what is a cardinality of the sets?

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leox

Posted: Mon May 08, 2006 2:44 am

1. What cardinality has a set of all monotone maps R to R? here R is
set of real numbers.
2. What cardinality has a set of all continuous maps [0,1] to [0,1]
which in all rational points takes on a rational values.

Thanks

Rupert

Posted: Mon May 08, 2006 3:24 am

leox wrote:
[quote:cb6c4788c4]1. What cardinality has a set of all monotone maps R to R? here R is
set of real numbers.
[/quote:cb6c4788c4]
The same cardinality as R.

[quote:cb6c4788c4]2. What cardinality has a set of all continuous maps [0,1] to [0,1]
which in all rational points takes on a rational values.

[/quote:cb6c4788c4]
This set is countable.

> Thanks

Denis Feldmann

Posted: Mon May 08, 2006 3:33 am

leox a écrit :
[quote:38e95f823c]1. What cardinality has a set of all monotone maps R to R? here R is
set of real numbers.
[/quote:38e95f823c]
The cardinality of R (assuming Axiom of Choice)

[quote:38e95f823c]2. What cardinality has a set of all continuous maps [0,1] to [0,1]
which in all rational points takes on a rational values.
[/quote:38e95f823c]
Idem

[quote:38e95f823c]
Thanks
[/quote:38e95f823c]

Dave Seaman

Posted: Mon May 08, 2006 4:19 am

On 8 May 2006 06:27:13 -0700, Martin wrote:

[quote:d5b140100f]Martin wrote:
leox wrote:
1. What cardinality has a set of all monotone maps R to R? here R is
set of real numbers.
2. What cardinality has a set of all continuous maps [0,1] to [0,1]
which in all rational points takes on a rational values.

Thanks

The proof of 2 you can find e.g. in the book Hrbacek, Jech:
Introduction to set theory (p.100, Thm 2.6).
Try
http://books.google.com/books?q=cardinality+%22continuous+functions%22
Basic idea is the fact that the rationals are dense in <0,1> and every
by choosing values for a dense set, the continuous function is uniquely
determined.
Sorry, I have just noticed the condition that the values in all
rational numbers are rational. Then the set is countable.
And the proof is in fact the same as without the conditions - the
cardinality si the same as the cardinality of all functions Q \cap
[0,1] to Q \cap [0,1].
(\cap stands here for the intersection)
[/quote:d5b140100f]
No, the set of mappings from Q to Q is uncountable, and the same goes for
Q[0,1]. The cardinality is (aleph_0)^(aleph_0) = 2^(aleph_0) = c.

--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>

Arturo Magidin

Posted: Mon May 08, 2006 4:21 am

In article <1147087593.878786.169540@i40g2000cwc.googlegroups.com>,
leox <leonid.uk@gmail.com> wrote:
[quote:196d3e66ce]Thanks but need any proof of it.
[/quote:196d3e66ce]
Proof of what?

http://groups.google.com/support/bin/answer.py?answer=14213&topic=250

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================

Arturo Magidin
magidin@math.berkeley.edu

leox

Posted: Mon May 08, 2006 5:26 am

Thanks but need any proof of it.

Martin

Posted: Mon May 08, 2006 8:29 am

Dave Seaman wrote:
[quote:5845f1fa9b]Sorry, I have just noticed the condition that the values in all
rational numbers are rational. Then the set is countable.
And the proof is in fact the same as without the conditions - the
cardinality si the same as the cardinality of all functions Q \cap
[0,1] to Q \cap [0,1].
(\cap stands here for the intersection)

No, the set of mappings from Q to Q is uncountable, and the same goes for
Q[0,1]. The cardinality is (aleph_0)^(aleph_0) = 2^(aleph_0) = c.

Of course you're right.[/quote:5845f1fa9b]
A silly mistake Sad

Martin

David Hartley

Posted: Mon May 08, 2006 8:30 am

In message <1147094833.721532.41840@j33g2000cwa.googlegroups.com>,
Martin <sleziak@fmph.uniba.sk> writes
[quote:f2a2e553dc]
Martin wrote:
leox wrote:
1. What cardinality has a set of all monotone maps R to R? here R is
set of real numbers.
2. What cardinality has a set of all continuous maps [0,1] to [0,1]
which in all rational points takes on a rational values.

Thanks

The proof of 2 you can find e.g. in the book Hrbacek, Jech:
Introduction to set theory (p.100, Thm 2.6).
Try
http://books.google.com/books?q=cardinality+%22continuous+functions%22
Basic idea is the fact that the rationals are dense in <0,1> and every
by choosing values for a dense set, the continuous function is uniquely
determined.
Sorry, I have just noticed the condition that the values in all
rational numbers are rational. Then the set is countable.
And the proof is in fact the same as without the conditions - the
cardinality si the same as the cardinality of all functions Q \cap
[0,1] to Q \cap [0,1].
(\cap stands here for the intersection)

Given any sequence <a_n> of 0's and 1's, define[/quote:f2a2e553dc]
f on [0,1] by

f(1/2^n) = (a_n)/(2^n) for n=1,2,...

f(0) = f(1) = 0

and elsewhere by interpolation.

You get a continuous function which is rational at all rational points
and different for each sequence. So there are at least 2^aleph_0 of
them. The set of all functions Q -> Q also has cardinality 2^aleph_0,
giving an upper bound. Thus the answer is 2^aleph_0.
--
David Hartley

Rupert

Posted: Mon May 08, 2006 5:58 pm

Denis Feldmann wrote:
[quote:936f73662f]leox a écrit :
1. What cardinality has a set of all monotone maps R to R? here R is
set of real numbers.

The cardinality of R (assuming Axiom of Choice)

[/quote:936f73662f]
I think I can do it without the axiom of choice.

[quote:936f73662f]
2. What cardinality has a set of all continuous maps [0,1] to [0,1]
which in all rational points takes on a rational values.

Idem

Thanks
[/quote:936f73662f]

Dave L. Renfro

Posted: Tue May 09, 2006 9:56 am

Anti_Dual wrote (in part):

[quote:f49c5f1d14]The cardinality of the set C_n of continuous functions
with exacty N discontinuities is c, wich implies that
the cardinality of the union {C_n}_n, is also c.

This set contains the set of the monotone functions.
[/quote:f49c5f1d14]
There are monotone functions with infinitely many
discontinuities, and none of these will be in the
union {C_n}_n.

However, it is true that the collection of all
functions with countably many discontinuities
has cardinality c, and this set contains (strictly,
by the way) all monotone functions. Much more is
true, in fact. The collection of all Baire one
functions (pointwise limits of sequences of
continuous functions) includes all functions
that have countably many discontinuities
(as well as some functions which have c many
discontinuities) and this collection has
cardinality c. In fact, the collection of
pointwise limits of Baire one functions (the
Baire 2 functions, which incidentally contains
the characteristic function of the rationals,
and hance contains some functions that have
no points of continuity), the collection of
pointwise limits of Baire 2 functions (the
Baire 3 functions), the collection of pointwise
limits of Baire 3 functions (the Baire 4 functions),
and so on all have cardinality c. The Baire class
w functions, the union of all these previous classes
(or is it the collection of pointwise limits of
of sequences whose terms are chosen from this union?),
has cardinality c, the Baire class w+1 functions
(pointwise limits of Baire class w functions) has
cardinality c, and so on. The Baire class 2w functions
(do the same thing for all the Baire class w+n
functions that one does for the Baire class n
functions), and so on through every countable ordinal,
has cardinality c. The collection of Baire functions
is the union of all these countable-ordinal indexed
Baire classes, and it too has cardinality c. Incidentally,
this last class is closed under the taking of pointwise
limits, so the operation of taking pointwise limits of
sequences of functions doesn't take you any further
(although you can always throw in up to c many new
functions and go through the entire process again
without ever getting more than c many functions
altogether).

Dave L. Renfro

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