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Science Forum Index » Mathematics Forum » For anyone who has "intro to topology and modern analysis" b
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Message |
| Justin Young |
 Posted: Fri Dec 19, 2003 8:05 pm |
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Guest
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on P. 20, problem 6.
The wording of the question is confusing to me:
Let X,Y be nonempty sets, and f:X->Y, if A is a subset of X, B a
subset of Y then prove:
(a) ff^-1(B) subset of B, and ff^-1(B) = B <=> f is onto;
now my question is,
let p : ff^-1(B) subset of B
q : ff^-1(B) = B
r : f is onto
am i trying to show p&q <=> r, or something else?
because since q => p, it seems strange to phrase it this way.
why not just say "prove q <=> r" ?
this strange phrasing makes me wonder if i am reading the question
wrong.
thanks for any assistance you can give. |
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| Eric J. Wingler |
| Posted: Fri Dec 19, 2003 10:10 pm |
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Guest
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"Justin Young" <x_static66@hotmail.com> wrote in message
news:bf317dec.0312191705.2275d727@posting.google.com...
Quote: on P. 20, problem 6.
The wording of the question is confusing to me:
Let X,Y be nonempty sets, and f:X->Y, if A is a subset of X, B a
subset of Y then prove:
(a) ff^-1(B) subset of B, and ff^-1(B) = B <=> f is onto;
now my question is,
let p : ff^-1(B) subset of B
q : ff^-1(B) = B
r : f is onto
am i trying to show p&q <=> r, or something else?
because since q => p, it seems strange to phrase it this way.
why not just say "prove q <=> r" ?
this strange phrasing makes me wonder if i am reading the question
wrong.
thanks for any assistance you can give.
**********************************************************
The question is asking you to prove two things:
1. ff^-1(B) subset of B
and
2. ff^-1(B) = B <=> f is onto
Eric J. Wingler
Department of Mathematics and Statistics
Youngstown State University |
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| Lot-o-fun |
| Posted: Fri Dec 19, 2003 10:31 pm |
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Guest
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The book is asking you to first show that
f f^{-1}(B) \subset B
Once you're done with that, show that
f f^{-1}(B) = B if and only if f is onto.
In article <bf317dec.0312191705.2275d727@posting.google.com>, Justin
Young <x_static66@hotmail.com> wrote:
Quote: on P. 20, problem 6.
The wording of the question is confusing to me:
Let X,Y be nonempty sets, and f:X->Y, if A is a subset of X, B a
subset of Y then prove:
(a) ff^-1(B) subset of B, and ff^-1(B) = B <=> f is onto;
now my question is,
let p : ff^-1(B) subset of B
q : ff^-1(B) = B
r : f is onto
am i trying to show p&q <=> r, or something else?
because since q => p, it seems strange to phrase it this way.
why not just say "prove q <=> r" ?
this strange phrasing makes me wonder if i am reading the question
wrong.
thanks for any assistance you can give. |
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| Back to top |
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