set theory/ topology question


by yxgao
Tags: theory or, topology
yxgao
yxgao is offline
#1
Mar16-04, 12:39 PM
P: 125
Let f be a real-valued 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.
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matt grime
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#2
Mar16-04, 01:30 PM
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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.
yxgao
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#3
Mar16-04, 01:56 PM
P: 125
How can you reach a conclusion by simply considering the case f(x) = 0?

matt grime
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#4
Mar16-04, 04:08 PM
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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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