of basic opens. 3 An open set is a set which is not closed. 4 U contained in R (standard top) is open iff for every x in U, there exists an E bigger than 0 so that (x-E, x+E) is contained in U. 5 If (X,d) is a metric space, then Bd(x,E) is open for every E>0. 6 If (X,d) is any metric space and and x is in X, then Bd(x,0) is not open.

Which of the following are true?

 1 A closed set C is a set whose complement X-C is open. 2 The arbitrary union of closed sets is closed. 3 The arbitrary intersection of closed sets is closed. 4 If (X,d) is a metric space, and D < E, then the closure of B_d(x,D) is contained in B_d(x,E). 5 If X is a topological space, A is contained in X, and there exists there exists a sequence of points of A that converge to x, then x is in the closure of A. If X is metrizable, then the converse holds.

Which of the following are true?

 1 In the lower limit topology on R, [a,b) is both open and closed. 2 In the standard topology on R, [a,b) is neither open, nor closed. 3 In the discrete topology on R, [a,b) is both open and closed. 4 In the cofinite topology on R (the same as the Zariski topology on R), [a,b) is neither open nor closed.

Which of the following are T1?

 1 R^n with the standard topology. 2 Any set with the discrete topology. 3 The Sierpinski space. 4 The natural numbers with the cofinite topology. 5 Any space with {x} closed for every x. 6 Any metrizable space with the topology induced from the metric. 7 The Zariski topology on R^n.

Which of the following are Hausdorff?

 1 R^n with the standard topology. 2 Any set with the discrete topology. 3 The Sierpinski space. 4 The natural numbers with the cofinite topology. 5 Any space with a sequence that converges to two different points. 6 Any metrizable space with the topology induced from the metric. 7 The Zariski topology on R^n.

Which of the following are true about continuity?

 1 If X has the discrete topology, then any function mapping X to some other space is continuous. 2 f: R-->R_l given by f(x)=x is continuous. 3 f: R_l-->R given by f(x)=x is continuous. 4 The standard Epsilon-delta definition of continuity from calculus is equivalent to the topological space definition of continuity on R with the standard metric. 5 If d is a metric on X, then T_d (the topology induced by d on X) is the coarsest topology so that d:   XxX   -->   R is continuous, where R has the standard topology.

Which of the following are true about continuity?

 1 If X has the discrete topology, then any function mapping X to some other space is continuous. 2 f: R-->R_l given by f(x)=x is continuous. 3 f: R_l-->R given by f(x)=x is continuous. 4 The standard Epsilon-delta definition of continuity from calculus is equivalent to the topological space definition of continuity on R with the standard metric. 5 If d is a metric on X, then T_d (the topology induced by d on X) is the coarsest topology so that d:   XxX   -->   R is continuous, where R has the standard topology.

Which of the following are true about homeomorphisms.

 1 (-1,1)   ~   R 2 R_l   ~   R 3 Sierpinski space   ~ {0,1}_discrete 4 X   ~   Y if there exists a continuous open map from X to Y 5 A circle and a square in R^2. 6 A donut and a mug. 7 [1,2)   ~   {0}U(1,2)

Match the following instructions for glueing to the resulting 2-dimensional space. Recall that WebCT changes the order of matching each time.

 1. a. 2. b. 3. c. 4. d. 5. e. 1 --> Choose Match abcde 2 --> Choose Match abcde 3 --> Choose Match abcde 4 --> Choose Match abcde 5 --> Choose Match abcde

Which of the following are true about the Heart of Math readings on the geometry of the earth and universe and the 4th dimension?

 1 Following the 40 degree latitude line from Chicago to Rome results in a path 5300 miles long, while the shortest distance path along the great circle between them is 4800 miles long. 2 Gauss tried to form a triangle using three mountain peaks near Gottingen. He had fires lit and used mirrors to reflect beams of light and measured the angles. His sum was within 1/180 of a degree of 180 degrees. This proves that the universe is Euclidean. 3 The fourth dimension is time. 4 Instead of trying to understand the 4th dimension - which appears too difficult, it is always wise to step back and try to address an easier but related question of how a 2-d creature could understand the 3rd dimension. 5 We shouldn't close our minds to ideas that first appear counterintuitive, since they can lead us to new frontiers. 6 Higher dimensional creatures can remove objects that seem completely closed within a lower dimensional safe. 7 One can unknot a rope in 4-d space without actually cutting it. 8 Higher dimensional analysis has applications to real life. For example, analyzing costs in a factory where there are four degrees of freedom (where our factory is located, how big it should be, how many workers we should hire, and how much capital we should borrow), we see that the problem of maximizing overall profit is a problem that lives within the fourth dimension.

Which of the following are true about the Heart of Math reading on rubber sheet geometry?

 1 The word "topology" was coined in 1847 and derives from the Greek word topos, which means surface. 2 Topology is a mind-stretching subject that frees us from conventional geometry and allows us to appreciate geometric characteristics of objects that remain unchanged even when the objects are stretched, shrunk, distorted, or contorted. 3 We cannot remove a topological vest without unbuttoning either the vest or the jacket. 4 We cannot turn a punctured torus inside out and obtain a punctured torus. 5 A theta curve is not the same as a circle, since there is a point on it that has three different directions to move in, but no corresponding point on the circle that has three directions to move in . 6 The sphere is a surface since every point has a neighborhood surrounding it that is equivalent by distortion to a neighborhood of a point in the flat plane. 7 A sphere and a torus are topologically different since we can find a loop on the torus that does not cut it into two different pieces, but for each loop on the sphere, it separates the sphere into two different pieces. 8 A knotted Jell-O solid two-holed torus is equivalent by distortion to the ordinary (unknotted) Jell-O solid two-holed torus.

Which of the following are true based on the Heart of Math reading on The Band that Wouldn't Stop Playing

 1 M.C. Escher represented the Mobius band in his work Mobius Strip II. 2 The Mobius band has one edge and one side. 3 If you cut the Mobius band down the middle, you end up with one piece that has two edges, two twists and two sides. 4 Cutting the Mobius band 1/3 of the way along the right edge results in a linked figure that has two pieces - one small Mobius band and one larger twisted cylinder. 5 In real life, Mobius bands appear as conveyor belts, which has double the life as a cylinder shaped band. 6 The projective plane is obtained by sewing a Mobius band onto a disk. The Klein bottle and the projective plane have only one side but no edges. 7 No one-sided surface without an edge can be constructed entirely in 3-dimensional space.

Which of the following are true about the Heart of Math reading on Feeling Edgy?

 1 For any connected graph in the plane, the Euler characteristic, V - E + F, where V is the number of vertices, E is the number of edges, and F is the number of faces, is 2. The proof of this is by induction. 2 There are only 5 Platonic solids - the tetrahedron, the cube, the octahedron, the dodecahedron, and the icosahedron. The proof is by contradiction, assuming that there is some Mysterahedron that we don't know about, and using properties of a regular solid and the Euler characteristic as 2 in order to solve for the all of the possibilities. 3 The polio virus has the shell of a cube. 4 Radiolaria are microorganisms whose skeletons often are regular solids.

Which of the following are true based on the Heart of Math reading on Knots and Links?

 1 The fiber of life itself is encoded in strands of DNA entwined intricately in its famous double helix. 2 During the replication phase of DNA, the two long strands stay individually connected. 3 Dewitt Sumners, a knot theorist, was able to prove in the abstract that only certain types of knots in DNA could result from the localized action of an enzyme. 4 Distinguishing one knot from another is easy. 5 Many questions about knots have yet to be answered, but other results about knots are known and some are quite surprising. 6 There exists a way of linking three rings such that the removal of any one will automaticaly unlink the other two.

Which of the following are true based on the Heart of Math reading on Fixed Points, Hot Loops, and Rainy Days?

 1 It is impossible to stretch, rotate, distort, fold and squash the red disk and place it back on top of the blue disk in such a manner that each point on the red disk moves. 2 The proof of the Brouwer fixed point theorem uses the concept of a winding number. 3 At every instant, there are two diametrically opposite places on Earth with identical temperatures and identical barometric pressures. 4 If we have a circle of variably heated wire, then the Hot Loop Theorem says that we have a pair of opposite points at which the temperatures are exactly the same. 5 The proof of the Hot Loop Theorem looks at the function f(x) = temp(x)-temp(x'), where x is a point on the circle, and x' is its diametric opposite. Since f(x) is positive in places, negative in others and continuous, then is must be 0 somewhere.

Which of the following are true about quotient spaces?

 1 Given a topological space X and a function f:X-->Y, the quotient topology on Y is the finest topology on Y s.t. f is continuous. 2 Given a topological space X and a function f:X-->Y, the quotient topology on Y is defined to be the sets U in P(Y) so that the inverse image of U is open in X. 3 Given a topological space X and a function f:X-->Y, if we send x to its equivalence class [x] via   x   ~f   y       iff   f(x)=f(y), then X/~ is homeomorphic to Y with the quotient topology on it. 4 f(t)=(cos(2Pi t), sin (2Pi t)) gives R/~ as S^1. 5 A quotient space X/~ is always homeomorphic to the original space X.

Which of the following are true regarding connected spaces?

 1 A space is connected if there does not exist a separation of X. 2 Every subspace of a connected space is connected. 3 The lower limit topology on R is not connected, but the standard topology on R is connected. 4 To prove that [0,1] is connected, we assume for contradiction that it is disconnected, and then arrive at a contradiction to the least upper bound property of one of the open sets separating [0,1]. 5 X is connected iff the only open subsets which are both open and closed are the empty set and the space itself. 6 The arbitrary union of a family of connected subspaces is connected. 7 If f:X-->Y is a continuous surjection and X is connected, then Y is connected. 8 A path from x to y is a continuous function mapping [0,1] to the space so that path(0)=x and path(1)=y. 9 A space is path connected if there is a path from x to y in X for all x and y. 10 Every path connected space is connected. 11 Every connected space is path connected. 12 R^2_Zar is not the same as the cofinite topology on R^2, yet both give connected topologies on R^2. 13 Connectedness can be used to show that spaces are not homeomorphic, since it is a topological property.

Which of the following are true regarding compactness?

 1 A space is compact if every open cover has a finite subcover. 2 R is not compact as {(n-2/3, n+2/3) | n in Z} is an open cover of R with no finite subcover, as every integer is in exactly 1 of these intervals. 3 {0} U {1/n} (for n in N) is not compact. 4 To show that [0,1] is compact, given an arbitrary cover, we look at E={x in [0,1] | [0,x) is covered by a finite subcollection of the cover}. We then look at the least upper bound of the set and show that it must be 1, making [0,1] compact. 5 A subspace of a compact Hausdorff space is compact iff it is closed. The Hausdorff condition is not needed in the reverse (and only if) direction. 6 The continuous image of a compact space is compact. 7 Any quotient of a compact space is compact. Hence S^1 is compact as it is a quotient of [0,1]. 8 Heine-Borel says that a subspace of R^n is compact iff it is closed and bounded. 9 There exists a space X and a subspace so that the statement compact iff closed and bounded does not hold. 10 If X is compact Hausdorff under two topologies, then they are equal or they are not comparable. 11 The finite union of compact sets is compact. 12 The intersection of two compact sets is compact.