Topology with hausdorff, quotient map.
mina_world
The Hausdorff property is not invariant even
under open quotient mapping.
example)
Let R be the real space with the usual topology.
Define an equivalence relation on the R as follows
x ~ y if x - y in Q.
Consider quotient mapping q : R -> R/~.
then R/~ is not Hausdorff.
----------------------------------------------------------
Maybe, q is not open mapping.
right ?
If right, repair ==> under quotient mapping.
If not right, of course, I need your advice.
Anyway, I will show that R/~ is not Hausdorff.
pf)
First, consider the R/~.
[1] = {x | x - 1 in Q} = Q
[2] = {x | x - 2 in Q} = Q
...
[sqrt(2)] = {x | x - sqrt(2) in Q} = {sqrt(2) + r | r in Q}
= sqrt(2) + Q (let)
[sqrt(3)] = ... = sqrt(3) + Q.
so, R/~ = Q U {i + Q | i in irrational}.
Maybe, R/~ = R/Q.
Since q^{-1}(U) is open in R <==> U is open in R/~,
it means that R/~ is an indiscrete topology.
Because,
example) U = {i + Q | i in irrational} = I + Q (let)
==> q^{-1}(U) = I + Q
==> I + Q is not open in R.
so, R/~ is not Hausdorff.
-------------------------------------------------------
OR)
Suppose that R/~ is Hausdorff.
For [a] =/= [b],
there is disjoint open set U, V such that [a] subset U,
[b] subset V.
q^{-1}(U) and q^{-1}(V) are open and disjoint.
Because,
c in [q^{-1}(U)] /\ [q^{-1}(V)]
==> q(c) in q[q^{-1}(U) /\ q^{-1}(V)]
==> q(c) in U /\ V. contradiction.
Since q^{-1}(U) and q^{-1}(V) are open,
there is a rational s, t such that s in q^{-1}(U),
t in q^{-1}(V).
Since s-t in Q, [s] = [t].
so, s in q^{-1}(U) ==> q(s) = [s] in U
and t in q^{-1}(V) ==> q(t) = [t] in V.
so, contradiction.
so, not Hausdorff.
José Carlos Santos
> The Hausdorff property is not invariant even
> under open quotient mapping.
>
> example)
> Let R be the real space with the usual topology.
> Define an equivalence relation on the R as follows
> x ~ y if x - y in Q.
>
> Consider quotient mapping q : R -> R/~.
> then R/~ is not Hausdorff.
>
> ----------------------------------------------------------
> Maybe, q is not open mapping.
> right ?
Yes, it is an open mapping. Trivially, because the restriction of _q_ to
_any_ non-empty open set is surjective.
Best regards,
Jose Carlos Santos
mina_world
>> under open quotient mapping.
>>
>> example)
>> Let R be the real space with the usual topology.
>> Define an equivalence relation on the R as follows
>> x ~ y if x - y in Q.
>>
>> Consider quotient mapping q : R -> R/~.
>> then R/~ is not Hausdorff.
>>
>> ----------------------------------------------------------
>> Maybe, q is not open mapping.
>> right ?
>
> Yes, it is an open mapping. Trivially, because the restriction of _q_ to
> _any_ non-empty open set is surjective.
Yes, it's really an open mapping.
Because, R/~ = Q U {i + Q | i in irrational}.
and these elements is dense in R.
It means that cl[Q] = R, cl[i + Q] = R.
Namely, for any non-empty open set U in R,
U /\ Q =/= emtpy, U /\ [i + Q] =/= R.
so, q(U) = R/~ by equivalent class mapping.
so, q is an open mapping.
José Carlos Santos
>>> The Hausdorff property is not invariant even
>>> under open quotient mapping.
>>>
>>> example)
>>> Let R be the real space with the usual topology.
>>> Define an equivalence relation on the R as follows
>>> x ~ y if x - y in Q.
>>>
>>> Consider quotient mapping q : R -> R/~.
>>> then R/~ is not Hausdorff.
>>>
>>> ----------------------------------------------------------
>>> Maybe, q is not open mapping.
>>> right ?
>> Yes, it is an open mapping. Trivially, because the restriction of _q_ to
>> _any_ non-empty open set is surjective.
>
> Yes, it's really an open mapping.
> Because, R/~ = Q U {i + Q | i in irrational}.
No. R/~ = {{Q + i} | i is irrational}.
> and these elements is dense in R.
R/~ is not a subset of R.
> It means that cl[Q] = R, cl[i + Q] = R.
> Namely, for any non-empty open set U in R,
> U /\ Q =/= emtpy, U /\ [i + Q] =/= R.
>
> so, q(U) = R/~ by equivalent class mapping.
> so, q is an open mapping.
What matters here is that every non-empty subset of R contains points
of all equivalence classes, which is equivalent to the fact that
q(U) = R/~.
William Elliot
> The Hausdorff property is not invariant even
> under open quotient mapping.
>
> example)
> Let R be the real space with the usual topology.
> Define an equivalence relation on the R as follows
> x ~ y if x - y in Q.
>
> Consider quotient mapping q : R -> R/~.
> then R/~ is not Hausdorff.
>
> Maybe, q is not open mapping. right ?
If U open subset R, then inverse of U/~ = \/{ u + Q | u in U }
= U + Q = \/{ q + U | q in Q } = union of open sets. Hm, = R.
Thus U/~ is open.
> Anyway, I will show that R/~ is not Hausdorff.
If x/~ in open { v/~ | v in some set A },
then q^-1(V) = \/{ v + Q | v in A } = A + Q is open, thus = R.
Hence the nonnul open sets of R/~ are
{ { v/~ : v in A } | int A nonnul }
Another way to see this is to prove
open continuous surjection f:X -> Y ==> topology of Y is
{ f(U) | U open }
For all x in R, some a in int A, q in Q with x = a + q.
Thus { v/~ | v in A } = R/~ and R/~ is indiscrete.
No R/~ /= R/Q because pi and pi+1 are distinct in R/Q.
If a < b, then some q in Q /\ (a - x, b - x)
x + q in (a,b); x = x+q - q.
mina_world
>>>> under open quotient mapping.
>>>>
>>>> example)
>>>> Let R be the real space with the usual topology.
>>>> Define an equivalence relation on the R as follows
>>>> x ~ y if x - y in Q.
>>>>
>>>> Consider quotient mapping q : R -> R/~.
>>>> then R/~ is not Hausdorff.
>>>>
>>>> ----------------------------------------------------------
>>>> Maybe, q is not open mapping.
>>>> right ?
>>> Yes, it is an open mapping. Trivially, because the restriction of _q_ to
>>> _any_ non-empty open set is surjective.
>>
>> Yes, it's really an open mapping.
>> Because, R/~ = Q U {i + Q | i in irrational}.
>
> No. R/~ = {{Q + i} | i is irrational}.
Where is Q ?
>> and these elements is dense in R.
>
> R/~ is not a subset of R.
Of course. R/~ is a partition of R.
>> It means that cl[Q] = R, cl[i + Q] = R.
>> Namely, for any non-empty open set U in R,
>> U /\ Q =/= emtpy, U /\ [i + Q] =/= R.
>>
>> so, q(U) = R/~ by equivalent class mapping.
>> so, q is an open mapping.
>
> What matters here is that every non-empty subset of R contains points
> of all equivalence classes, which is equivalent to the fact that
> q(U) = R/~.
What matters ?
q(r) = Q for r in Q.
q(i) = i + Q for i in Q^c.
q^{-1}(Q) = Q
q^{-1}(i + Q) = i + Q.
so, q(U) = R/~.