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Set theory/ topology question 
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#1
Mar1604, 12:39 PM

P: 124

Let f be a realvalued function defined and continuous on the set of real numbers R. Which of the following must be true of the set S = {f(c): 0<c<1}?
I. S is a connected subset of R II. S is an open subset of R III. S is a bounded subset of R The answer is I and III only. I understand why I is true. But, why is is bounded, and why is it not an open subset? Thanks. 


#2
Mar1604, 01:30 PM

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P: 9,397

it is bounded because it is continuous on the compact set [0,1] and the continuous image of a compact (closed and bounded set) is closed and bounded, the image of (0,1) is a subsert of this bounnded set and is hence bounded.
define f(x) = 0 for all x. the image of an open set is then closed. 


#3
Mar1604, 01:56 PM

P: 124

How can you reach a conclusion by simply considering the case f(x) = 0?



#4
Mar1604, 04:08 PM

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P: 9,397

Set theory/ topology question
Because the question asks if it MUST be true that the image of an open set is open. I just showed that it isn't necessarily true. To disprove a statement it suffices to provide ONE counter example.
The negation of the statement 'for all continuous f (on R) the restriction to (0,1) is an open map (ie the image is open)' is 'there exists A continuous map on r R whose restriction to (0,1) is not an open map'. 


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