Re: Topology with topological property.


"William Elliot" <marsh@xxxxxxxxxxxxxxxxxx> wrote in message
news:Pine.BSI.4.58.0709132216440.26819@xxxxxxxxxxxxxxxxxxxx
On Thu, 13 Sep 2007, mina_world wrote:
"William Elliot" <marsh@xxxxxxxxxxxxxxxxxx> wrote in message
On Wed, 12 Sep 2007, mina_world wrote:

X : metrizable space.
X~Y : homeomorphic.

then Y : metrizable space.

is this possible ?

Homeomorphism f:(X,d) -> Y ==>
D(x,y) = d(f^-1(x), f^-1(y) metric for Y.

Yes, I see.
1) D(x,y) >= 0,
D(x,y) =0 <=> x=y.
2) D(x,y) = D(y,x)
3) D(x,z) <= D(x,y) + D(y,z)

Yes, D is a metric.

But wait, there's more to do. Besides noting g = f^-1,
is a bijection from Y to X. You have to show
f:(X,d) -> (Y,D)
g:(Y,D) -> (X,d)

are contiguous, that f is bicontinuous bijection.

Yes, exact...
Since X~Y,
there is a bijection function f : X -> Y with (X,d).

Define D(x,y) = d(f^-1(x), f^-1(y) metric for Y.

I must show that f and f^-1 are continuous.

Since B_d(x, e) subset f^-1{B_D(f(x), e)} for any e>0,
Since B_D(f(x), e) subset f{B_d(x, e)} for any e>0,
f and f^-1 are continuous.

Please excuse me, I have mislead you.

By the above with = instead of subset,
you'd shown X and (Y,D) are homeomorphic.
Since X and Y are homeomorphic,
Y is homeomorphic to (Y,D).
Consequently to conclude the problem that Y is metrizable,
we need to use the result of the problem, that a space
homeomorphic to a metric space is metrizable. Whoops!

--
A space (X,tau) is metrizable when there's a metric d
such that (X,tau) and (X,d) have the same topology.

Thus what's needed to be shown is the topological space
Y and the metric space (Y,D) have the same topology.

Yes, I try again.
1)
X is said to be metrizable
if there exists a metric d on the set X that induces the topology of X.

2)
If d is a metric on the set X,
then the collection of all e-balls B_d(x,e) for x in X and e>0,
is a basis for a topology on X, called the metric topology induced by d.

so, I must show that Y is the topology that induced by metric D.
D(x,y) = d(f^-1(x), f^-1(y)) metric for Y.

Ok, Let's go.

Since X~Y, there is a homeomorphism f : (X,d) -> Y.
Define D(x,y) = d(f^-1(x), f^-1(y)) metric for Y.

For any open set V of Y,
f^-1(V) = U is a open set of X.
I can express that U = Union_{x in A} B_d(x, e_x) for some set A by basis
definition.

so, V = f(U) = Union_{x in A} B_D(f(x), e_x).
Because, f(B_d(x,e)) = B_D(f(x),e).

so, B = {B_D(y, e) | y in Y, e >0}is a basis of Y.
so, Y is the metric topology induced by D.
so, there exists a metric D on the set Y that induces the topology of Y.
so, Y is metrizable.

Really End ?


.

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