Homeomorphic Measures In Metric Spaces Homework

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From: Henno Brandsma
Date: April 29, 2004
Subject: Re: homeomorphisms in metric spaces (lots of parts)

In reply to "homeomorphisms in metric spaces (lots of parts)", posted by mack on April 28, 2004:
>i haven't studies homeomorphisms since abstract algebra (which was forever ago)... please help!!!
>cant seem to figure any of these out.. :( thx
>
>Mackenzie
>
>Let f be a one-to-one function from M1 into M2, both metric spaces. If f and
>f_inv are continuous, we say that M1 and M2 are homeomorphic.
>
>i) Prove that any two closed intervals of the real number line are homeomorphic.
A linear map will do (of the form x --> ax+b...)

>
>ii) Prove that a closed interval is not homeomorphic to either an open interval
>or a half-open interval.
>
If f: X --> Y is a homeomorphism and f(x) = y
then f: X\{x} --> Y\{y} is also a homeomorphism.
Now use connectedness...

Alternatively: an (half-)open interval is not compact so cannot
be the continuous image of a (compact!) closed interval...

>iii) Let M be a metric space; Let G(M) denoted the set of homeomorphisms from
>M onto M. Identify the groups G(M) in case G is finite; , also show that if M1 and M2
>are homeomorphic, then G(M1) is isomorphic to G(M2).
If M is finite, then M = {x_1,...,x_n}, and all points are isolated
points, so any map from M to M is continuous.
So homeomorphism == bijection for finite spaces and we have S_n
(the symmetric group on n points).
As to the second: if h is the homeomorphism between M1 and M2, with
inverse h', then the map F: G(M1) --> G(M2)
defined by F(f) = h o f o h' (o = composition) will work...

>
>iv) Prove that any metric space M is homeomorphic to a metric space (M*, d),
>where d is bounded by 1.
If d' is the original metric, then d(x,y) = min(d'(x,y),1) will work.
Show that the identity is continuous both ways.

>
>v) Let M now be a sepearable metric space. Prove that there exists a one-to-one
>continuous function f from M into the Hilbert cube.

Let U_n (n in N) be a countable base. Then the map i that sends
x to (min(d(x,X\U_n),1) )_n in [0,1]^N is continuous and 1-1...
>
>vi)Prove: A metric space M is compact iff M is homeomorphic to a closed subset of the Hilbert cube.
>
If M is compact, then the above map i (from v)) will be a homeomorphism
between M and i[M] (because a 1-1 continuous map from a compact metric space
to another automatically has a continuous inverse, as it maps closed==compact
sets to closed==compact sets...)
And i[M] is compact, so closed in the Hilbert cube.

The other way is trivial because the Hilbert cube is compact, and
so are all its closed subspaces...

Henno


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