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seen Oct 20 at 3:19

escher


Jul
2
awarded  Curious
Dec
30
accepted Axiom of Choice and Recursion Theorem exercise
Dec
30
revised Axiom of Choice and Recursion Theorem exercise
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Dec
30
asked Axiom of Choice and Recursion Theorem exercise
Dec
30
comment A property of functions from $\mathbb{R}\rightarrow \mathbb{R}$
@DanielFischer, I may assume the continuum hypothesis. Thank you for helping out.
Dec
30
comment A property of functions from $\mathbb{R}\rightarrow \mathbb{R}$
@DanielFischer The countable union of the sets $M_k$ must add up to the domain of $f$ which is $\mathbb{R}$. If all $M_k$ are countable then their union ia also countable, a contradiction. Thus at least one has the cardinality of $\mathbb{R}$. Did I got it right?
Dec
30
asked A property of functions from $\mathbb{R}\rightarrow \mathbb{R}$
Dec
24
comment Is my claim wrong?
@AndreasBlass you're right of course... I really liked my proof lol I'll try and prove your claim (more homework lol) Don't be too surprised if you see me posting again in few days:) . You've been most helpfull, thank you again.
Dec
24
comment Is my claim wrong?
@AndreasBlass After a long night I think I managed to prove the following (which actually proves the claim about the existance of a monomorphism): If $U,V$ are well ordered sets and $\pi:U\rightarrow V$ a monomorphism which respects the ordering ($x\leq_U y\iff \pi(x)\leq_V \pi(y)$) then $p[U]=seg_V(w)=\{y\in V| y<w\}$. So yes you're right there is a $a_0\in\chi(\mathbb{N})$ such that for all $a>a_0$ I don't have $M_a$ for most $a\in\chi(\mathbb{N})$. Andreas thank you for your time. If you post your comment as an answer I'll accept it.
Dec
23
comment Hartogs space of $\mathbb{N}$
@CameronBuie $A\leq_0 B$ means there exists a monomorphism $\pi$ from $A$ to $B$ which respects the ordering $(x\leq_{A} y\iff \pi(x)\leq_{B} \pi(y))$
Dec
23
answered Order of elements within a group
Dec
23
asked Is my claim wrong?
Dec
23
accepted Hartogs space of $\mathbb{N}$
Dec
23
comment Hartogs space of $\mathbb{N}$
@AsafKaragila thank you for helping out, I'll work a bit more on the details.
Dec
23
comment Hartogs space of $\mathbb{N}$
@AsafKaragila yes but I'm still working on a few things. Following your hint I did the following (I'm not really sure though about my solution): Since $\chi(\mathbb{N})$ is well ordered it has a least element let's call it $0$ and then $M_0$ is also the first set in the sequence defined above. Assuming the set sequence doesn't become 'constant' I can find $b,c\in \chi(\mathbb{N})$ with $b<c$ such that $M_c\subset M_b$. I can pick $x_0\in\M_0$ such that $x_0\in M_b\setminus M_c$. I then move to the next set etc.Picking different elements from each set I have a contradiction $M_0>_c \mathbb{N}$
Dec
23
revised Hartogs space of $\mathbb{N}$
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Dec
22
revised Hartogs space of $\mathbb{N}$
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Dec
22
comment Hartogs space of $\mathbb{N}$
@AndresCaicedo I'm not sure this is false. This is an exam exercise and it is very unlikely.
Dec
22
revised Hartogs space of $\mathbb{N}$
added 527 characters in body
Dec
22
revised Hartogs space of $\mathbb{N}$
added 223 characters in body