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Definition of preimage or inverse image
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Author Topic: Definition of preimage or inverse image  (Read 40545 times)
deepnorth
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« on: August 10, 2009, 02:54:43 AM »

For discussion.

Let A and B be two sets and f a function from A to B. If C is a subset of the range B then the preimage, or inverse image, of C under the function f is the set defined as f -1(C) = {x belong to A : f(x) belong to C }

http://www.mathcs.org/analysis/reals/logic/relation.html

I think C is not necessary a subset of the range B. C can be any set as long as" f -1(C) = {x belong to A : f(x) belong to C } "

Otherwise, we can not solve this question by "http://www.mathcs.org/analysis/reals/cont/answers/topcont2.html"
Let f(x) = 1 if x > 0 and f(x) = -1 if x  0. Show that f is not continuous by
1.finding an open set whose inverse image is not open.
2.finding a closed set whose inverse image is not closed.

the range of f(x) is 1 and -1, and are only these two numbers, but (-2,0) is not a subset of B. If we use above inverse image definition even the question "find an open set" itself is not correct.
So I think the domain of inverse image is not necessarily a subset of rang f.
« Last Edit: October 30, 2009, 03:33:55 PM by webmaster » Logged
GuitarCrazyo
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« Reply #1 on: October 29, 2009, 09:42:54 AM »

Easier to just use html format old style, where you use img src tags for each individual image you are posting. this is how the old forum was

Tim
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GuitarCrazyo
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« Reply #2 on: November 05, 2009, 11:13:53 AM »

I posted earlier, but it doesnt show the large format, just a smaller jpeg which is clicked upon to open the image in another window .....HMMMFFF
Please help me out if you can.
THX John
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ste
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« Reply #3 on: December 22, 2009, 03:38:21 PM »

The question is very vague, but it sounds like f(x) is defined on the real number line where its range is the two point set {-1,1}, so f: A -> B where A = {the real line} and B = {-1,1}.

If the set A has the usual topology associated with the real line then open sets are of the form (a,b) where a < b and closed sets are of the form [a,b] and [a,a], where the latter denotes the singleton set {a}. If the set B is equipped with the discrete topology, then every singleton in B is an open set i.e. {-1} and {1} are both open in B, which implies that {-1} and {1} are also both closed in B (because the complement of an open set is closed and the complement of a closed set is open). So in this case, the complement of {-1} in B is {1} and the complement of {1} in B is {-1}.

Let U be a subset of the interval (0, infinity). The image set, f(U), regardless of whether U is open or closed, is {1} which is both open and closed.

This is not a complete solution, but look over the finer details of the question again and see if it helps.

Ste.
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mixmastermike00
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« Reply #4 on: March 12, 2010, 05:37:51 AM »

I prefer to stick with the old html method. Inverse imagery should not be so complicated if you ask me.  Just using img source tags is way simpler. Unless that stop popping up, i wouldn't mess with it.  pop up tv lift
« Last Edit: February 17, 2011, 07:26:03 AM by mixmastermike00 » Logged
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