Homeomorphic Spaces
Maury Barbato
do you know some example of two topological spaces S, T
such that:
(I) there exist a continuous bijection f:S->T
(II) there exist a continuous bijection g:T->S
but S and T are not homeomorphic?
Thank you very much for your help.
My Best Regards,
Maury
Jules
think of. Let S and T both denote the set of real numbers, with the
following topologies: A < S is open iff A intersect [0, 1] is open in
[0, 1]. B < T is open iff b intersect [0, 1) is open in [0, 1). It is
easy to see that S and T are not homeomorphic, because the connected
components of S are a bunch of singleton sets, along with the the set
[0, 1], and the components of T are a bunch of singletons, along with
the set [0, 1), and these sets are not homeomorphic (note: in the
subspace topology, [0, 1] < S and [0, 1) < T have the usual topology).
Now, take f: S --> T defined by f(s) = s / 2, and g: T --> S defined by
g(t) = t / 2. Both of these functions are certainly bijections, and it
can also be checked that they are both continuous.
William Elliot
Newsgroups: sci.math
Subject: Re: Homeomorphic Spaces
Maury Barbato wrote:
> > do you know some example of two topological spaces S, T
> > such that:
> > (I) there exist a continuous bijection f:S->T
> > (II) there exist a continuous bijection g:T->S
> > but S and T are not homeomorphic?
> There is probably a simpler example, but this is the first one I
> could think of. Let S and T both denote the set of real numbers,
> with the following topologies: A < S is open iff A intersect [0, 1]
> is open in [0, 1]. B < T is open iff b intersect [0, 1) is open in
> [0, 1). It is easy to see that S and T are not homeomorphic, because
> the connected components of S are a bunch of singleton sets, along
> with the the set [0, 1], and the components of T are a bunch of
> singletons, along with the set [0, 1), and these sets are not
> homeomorphic (note: in the subspace topology, [0, 1] < S and [0, 1) <
> T have the usual topology). Now, take f: S --> T defined by f(s) = s
> / 2, and g: T --> S defined by g(t) = t / 2. Both of these functions
> are certainly bijections, and it can also be checked that they are
> both continuous.
Simpler would be to use [0,oo) instead of R and for a simpler description
of the topologies, the topology of S has base of all the open subsets of
[0,1] and all singletons out side of [0,1]. Similar with T, using [0,1)
instead. Nice example.
Does < mean proper subset or subset?
Why are you over using less than < ?
----
Maury Barbato
>
> Why are you over using less than < ?
>
William, I can understand your mathematical explanations,
with some greater or less effort, but I can't understand
your english!
Forgive my ignorance of English languaga, but what did
you mean by this phrase?
Friendly Regards by Italy,
Maury
William Elliot
> William Elliot wrote:
>
> > Why are you over using less than < ?
>
> William, I can understand your mathematical explanations,
> with some greater or less effort, but I can't understand
> your english!
> Forgive my ignorance of English languaga, but what did
> you mean by this phrase?
Whoops. Of little important except perhaps if < means "proper subset", as
you expect it to do, or if it actually means "subset", that is "proper
subset of or equal to". Any way to properly punctuate my blurb
Why are you over using "less than <"?
or to make it clear
Why are you over using the less than symbol "<"?
> Friendly Regards by Italy,
> Maury
>
As forgiving your ignorance of English, there is little to forgive,
a few misspellings perhaps. In all your English is amply good.
Jules
I used < to denote a not neccesarily proper subset. I used this symbol
because there is no subset symbol on the keyboard. I should have
clarified my notation beforehand.
William Elliot
William Elliot wrote:
> > Does < mean proper subset or subset?
> > Why are you over using less than < ?
> I used < to denote a not neccesarily proper subset. I used this
> symbol because there is no subset symbol on the keyboard. I should
> have clarified my notation beforehand.
As the notation indicates, < would be proper subset and <= subset.
I use
A proper subset B, A subset B
for
A < B and A <= B.
There are other more useful ways to overwork <=. For example,
x <= A when for all a in A, x <= a
ie x is a lower bound of A.
----
Dave L. Renfro
> do you know some example of two topological spaces S, T
> such that:
> (I) there exist a continuous bijection f:S->T
> (II) there exist a continuous bijection g:T->S
> but S and T are not homeomorphic?
Others have replied, but I thought I'd mention that this
property (without the non-homeomorphic part) was studied
by several people in the early days of topology (Banach,
Kuratowski, Hausdorff, Sierpinski, etc. in the 1920's).
I believe the notion originates from Frechet (1910), who
called it "type de dimensions". Note that any two intervals
in R have this property. There is a lot about this relation
(an "equivalence" relation on topological spaces that's
strictly weaker than being homeomorphic) in Sierpinski's
"General Topology" (where it's called "dimensional type",
see below) and Kuratowski's "Topology" (Academic Press,
1966 -- where it's called "topological rank").
"General Topology" by Waclaw Sierpinski, Dover Publications,
1956/2000. Use 'Search in this book' = "dimensional type"
(include quotes), which gives pp. 130-133, 137, 141, 142,
144, 145, 163, 165.
http://books.google.com/books?vid=ISBN0486411486
Dave L. Renfro