Re: No homeomorphism (0,1) <--> [0,1]

On Thu, 8 Sep 2005 12:54:04 -0500, mstemper@xxxxxxxxxxxxxxxx (Michael
Stemper) wrote:

>In article <3ob7q4F54kquU1@xxxxxxxxxxxxxx>, Jose Carlos Santos writes:
>>On 08-09-2005 17:34, Michael Stemper wrote:
>>
>>> I'm trying to prove that there isn't a homeomorphism between [0,1]
>>> and (0,1), and I'm stuck. What I know so far is the definitions of
>>> homeomorphism and continuity.
>>
>>Suppose that there was a homeomorphism f:[0,1] --> ]0,1[. It's
>>continuous, and so, by Weierstrass' theorem,
>
>Thanks for the reply. However, all that I've learned so far is the
>definitions of metric space, continuity, and homeomorphism. I certainly
>haven't encountered this theorem yet.
>
>For anybody playing along at home, this exercise is a post script
>to Exercise C, Chapter III, in _Topology_, by George McCarty.

Well, since Advanced Calculus is usually a prerequisite to Topology,
there's no reason why you can't use the basic theorems of Calculus.

For example, let's assume you can use the intermediate value theorem:

If f is continuous on a closed interval [a,b] then f([a,b])
contains [f(a),f(b)].

So suppose f is a homeomorphism from [0,1] to (0,1), and let f(0)=a,
f(1)=b. Since f is 1-1, either a<b or b<a. Let c=min(a,b), d=max(a,b).
Since f^(-1) is continuous, then by the intermediate value
f^(-1)([c,d]) contains [0,1], hence equals [0,1]. By the definition of
c,d and since c,d are in (0,1), we have 0<c<d<1, so [c,d] is a proper
subset of (0,1). Let r be any element of (0,1) that is not in [c,d].
Then f^(-1)(r) is stuck -- it can't go anywhere, since f^(-1)([c,d])
has already used all of the values from 0 to 1 inclusive, so any value
for f^(-1)(r) would make f^(-1) not 1-1.

quasi
.

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