More questions before Nov 12th, hope you can help me ASAP.

[mhyyh]

1. S is a compact connected set in R^n, and how many connected components can the complement of S have?

I totally don't know how to deal with this.....
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2. X is a set, f: X->P(x) [power set of X] is a function. Z={s: s belongs to X, and s doesn't belong to f(X)}
A. Z is empty
B. Z is not empty
C. Z=X
D. Z doesn't belong to P(X)
E. the complement of Z belongs to P(X)

I think the answer might be B, coz f(X) is a set of set..
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3. Which are equal to "map f is continuous"?
I. For every set A, f^(-1)(in(A))=in(f^(-1)(A))
II. For every set B, cl(f^(-1)(B)) contains f^(-1)(cl(B))
III. f is a open map

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4. a,b belong to group G, both have finite orders
I. If ab=ba, then ab has finite order
II. If ab has finite order,then ba has finite order
III. If ab has finite order, then a^(-1)b^(-1) has finite order

still Which are correct?....
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5.it is a linear algebra question.
Tu=u,Tv=2v,Tw=3w. T is the linear transformer on R^3. which are right?
I. detT=6;
II. u,v,w is a basis of R^3
III. character poly |xI-T|=(x-1)(x-2)(x-3)

I think all of them are correct, how about you?....

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6. T is an orthogonal matrix, which of the following is right?[. means dot product]
I. Tu . Tv = u . v
II. If v . (Sqrt[2]/2,Sqrt[2]/2,0)=0, Tv . v=0
III. T^2=1

A friend think they are all right.
But i think T is not a normed orthogonal matrix, coz i remember only orthogonal property cannot ensure its determinant is 1.
Ironically, http://en.wikipedia.org/wiki/Orthogonal_matrix
orthogonal matrix has been normed from wiki. Am I wrong?...
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thanks, and please enlighten me before my test.....=_____=

[blitzer6266]

1. Connected compact in R1 means a closed interval. The complement has 2 components. In R2 a compact set can have lots of holes so it is unlimited.
2. D- Pretend it is and you'll get a contradiction (Z is empty if X is empty)
3. Your question doesn't make sense (maybe which are true if f is continuous?)
4. All 3 are true
5. All 3 are false (unless you assume u, v, and w are all nonzero)
6. 1 and 3 are right. 2 is false (consider T = identity and v=(0,0,1)) (orthogonal matrix means orthonormal basis)

[blitzer6266]

Actually I think 6. 3 is false as well. It needs to be T transpose times T, not T^2

[fireandgladstone]

For 3, none of them are equivalent conditions. For (1), consider a triangle type function (i.e. something that looks like ---^---). For (2), consider something like f(x) = x for x <= 0, f(x) = 0 for 0<x<1, and f(x) = x-1 for x>=1. For (3), consider f(x) = x^2.

Where are you getting these questions?

[mhyyh]

For 3, none of them are equivalent conditions. For (1), consider a triangle type function (i.e. something that looks like ---^---). For (2), consider something like f(x) = x for x <= 0, f(x) = 0 for 0<x<1, and f(x) = x-1 for x>=1. For (3), consider f(x) = x^2.

Where are you getting these questions?

still thx, even i think your solution made me more confused....
anyway, i've figured it out myself.
let f be dirchlet funtion, if A is {1} & B is{0}, both I&II are correct, while function is not continuous. and open map does not necessary need it to be continuous.

and these questions are from previous students took Mtest in Asia. im not sure it can help u, if u are in euro or NA, and most of the questions are not complete. if u dont mind its flaw, just send me your email address.

[mhyyh]

1. Connected compact in R1 means a closed interval. The complement has 2 components. In R2 a compact set can have lots of holes so it is unlimited.
2. D- Pretend it is and you'll get a contradiction (Z is empty if X is empty)
3. Your question doesn't make sense (maybe which are true if f is continuous?)
4. All 3 are true
5. All 3 are false (unless you assume u, v, and w are all nonzero)
6. 1 and 3 are right. 2 is false (consider T = identity and v=(0,0,1)) (orthogonal matrix means orthonormal basis)

u r right. i just assumed that S is a ball in Rn.

brilliant solutions of 2,4,5,6
thank u , i've solved 3 by myself. it's not too hard. just a little bit afraid of topological stuff.

[fireandgladstone]

For 3, none of them are equivalent conditions. For (1), consider a triangle type function (i.e. something that looks like ---^---). For (2), consider something like f(x) = x for x <= 0, f(x) = 0 for 0<x<1, and f(x) = x-1 for x>=1. For (3), consider f(x) = x^2.

Where are you getting these questions?

still thx, even i think your solution made me more confused....
anyway, i've figured it out myself.
let f be dirchlet funtion, if A is {1} & B is{0}, both I&II are correct, while function is not continuous. and open map does not necessary need it to be continuous.

and these questions are from previous students took Mtest in Asia. im not sure it can help u, if u are in euro or NA, and most of the questions are not complete. if u dont mind its flaw, just send me your email address.

Sure, that'd be great. I PM-ed you my email address.

EDIT: I'm not actually sure your answer to question 3 is correct (it's true that none of them are equivalent but your reasoning doesn't seem correct). It has to be true for every set A/B, not just a particular set. Moreover, condition I actually implies continuity. It's just that continuity doesn't imply condition 1. An easier example than the one I gave to see this (but the same idea) is f(x) = |x|. Then the inverse image of [0,1) is (-1,1) and the interior of (-1, 1) is just (-1, 1). However, the interior of [0,1) is (0,1) and the inverse image of this is (-1,0) union (0, 1).

Also, notice that if you flip the containment in II you get an equivalent condition.

[Jnes]

1. Connected compact in R1 means a closed interval. The complement has 2 components. In R2 a compact set can have lots of holes so it is unlimited.
2. D- Pretend it is and you'll get a contradiction (Z is empty if X is empty)
3. Your question doesn't make sense (maybe which are true if f is continuous?)
4. All 3 are true
5. All 3 are false (unless you assume u, v, and w are all nonzero)
6. 1 and 3 are right. 2 is false (consider T = identity and v=(0,0,1)) (orthogonal matrix means orthonormal basis)

->For number 2 I'm not sure how having Z element of P(X) causes contradiction =(
If Z is not emtpy (since blizter said Z empty => X emtpy), Z must be some combination of elements of X right?
Then shouldn't Z be element of P(X)? I think I might be missing something crucial from the problem but I'm not seeing it!
This problem has been bothering me since yesterday!! urg.

[PNT]

For 2. what am i misunderstanding? isn't the answer E?

Since Z is a subset of X, Z is necessarily in P(X). The empty set is in P(X) so if X is empty it still holds. The compliment of Z is also a subset of X so should be in P(X)?

Also a little more explanation for 4. For questions like this it's just a matter of playing around with words ie

1. if ab=ba then (ab)^n=abab....ab=a^2b^2ab...ab=a^n b^n=1
2. if (ab)^n=1 then 1=abab...abab and a^-1 b^-1=(ba)^(n-1) so 1=(ba)^(n-1)ba
3. if (ab)^n=1 then ab...ab=1 and ba...ba=a^-1 b^-1 so multiplying inverses on the right gives (a^-1 b^-1)^n=1

[blitzer6266]

Yup, I completely messed that one up and I thought I read something else. My initial reading was

Z={s in X such that s is not in f(s)}
and for some reason as well I thought D said
Z doesn't belong to f(X)

The reason I was tricked into thinking I saw this was because this is the standard trick used to show that card(X)< card(P(X)).

Sorry for the confusion... this problem is much more simple than that since the complement of any set within X (where the complement is taken with respect to X) is clearly in P(X)

[deckoff8]

how number 2, which isnt a also a correct answer? I mean if an element is inside X it's obviously inside the power set of X. or is empty here considered the empty set so empty is also inside P(X)?

[Jnes]

Choices A/B/C depends on the function I guess.
If f(x) is identity function, then Z would be empty. We can make some other function so that Z=f(X) as well.
I kinda suspected that the previous answer (D) came from the cardinality proof since I remember seeing something similar too =) but didn't make any sense since in the proof, we had to assume that f was onto and bring out the contradiction.
D is always false I guess?
E is always true so E would be the answer?
(I'm not sure about this tho! someone should think it over I think. I'd go with E.)

[PNT]

Z is not necessarily empty because f can map any element of X into any subset of P(X), so you can choose an f such that some element of X is not in f(X). ie X={a,b} then define f(a)={a}, f(b)={a,b} then Z={b}.

[deckoff8]

ohhh. i was totally confused. I thought the function f mapped the set X to the set P(X), not elements. thanks.