« first day (1855 days earlier)   

6:23 AM
This discussion from the main chatroom seems to be related to descriptive set theory:
in Mathematics, 25 mins ago, by DHMO
If $|\Bbb R| = \aleph_2$, can a closed set have cardinality $\aleph_1$?
in Mathematics, 24 mins ago, by Martin Sleziak
@DHMO Do I remember correctly that every infinite Borel set has cardinality either $\aleph_0$ or $\mathfrak c$?
in Mathematics, 21 mins ago, by Martin Sleziak
Every uncountable closed has cardinality $\mathfrak c$. This is (consequence of) Cantor-Bendixson theorem.
in Mathematics, 21 mins ago, by Martin Sleziak
But I think the same is true for Borel sets. (I might have misrememberd this.)
in Mathematics, 25 mins ago, by DHMO
can you give me a Borel set that has cardinality $\aleph_0$?
in Mathematics, 21 mins ago, by Martin Sleziak
You can have a closed set with cardinality $\aleph_0$, for example the convergent sequence: $\{0\} \cup \{1/n; n=1,2,\dots\}$.
I searched a bit for some reference for the claim about Bore sets.
This is stated in Kechris (Descriptive Set Theory) as Theorem 13.6 (The Perfect Set Theorem for Borel Sets; Alexandrov, Hausdorff).
Let $X$ be Polish and $A\subseteq X$ be Borel. Then either $A$ is countable or else it contains a Cantor set. In particular, every uncountable standard Borel space has cardinality $2^{\aleph_0}$.
The convention used there is: For convenience we will say that a subset $C$ of a topological space is \emph{a Cantor set} if it is homeomorphic to the Cantor space $\mathcal C$.
I guess the proof is probably not that easy, since it seems likely that some facts about Borel sets and Polish spaces derived earlier in the book are used.
 
6:46 AM
And this is related to descriptive set theory, too:
in Mathematics, 5 mins ago, by DHMO
@MartinSleziak can you give me an example of something which is not a Borel set?
in Mathematics, 2 mins ago, by Martin Sleziak
@DHMO Didn't we talk about something similar already? Or was it with secret? Existence of such sets can be shown with cardinality argument. But since it cannot be shown in ZF, we cannot expect "explicit" example.
in Mathematics, 2 mins ago, by Martin Sleziak
Probably for any reasonable interpretation of the word explicit. Any proof of existence of non-Borel set should contain a step which is not effective (cannot be proved in ZF).
in Mathematics, 45 secs ago, by DHMO
@MartinSleziak the most humbling thing I've read (paraphrase): there's only countably many definable objects, because there's only countably many finite strings to define objects
in Mathematics, 58 secs ago, by DHMO
thus, many Borel sets are not even definable
 
7:26 AM
And also a topic of chains in $\mathcal P(X)$ resurfaced again.
in Mathematics, 9 mins ago, by DHMO
@MartinSleziak a chain of inclusion of R can be as long as 2^R?
in Mathematics, 7 mins ago, by Martin Sleziak
So you are asking about maximal cardinality of a chain in $(\mathcal P(\mathbb R),\subseteq)?
in Mathematics, 5 mins ago, by Martin Sleziak
Something very similar was discussed not that long ago. This answer to a more general question says that the answer is yes if you assume continuum hypothesis.
 
 
3 hours later…
10:35 AM
Thanks for your answer from a couple of days ago @Martin, I haven't had time to read it in detail, but it already helped me with those doubts I had
 
10:46 AM
I am glad to hear it.
To be honest, I did not have time to check all the details at the time. But I mentioned explicitly that there might be some gaps or inaccuracies.
But since you write that you have been able to figure it out, it probably does not matter that much.
Some time ago with some friends we decided to go through Oxtoby's book. But we did not finish it. I believe we finished with the chapter about Banach category theorem. (And I am not sure whether there are some parts before this chapter that we skipped.) So we did not get to the chapter about Sierpinski-Erdos duality theorem.
 

« first day (1855 days earlier)