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-At first I thought of a counterexample like f=sinx but it seems that its range is not R. So will the answer be yes? And how can we prove it? Will the preimage have to be bounded in this case?

Thanks.

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- Thread starter sapporozoe
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- #1

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-At first I thought of a counterexample like f=sinx but it seems that its range is not R. So will the answer be yes? And how can we prove it? Will the preimage have to be bounded in this case?

Thanks.

- #2

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false, take f=1 for all x in R.

then preimage of the closed unit interval is R which is not compact.

then preimage of the closed unit interval is R which is not compact.

- #3

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but the range of this f is not R, is it?

false, take f=1 for all x in R.

then preimage of the closed unit interval is R which is not compact.

- #4

Hurkyl

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(But it's easy enough to modify yours, or SiddharthM's, suggestion so that the image of the function really is all of

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(But it's easy enough to modify yours, or SiddharthM's, suggestion so that the image of the function really is all ofR)

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Hurkyl

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But the image of this function is the set {0}.

- #9

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Perhaps we are familiar with a statement like this:＂if f:Rn to Rm is continuous and B in Rn is bounded, then f(B) is Rm is bounded". We know that boundedness is not preserved under continuous mapping. But this statement is true in that Rn is its domain... So I am confused about the exact meaning "f is Rn to Rm".

RtoR.

But the image of this function is the set {0}.

- #10

Hurkyl

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There exist continuous functions that don't preserve boundedness. That doesn't imply all continuous functions don't preserve boundedness.We know that boundedness is not preserved under continuous mapping.

In the case you cited, the theorem follows from the fact that continuous maps preserve compactness. (And the properties of Euclidean space)

I'm not sure why you think this relates to the notion of range, though.

- #11

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The statement is true because we exclude cases like: f(x)=tanx, x in (-pi/2, pi/2). We can exclude such cases as we require the domain of f(x) is Rn.

So my question becomes: if we say "f is from Rn to Rm", do we mean "the domain of f is Rn but the range of f is a subset of Rm"?

Thanks again:)

- #12

Hurkyl

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The statement is true because we exclude cases like: f(x)=tanx, x in (-pi/2, pi/2). We can exclude such cases as we require the domain of f(x) is Rn.

So my question becomes: if we say "f is from Rn to Rm", do we mean "the domain of f is Rn but the range of f is a subset of Rm"?

Thanks again:)

I have seen the word "range" used in two different ways:

(1) The target of a function. (R

(2) The image of a function. (f(R

It appears to me like your issue is entirely due to mixing up the two usages of the word "range".

(the target is also called the codomain)

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i almost get it:) Thanks.

- #15

mathwonk

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the interesting part is: when is the preimage of a compact set compact? continuos functions with this property are called proper and are very important.

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